Samacheer Kalvi Class 8 Maths Solutions Chapter 3 Algebra InText Questions

Recap (Text Book Page No. 74 & 75)

Question 1. Write the number of terms in the following expressions
(i) \( x + y + z - xyz \)
(ii) \( m^2 n^2 c^2 \)
(iii) \( a^2 b^2 c - ab^2 c^2 + a^2 bc^2 + 3abc \)
(iv) \( 8x^2 - 4xy + 7xy^2 \)
Answer:
(i) The expression \( x + y + z - xyz \) has four terms. Each part connected by a plus or minus sign is a term. The different terms are \( x \), \( y \), \( z \), and \( -xyz \).
(ii) The expression \( m^2 n^2 c^2 \) has only one term. This is because all parts are multiplied together, forming a single group.
(iii) The expression \( a^2 b^2 c - ab^2 c^2 + a^2 bc^2 + 3abc \) contains four terms. These are \( a^2 b^2 c \), \( -ab^2 c^2 \), \( a^2 bc^2 \), and \( 3abc \).
(iv) The expression \( 8x^2 - 4xy + 7xy^2 \) has three terms. These terms are \( 8x^2 \), \( -4xy \), and \( 7xy^2 \).
In simple words: Count how many parts are separated by plus or minus signs. Each part is a "term".

🎯 Exam Tip: Remember that terms are separated by addition or subtraction. Multiplication or division within a term does not create new terms.

Question 2. Identify the numerical co-efficient of each term in the following expressions.
(i) \( 2x^2 - 5xy + 6y^2 + 7x - 10y + 9 \)
(ii) \( \frac{x}{3}+\frac{2y}{5} - xy + 7 \)
Answer:
(i) In the expression \( 2x^2 - 5xy + 6y^2 + 7x - 10y + 9 \):
The numerical coefficient of \( 2x^2 \) is \( 2 \).
The numerical coefficient of \( -5xy \) is \( -5 \).
The numerical coefficient of \( 6y^2 \) is \( 6 \).
The numerical coefficient of \( 7x \) is \( 7 \).
The numerical coefficient of \( -10y \) is \( -10 \).
The numerical coefficient of \( 9 \) is \( 9 \). This is called a constant term, where the number itself is the coefficient.
(ii) In the expression \( \frac{x}{3}+\frac{2y}{5} - xy + 7 \):
The numerical coefficient of \( \frac{x}{3} \) (which is \( \frac{1}{3}x \)) is \( \frac{1}{3} \).
The numerical coefficient of \( \frac{2y}{5} \) is \( \frac{2}{5} \).
The numerical coefficient of \( -xy \) is \( -1 \) (since \( -xy \) is the same as \( -1xy \)).
The numerical coefficient of \( 7 \) is \( 7 \) (it is a constant term).
In simple words: The number part of a term is its numerical coefficient. If there is no number written, it is \( 1 \) or \( -1 \).

🎯 Exam Tip: Pay close attention to the sign in front of the number; it is part of the coefficient.

Question 3. Pick out the like terms from the following:
\( x^2, 3y, 3a^2b^2, 4x, x^2y, 9p^2, -3y, 4ba^2, 9ab, 7q, 8p, -25x, -5x^2, 2x, 9x^2y, -9p^2, qp^2, 2xy^2, -10p^2, q, 3y^2x, \frac{2}{5}p, -x^2, a^2b^2, -ab, a^2b, 2ba, mn^2, 5m^3n^2 \)
Answer:
Like terms are terms that have the same variables raised to the same powers. The order of variables does not matter (e.g., \( x^2y \) is like \( yx^2 \)).
Here are the like terms grouped together:
\( \mathbf{x^2} \): \( x^2, -5x^2, -x^2 \)
\( \mathbf{y} \): \( 3y, -3y \)
\( \mathbf{a^2b^2} \): \( 3a^2b^2, a^2b^2 \)
\( \mathbf{x} \): \( 4x, -25x, 2x \)
\( \mathbf{x^2y} \): \( x^2y, 9x^2y \)
\( \mathbf{p^2} \): \( 9p^2, -9p^2, -10p^2 \)
\( \mathbf{p} \): \( 8p, \frac{2}{5}p \)
\( \mathbf{ab} \): \( 9ab, -ab, a^2b, 2ba \)
\( \mathbf{q} \): \( 7q, q, p^2q \)
\( \mathbf{y^2x} \): \( 2xy^2, 3y^2x \)
\( \mathbf{mn^2} \): \( mn^2, 5m^3n^2 \) (Note: \( mn^2 \) and \( 5m^3n^2 \) are NOT like terms as the power of \( m \) is different. The source groups them, but mathematically they are not like terms. I will list them as they appear in the source's visual grouping, with this note. However, for like terms, they must have identical variable parts. I will correct this grouping to be strict.)

Corrected grouping:
\( \mathbf{x^2} \): \( x^2, -5x^2, -x^2 \)
\( \mathbf{y} \): \( 3y, -3y \)
\( \mathbf{a^2b^2} \): \( 3a^2b^2, a^2b^2, 4ba^2 \) (since \( ba^2 = a^2b \))
\( \mathbf{x} \): \( 4x, -25x, 2x \)
\( \mathbf{x^2y} \): \( x^2y, 9x^2y \)
\( \mathbf{p^2} \): \( 9p^2, -9p^2, -10p^2 \)
\( \mathbf{p} \): \( 8p, \frac{2}{5}p \)
\( \mathbf{ab} \): \( 9ab, -ab, 2ba \)
\( \mathbf{q} \): \( 7q, q \)
\( \mathbf{pq^2} \): \( qp^2 \)
\( \mathbf{xy^2} \): \( 2xy^2, 3y^2x \)
\( \mathbf{mn^2} \): \( mn^2 \)
\( \mathbf{m^3n^2} \): \( 5m^3n^2 \)
In simple words: Like terms have the same letters with the same small numbers (powers) on them. We can group them together. For example, all \( x^2 \) terms are like terms.

🎯 Exam Tip: To identify like terms, always check both the variables and their exponents. The numerical coefficient does not affect whether terms are "like" or not, only their value.

Question 4. Add: \( 2x, 6y, 9x - 2y \)
Answer:
To add the given expressions, we combine the like terms.
\( (2x) + (6y) + (9x - 2y) \)
First, remove the parentheses: \( 2x + 6y + 9x - 2y \)
Next, group the like terms together: \( 2x + 9x + 6y - 2y \)
Now, add the coefficients of the like terms:
\( (2 + 9)x + (6 - 2)y \)
\( = 11x + 4y \)
So, the sum of the expressions is \( 11x + 4y \).
In simple words: To add these, put the \( x \) parts together and the \( y \) parts together. Add the numbers in front of the \( x \)'s and the numbers in front of the \( y \)'s.

🎯 Exam Tip: Always make sure to group all like terms correctly before adding or subtracting their coefficients to avoid errors.

Question 5. Simplify: \( (5x^3y^3 - 3x^2y^2 + xy + 7) + (2xy + x^3y^3 - 5 + 2x^2y^2) \)
Answer:
To simplify the expression, we need to combine the like terms after removing the parentheses.
\( (5x^3y^3 - 3x^2y^2 + xy + 7) + (2xy + x^3y^3 - 5 + 2x^2y^2) \)
First, remove the parentheses: \( 5x^3y^3 - 3x^2y^2 + xy + 7 + 2xy + x^3y^3 - 5 + 2x^2y^2 \)
Now, group the like terms:
\( (5x^3y^3 + x^3y^3) + (-3x^2y^2 + 2x^2y^2) + (xy + 2xy) + (7 - 5) \)
Combine the coefficients of each group:
\( (5 + 1)x^3y^3 + (-3 + 2)x^2y^2 + (1 + 2)xy + (2) \)
\( = 6x^3y^3 - x^2y^2 + 3xy + 2 \)
The simplified expression is \( 6x^3y^3 - x^2y^2 + 3xy + 2 \).
In simple words: Take off the brackets. Then, find all the terms that have the same letters and powers, and add or subtract their numbers.

🎯 Exam Tip: Be careful with signs when combining terms. Remember that \( x^3y^3 \) is different from \( x^2y^2 \), so they are not like terms.

Question 6. The sides of a triangle are \( 2x - 5y + 9 \), \( 3y + 6x - 7 \) and \( -4x + y + 10 \). Find the perimeter of the triangle.
Answer:
The perimeter of a triangle is found by adding the lengths of all three of its sides.
Perimeter \( = \text{Sum of three sides} \)
Perimeter \( = (2x - 5y + 9) + (3y + 6x - 7) + (-4x + y + 10) \)
Remove the parentheses:
Perimeter \( = 2x - 5y + 9 + 3y + 6x - 7 - 4x + y + 10 \)
Group the like terms together:
Perimeter \( = (2x + 6x - 4x) + (-5y + 3y + y) + (9 - 7 + 10) \)
Combine the coefficients of each group:
Perimeter \( = (2 + 6 - 4)x + (-5 + 3 + 1)y + (9 - 7 + 10) \)
Perimeter \( = 4x + (-1)y + (12) \)
Perimeter \( = 4x - y + 12 \)
Thus, the perimeter of the triangle is \( 4x - y + 12 \) units.
In simple words: Add up all the side lengths. Put all the \( x \) parts together, all the \( y \) parts together, and all the plain numbers together.

🎯 Exam Tip: When calculating perimeter, make sure to add all sides completely. Also, be very careful with negative signs when grouping and combining like terms.

Question 7. Subtract \( -2mn \) from \( 6mn \).
Answer:
When we subtract one expression from another, we write the expression to be subtracted second, inside parentheses.
\( 6mn - (-2mn) \)
When there are two negative signs together, they become a positive sign:
\( = 6mn + 2mn \)
Now, combine the like terms by adding their numerical coefficients:
\( = (6 + 2)mn \)
\( = 8mn \)
So, subtracting \( -2mn \) from \( 6mn \) gives \( 8mn \).
In simple words: "Subtracting a negative" is the same as "adding a positive". So, \( 6mn \) plus \( 2mn \) equals \( 8mn \).

🎯 Exam Tip: Remember the rule: subtracting a negative number is equivalent to adding a positive number, e.g., \( a - (-b) = a + b \).

Question 8. Subtract \( 6a^2 - 5ab + 3b^2 \) from \( 4a^2 - 3ab + b^2 \).
Answer:
To subtract the first expression from the second, we write the second expression first and then subtract the first expression (in parentheses).
\( (4a^2 - 3ab + b^2) - (6a^2 - 5ab + 3b^2) \)
Distribute the negative sign to each term inside the second parenthesis:
\( = 4a^2 - 3ab + b^2 - 6a^2 + 5ab - 3b^2 \)
Group the like terms together:
\( = (4a^2 - 6a^2) + (-3ab + 5ab) + (b^2 - 3b^2) \)
Combine the coefficients of each group:
\( = (4 - 6)a^2 + (-3 + 5)ab + (1 - 3)b^2 \)
\( = -2a^2 + 2ab - 2b^2 \)
The result of the subtraction is \( -2a^2 + 2ab - 2b^2 \).
In simple words: When subtracting, change the sign of every term in the second bracket, then combine all the like terms.

🎯 Exam Tip: The most common error in subtraction of polynomials is forgetting to distribute the negative sign to *all* terms inside the parenthesis being subtracted.

Question 9. The length of a log is \( 3a + 4b - 2 \) and a piece \( (2a - b) \) is removed from it. What is the length of the remaining log?
Answer:
To find the remaining length, we subtract the length of the removed piece from the original length of the log.
Original length of the log \( = 3a + 4b - 2 \)
Length of the piece removed \( = 2a - b \)
Remaining length of the log \( = (3a + 4b - 2) - (2a - b) \)
Distribute the negative sign:
\( = 3a + 4b - 2 - 2a + b \)
Group the like terms:
\( = (3a - 2a) + (4b + b) - 2 \)
Combine the coefficients:
\( = (3 - 2)a + (4 + 1)b - 2 \)
\( = 1a + 5b - 2 \)
\( = a + 5b - 2 \)
The remaining length of the log is \( a + 5b - 2 \) units. This is similar to how we calculate the change when we take a part away from a whole.
In simple words: Start with the log's total length. Take away the length of the piece cut off. Group letters that are the same and subtract their numbers.

🎯 Exam Tip: Remember to change the sign of every term inside the parentheses when subtracting an entire expression, especially when dealing with variables.

Question 10. A tin had 'x' litre oil. Another tin had \( (3x^2 + 6x - 5) \) litre of oil. The shopkeeper added \( (x + 7) \) litre more to the second tin. Later he sold \( (x^2 + 6) \) litres of oil from the second tin How much oil was left in the second tin?
Answer:
Let's break down the problem step by step to find the oil left in the second tin.
1. Initial quantity of oil in the second tin \( = (3x^2 + 6x - 5) \) litres.
2. Quantity of oil added to the second tin \( = (x + 7) \) litres.
3. Total quantity of oil in the second tin after adding:
\( = (3x^2 + 6x - 5) + (x + 7) \)
Group like terms:
\( = 3x^2 + (6x + x) + (-5 + 7) \)
\( = 3x^2 + 7x + 2 \) litres.
This is the new total amount of oil before selling any.
4. Quantity of oil sold from the second tin \( = (x^2 + 6) \) litres.
5. Quantity of oil left in the second tin:
\( = (3x^2 + 7x + 2) - (x^2 + 6) \)
Distribute the negative sign:
\( = 3x^2 + 7x + 2 - x^2 - 6 \)
Group like terms:
\( = (3x^2 - x^2) + 7x + (2 - 6) \)
Combine coefficients:
\( = (3 - 1)x^2 + 7x + (-4) \)
\( = 2x^2 + 7x - 4 \) litres.
Therefore, \( 2x^2 + 7x - 4 \) litres of oil were left in the second tin.
In simple words: Start with the oil in the second tin. Add the oil the shopkeeper put in. Then, take away the oil he sold. Group the \( x^2 \) parts, the \( x \) parts, and the numbers separately.

🎯 Exam Tip: Break down word problems into smaller algebraic steps. Always remember to distribute negative signs correctly when subtracting polynomials.

Think (Text Book Page No. 77)

Question 1. Every algebraic expression is a polynomial. Is this statement true? Why?
Answer:
No, this statement is not true. Not every algebraic expression is a polynomial.
A polynomial is a special type of algebraic expression where the powers (exponents) of the variables must only be whole numbers (0, 1, 2, 3,...). Algebraic expressions can have variables with fractional or negative powers, which means they are not polynomials.
For example, \( 2y^2 + 5y^{-1} - 3 \) is an algebraic expression, but it is not a polynomial because the variable \( y \) has a negative power of \( -1 \). Understanding this difference is key in algebra.
In simple words: No, not all algebraic expressions are polynomials. For something to be a polynomial, all the small numbers (powers) on the letters must be positive whole numbers or zero.

🎯 Exam Tip: Clearly state that polynomial exponents must be non-negative integers. Giving a counterexample like \( x^{-1} \) or \( \sqrt{x} \) (which is \( x^{1/2} \)) strengthens your explanation.

Try These (Text Book Page No. 78)

Question. (i) Find the product of \( 3ab^2 \) and \( -2a^2b^3 \)
Answer:
To find the product, multiply the numerical coefficients and then multiply the variable parts.
\( (3ab^2) \times (-2a^2b^3) \)
Multiply the signs: \( (+) \times (-) = (-) \)
Multiply the numbers: \( 3 \times 2 = 6 \)
Multiply the variable \( a \) parts: \( a \times a^2 = a^{1+2} = a^3 \)
Multiply the variable \( b \) parts: \( b^2 \times b^3 = b^{2+3} = b^5 \)
Combine these results:
\( = -6a^3b^5 \)
The product is \( -6a^3b^5 \).
In simple words: Multiply the numbers first, then multiply the same letters by adding their small power numbers. If there are different signs, the answer is negative.

🎯 Exam Tip: When multiplying terms, multiply the coefficients and add the exponents of the same variables. Pay close attention to the signs.

Question. (ii) Find the product of \( 4xy \), \( 5y^2x \), and \( -x^2 \)
Answer:
To find the product of these three terms, multiply their coefficients and then their variable parts.
\( (4xy) \times (5y^2x) \times (-x^2) \)
Multiply the signs: \( (+) \times (+) \times (-) = (-) \)
Multiply the numbers: \( 4 \times 5 \times 1 = 20 \)
Multiply the variable \( x \) parts: \( x \times x \times x^2 = x^{1+1+2} = x^4 \)
Multiply the variable \( y \) parts: \( y \times y^2 = y^{1+2} = y^3 \)
Combine these results:
\( = -20x^4y^3 \)
The product is \( -20x^4y^3 \).
In simple words: Multiply the numbers and letters together. For letters, add their powers. If there is one negative sign, the final answer will be negative.

🎯 Exam Tip: Remember that \( x \) is the same as \( x^1 \). If a variable appears without a visible exponent, its exponent is \( 1 \).

Question. (iii) Find the product of \( 2m \), \( -5n \), and \( -3p \)
Answer:
To find the product, multiply the numerical coefficients and then the variable parts.
\( (2m) \times (-5n) \times (-3p) \)
Multiply the signs: \( (+) \times (-) \times (-) = (+) \times (+) = (+) \)
Multiply the numbers: \( 2 \times 5 \times 3 = 30 \)
Multiply the variable parts: \( m \times n \times p = mnp \)
Combine these results:
\( = 30mnp \)
The product is \( 30mnp \).
In simple words: Multiply the numbers. If there are two negative signs, the answer is positive. Then put all the letters together.

🎯 Exam Tip: An even number of negative signs in multiplication results in a positive product, while an odd number results in a negative product.

Think (Text Book Page No. 79)

Question. why \( 3 + (4x - 7y) \neq 12x - 21y \)?
Answer:
The statement \( 3 + (4x - 7y) \neq 12x - 21y \) is true because addition and multiplication are different operations and follow different rules.
On the left side, \( 3 + (4x - 7y) = 3 + 4x - 7y \). We can only add or subtract like terms. Since \( 3 \), \( 4x \), and \( -7y \) are not like terms, we cannot simplify this expression further.
On the right side, \( 12x - 21y \). This expression would be obtained if we multiplied \( 3 \) by \( (4x - 7y) \) in a specific way, or if \( 3 \) was a common factor of \( 12x \) and \( 21y \). However, the operation shown on the left is addition. The number \( 3 \) cannot combine with \( 4x \) or \( -7y \) through addition because they have different variable parts. For example, you can't add 3 apples to 4 bananas to get 7 "apple-bananas".
In simple words: We cannot add a plain number like \( 3 \) to terms with letters like \( 4x \) or \( 7y \). They are not "like terms." We can only multiply numbers into brackets, not add them into brackets like that.

🎯 Exam Tip: Clearly distinguish between operations: addition/subtraction require like terms, while multiplication applies to all terms. A number outside parentheses multiplies everything inside.

Question 1. Multiply
(i) \( (5x^2 + 7x - 3) \text{ by } -4x^2 \)
(ii) \( (10x - 7y + 5z) \text{ by } 6xyz \)
(iii) \( (ab + 3bc - 5ca) \text{ by } 3a^2bc \)
(iv) \( (4m^2 - 3m + 7) \text{ by } -5m^3 \)
Answer:
(i) To multiply \( (5x^2 + 7x - 3) \) by \( -4x^2 \), we distribute \( -4x^2 \) to each term inside the parentheses.
\( (5x^2 + 7x - 3) \times (-4x^2) \)
\( = (5x^2)(-4x^2) + (7x)(-4x^2) + (-3)(-4x^2) \)
\( = -20x^{2+2} - 28x^{1+2} + 12x^2 \)
\( = -20x^4 - 28x^3 + 12x^2 \)

(ii) To multiply \( (10x - 7y + 5z) \) by \( 6xyz \), we distribute \( 6xyz \) to each term inside the parentheses.
\( (10x - 7y + 5z) \times (6xyz) \)
\( = (10x)(6xyz) + (-7y)(6xyz) + (5z)(6xyz) \)
\( = 60x^{1+1}yz - 42xy^{1+1}z + 30xyz^{1+1} \)
\( = 60x^2yz - 42xy^2z + 30xyz^2 \)

(iii) To multiply \( (ab + 3bc - 5ca) \) by \( 3a^2bc \), we distribute \( 3a^2bc \) to each term inside the parentheses.
\( (ab + 3bc - 5ca) \times (3a^2bc) \)
\( = (ab)(3a^2bc) + (3bc)(3a^2bc) + (-5ca)(3a^2bc) \)
\( = 3a^{1+2}b^{1+1}c + 9a^2b^{1+1}c^{1+1} - 15a^{1+2}b c^{1+1} \)
\( = 3a^3b^2c + 9a^2b^2c^2 - 15a^3bc^2 \)

(iv) To multiply \( (4m^2 - 3m + 7) \) by \( -5m^3 \), we distribute \( -5m^3 \) to each term inside the parentheses.
\( (4m^2 - 3m + 7) \times (-5m^3) \)
\( = (4m^2)(-5m^3) + (-3m)(-5m^3) + (7)(-5m^3) \)
\( = -20m^{2+3} + 15m^{1+3} - 35m^3 \)
\( = -20m^5 + 15m^4 - 35m^3 \)
In simple words: Take the term outside the bracket and multiply it by every single term inside the bracket. Remember to add the powers of the same letters.

🎯 Exam Tip: When multiplying a monomial by a polynomial, ensure the monomial is multiplied by *every* term within the polynomial, applying the rules of signs and exponents correctly.

Try These (Text Book Page No. 81)

Question. (i) Multiply \( (a - 5) \) and \( (a + 4) \)
Answer:
To multiply \( (a - 5) \) and \( (a + 4) \), we use the distributive property (FOIL method).
\( (a - 5)(a + 4) = a(a + 4) - 5(a + 4) \)
\( = (a \times a) + (a \times 4) + (-5 \times a) + (-5 \times 4) \)
\( = a^2 + 4a - 5a - 20 \)
Combine the like terms \( 4a \) and \( -5a \):
\( = a^2 - a - 20 \)
The product is \( a^2 - a - 20 \). This method helps ensure all terms are multiplied correctly.
In simple words: Multiply each part of the first bracket by each part of the second bracket. Then, add any similar terms together.

🎯 Exam Tip: When multiplying two binomials, ensure each term in the first binomial is multiplied by each term in the second binomial. The FOIL method (First, Outer, Inner, Last) is a helpful mnemonic.

Question. (ii) Multiply \( (a + b) \) and \( (a - b) \)
Answer:
To multiply \( (a + b) \) and \( (a - b) \), we use the distributive property.
\( (a + b)(a - b) = a(a - b) + b(a - b) \)
\( = (a \times a) + (a \times -b) + (b \times a) + (b \times -b) \)
\( = a^2 - ab + ab - b^2 \)
Combine the like terms \( -ab \) and \( +ab \):
\( = a^2 - b^2 \)
The product is \( a^2 - b^2 \). This is a very important algebraic identity, often called the "difference of squares" formula.
In simple words: Multiply each part of the first bracket by each part of the second. The middle terms \( -ab \) and \( +ab \) cancel each other out, leaving only \( a^2 - b^2 \).

🎯 Exam Tip: Recognize \( (a+b)(a-b) = a^2-b^2 \) as a standard identity. Using this shortcut can save time and reduce calculation errors.

Question. (iii) Multiply \( (m^4 + n^4) \) and \( (m - n) \)
Answer:
To multiply these expressions, distribute each term from the first parenthesis to each term in the second.
\( (m^4 + n^4)(m - n) = m^4(m - n) + n^4(m - n) \)
\( = (m^4 \times m) + (m^4 \times -n) + (n^4 \times m) + (n^4 \times -n) \)
\( = m^{4+1} - m^4n + mn^4 - n^{4+1} \)
\( = m^5 - m^4n + mn^4 - n^5 \)
The product is \( m^5 - m^4n + mn^4 - n^5 \). In this case, there are no like terms to combine further.
In simple words: Multiply \( m^4 \) by both parts of the second bracket, then multiply \( n^4 \) by both parts of the second bracket. Add the results.

🎯 Exam Tip: Remember to add exponents only when multiplying terms with the same base. If the bases are different, like \( m^4 \) and \( n \), they remain separate in the product.

Question. (iv) Multiply \( (2x + 3)(x + 4) \)
Answer:
To multiply \( (2x + 3) \) and \( (x + 4) \), use the distributive property.
\( (2x + 3)(x + 4) = 2x(x + 4) + 3(x + 4) \)
\( = (2x \times x) + (2x \times 4) + (3 \times x) + (3 \times 4) \)
\( = 2x^2 + 8x + 3x + 12 \)
Combine the like terms \( 8x \) and \( 3x \):
\( = 2x^2 + (8 + 3)x + 12 \)
\( = 2x^2 + 11x + 12 \)
The product is \( 2x^2 + 11x + 12 \). This method is widely used for multiplying binomials.
In simple words: Multiply the first term of the first bracket by both terms in the second bracket. Then do the same with the second term of the first bracket. Finally, add any terms that are alike.

🎯 Exam Tip: Double-check that you've multiplied all four pairs of terms (First, Outer, Inner, Last) and correctly combined the middle like terms.

Question. (v) Multiply \( (x - 5)(3x + 7) \)
Answer:
To multiply \( (x - 5) \) and \( (3x + 7) \), use the distributive property.
\( (x - 5)(3x + 7) = x(3x + 7) - 5(3x + 7) \)
\( = (x \times 3x) + (x \times 7) + (-5 \times 3x) + (-5 \times 7) \)
\( = 3x^2 + 7x - 15x - 35 \)
Combine the like terms \( 7x \) and \( -15x \):
\( = 3x^2 + (7 - 15)x - 35 \)
\( = 3x^2 - 8x - 35 \)
The product is \( 3x^2 - 8x - 35 \). Always be careful with negative signs during multiplication and combination.
In simple words: Multiply each term in the first bracket by each term in the second bracket. Then, combine the \( x \) terms by adding their numbers.

🎯 Exam Tip: Watch out for negative signs, especially when multiplying terms and combining like terms. A small sign error can lead to a completely different answer.

Question. (vi) Multiply \( (x - 2)(6x - 3) \)
Answer:
To multiply \( (x - 2) \) and \( (6x - 3) \), use the distributive property.
\( (x - 2)(6x - 3) = x(6x - 3) - 2(6x - 3) \)
\( = (x \times 6x) + (x \times -3) + (-2 \times 6x) + (-2 \times -3) \)
\( = 6x^2 - 3x - 12x + 6 \)
Combine the like terms \( -3x \) and \( -12x \):
\( = 6x^2 + (-3 - 12)x + 6 \)
\( = 6x^2 - 15x + 6 \)
The product is \( 6x^2 - 15x + 6 \). This process of expanding and combining is fundamental in algebra.
In simple words: Multiply the first term of the first bracket by both terms of the second. Then do the same with the second term of the first. Add the \( x \) terms together.

🎯 Exam Tip: Pay extra attention to the multiplication of two negative numbers, which always results in a positive number.

Think (Text Book Page No. 81)

Question. (i) In \( 3x^2(x^4 - 7x^3 + 2) \) what is the highest power in the expression?
Answer:
To find the highest power, first expand the expression by multiplying \( 3x^2 \) by each term inside the parentheses.
\( 3x^2(x^4 - 7x^3 + 2) \)
\( = (3x^2)(x^4) + (3x^2)(-7x^3) + (3x^2)(2) \)
\( = 3x^{2+4} - 21x^{2+3} + 6x^2 \)
\( = 3x^6 - 21x^5 + 6x^2 \)
Now, identify the powers of \( x \) in each term: \( 6 \), \( 5 \), and \( 2 \).
The highest power among these is \( 6 \). This is the degree of the polynomial.
In simple words: Multiply the number outside the bracket by everything inside. For the \( x \) terms, add their small power numbers. The biggest power you get for \( x \) is the highest power.

🎯 Exam Tip: The highest power in an expression is found *after* all multiplications are performed and like terms (if any) are combined. This is also known as the degree of the polynomial.

Question. (ii) Is \( -5y^2 + 2y - 6 = -(5y^2 + 2y - 6) \)? If not, correct the mistake.
Answer:
No, the statement \( -5y^2 + 2y - 6 = -(5y^2 + 2y - 6) \) is not correct.
Let's evaluate the right side of the equation:
\( -(5y^2 + 2y - 6) \)
Distribute the negative sign to each term inside the parentheses:
\( = -5y^2 - 2y + 6 \)
Comparing this to the left side, \( -5y^2 + 2y - 6 \), we can see they are not equal. The signs for the \( 2y \) term and the constant term are different.
To correct the mistake, the equation should be:
\( -5y^2 + 2y - 6 \neq -5y^2 - 2y + 6 \)
This clearly shows that the expressions are not identical.
In simple words: No, they are not equal. When you put a minus sign in front of a bracket, it changes the sign of every single thing inside the bracket. The second part should be \( -5y^2 - 2y + 6 \), which is not the same as the first part.

🎯 Exam Tip: A negative sign outside a parenthesis changes the sign of *every* term inside when the parenthesis is removed. This is a common error to avoid.

Think (Text Book Page No. 83)

Question. Are the following correct?
(i) \( \frac{x^3}{x^8} = x^{8-3} = x^5 \)
Answer:
No, the given statement \( \frac{x^3}{x^8} = x^{8-3} = x^5 \) is not correct.
When dividing terms with the same base, we subtract the exponent of the denominator from the exponent of the numerator. So, \( \frac{x^a}{x^b} = x^{a-b} \).
Applying this rule:
\( \frac{x^3}{x^8} = x^{3-8} = x^{-5} \)
Alternatively, if we want a positive exponent, we can write it as:
\( \frac{x^3}{x^8} = \frac{1}{x^{8-3}} = \frac{1}{x^5} \)
Both \( x^{-5} \) and \( \frac{1}{x^5} \) are the correct forms. The statement \( x^5 \) is incorrect because the subtraction of exponents should be \( 3-8 \), not \( 8-3 \). Understanding exponent rules is crucial for simplification.
In simple words: No, it's wrong. When you divide letters with powers, you subtract the bottom power from the top power. So it should be \( x \) to the power of \( (3-8) \), which is \( x^{-5} \) or \( \frac{1}{x^5} \), not \( x^5 \).

🎯 Exam Tip: Always remember that for division, the exponent of the denominator is subtracted from the exponent of the numerator: \( \frac{x^a}{x^b} = x^{a-b} \).

Question. (ii) Are the following correct? \( \frac{10m^4}{10m^4}=0 \)
Answer:
No, the given statement \( \frac{10m^4}{10m^4}=0 \) is not correct.
When any non-zero number or expression is divided by itself, the result is always \( 1 \).
Let's apply the rules of exponents for division:
\( \frac{10m^4}{10m^4} = \frac{10}{10} \times m^{4-4} \)
\( = 1 \times m^0 \)
We know that any non-zero base raised to the power of \( 0 \) is \( 1 \) (i.e., \( m^0 = 1 \)).
\( = 1 \times 1 \)
\( = 1 \)
Therefore, the correct answer is \( 1 \), not \( 0 \). This is a basic rule of division and exponents.
In simple words: No, it's wrong. Any number or expression divided by itself is always \( 1 \), not \( 0 \). When you divide the same letters with the same powers, they become \( 1 \) (because \( m^0=1 \)).

🎯 Exam Tip: Remember that any non-zero number or variable raised to the power of zero is \( 1 \). Also, anything divided by itself is \( 1 \).

Question. (iii) Are the following correct? When a monomial is divided by itself, we will get 1. ?
Answer:
Yes, the given statement is correct.
When a monomial (a single-term algebraic expression) is divided by itself, the result is always \( 1 \).
For example, consider the monomial \( x \). When we divide it by itself:
\( \frac{x}{x} = x^{1-1} = x^0 = 1 \)
Similarly, if we have a monomial like \( 5a^2b^3 \), dividing it by itself gives:
\( \frac{5a^2b^3}{5a^2b^3} = 1 \)
This fundamental rule applies to all non-zero monomials, as long as the denominator is not zero. This is a key concept in simplifying algebraic fractions.
In simple words: Yes, this is true. When you divide a single algebra term by the exact same term, you always get \( 1 \). Like \( 5 \div 5 = 1 \).

🎯 Exam Tip: Reinforce that any non-zero expression divided by itself equals \( 1 \). This applies broadly in algebra, not just to monomials.

Try These (Text Book Page No. 83)

Question. (i) Divide \( 12x^3y^2 \) by \( x^2y \)
Answer:
To divide \( 12x^3y^2 \) by \( x^2y \), we divide the numerical coefficients and subtract the exponents of the same variables.
\( \frac{12x^3y^2}{x^2y} \)
Divide the numbers: \( \frac{12}{1} = 12 \)
Divide the \( x \) terms: \( \frac{x^3}{x^2} = x^{3-2} = x^1 \)
Divide the \( y \) terms: \( \frac{y^2}{y^1} = y^{2-1} = y^1 \)
Combine the results:
\( = 12x^1y^1 \)
\( = 12xy \)
The result of the division is \( 12xy \). This is how terms are simplified in algebraic division.
In simple words: Divide the numbers first. Then, for each letter, subtract its power in the bottom part from its power in the top part.

🎯 Exam Tip: When a variable appears without an exponent, its exponent is \( 1 \). Always subtract the exponents carefully, especially when some terms might become \( x^0 \).

Question. (ii) Divide \( -20a^5b^2 \) by \( 2a^3b^7 \)
Answer:
To divide \( -20a^5b^2 \) by \( 2a^3b^7 \), we divide the numerical coefficients and subtract the exponents of the same variables.
\( \frac{-20a^5b^2}{2a^3b^7} \)
Divide the numbers: \( \frac{-20}{2} = -10 \)
Divide the \( a \) terms: \( \frac{a^5}{a^3} = a^{5-3} = a^2 \)
Divide the \( b \) terms: \( \frac{b^2}{b^7} = b^{2-7} = b^{-5} \)
Combine the results:
\( = -10a^2b^{-5} \)
To express with positive exponents, move \( b^{-5} \) to the denominator:
\( = \frac{-10a^2}{b^5} \)
The result of the division is \( \frac{-10a^2}{b^5} \). This shows how negative exponents are handled in division.
In simple words: Divide the numbers. For each letter, subtract the bottom power from the top power. If you get a negative power, move that letter to the bottom of a fraction and make its power positive.

🎯 Exam Tip: Remember that \( x^{-n} = \frac{1}{x^n} \). If the exponent in the denominator is larger, the variable will end up in the denominator with a positive exponent.

Question. (iii) Divide \( 28a^4c^2 \) by \( 21ca^3 \)
Answer:
To divide \( 28a^4c^2 \) by \( 21ca^3 \), we divide the numerical coefficients and subtract the exponents of the same variables.
\( \frac{28a^4c^2}{21ca^3} \)
Simplify the numerical fraction: \( \frac{28}{21} = \frac{4 \times 7}{3 \times 7} = \frac{4}{3} \)
Divide the \( a \) terms: \( \frac{a^4}{a^3} = a^{4-3} = a^1 \)
Divide the \( c \) terms: \( \frac{c^2}{c^1} = c^{2-1} = c^1 \)
Combine the results:
\( = \frac{4}{3} a^1 c^1 \)
\( = \frac{4}{3} ac \)
The result of the division is \( \frac{4}{3}ac \). Simplifying the numerical fraction is an important first step.
In simple words: First, reduce the numbers like a regular fraction. Then, for each letter, subtract the power from the bottom from the power at the top.

🎯 Exam Tip: Always simplify the numerical coefficients by finding their greatest common divisor before dealing with the variables and their exponents.

Question. (iv) Divide \( (3x^2y)^3 \) by \( 6x^2y^3 \)
Answer:
First, simplify the numerator \( (3x^2y)^3 \). When raising a product to a power, raise each factor to that power.
\( (3x^2y)^3 = 3^3 \times (x^2)^3 \times y^3 = 27x^{2 \times 3}y^3 = 27x^6y^3 \)
Now, perform the division:
\( \frac{27x^6y^3}{6x^2y^3} \)
Simplify the numerical fraction: \( \frac{27}{6} = \frac{9 \times 3}{2 \times 3} = \frac{9}{2} \)
Divide the \( x \) terms: \( \frac{x^6}{x^2} = x^{6-2} = x^4 \)
Divide the \( y \) terms: \( \frac{y^3}{y^3} = y^{3-3} = y^0 = 1 \)
Combine the results:
\( = \frac{9}{2}x^4(1) \)
\( = \frac{9}{2}x^4 \)
The result is \( \frac{9}{2}x^4 \). Remember to apply the power rule to all parts within the parentheses.
In simple words: First, cube everything in the top part. Then, divide the numbers and subtract the powers of the same letters. If a letter's power becomes zero, that letter becomes \( 1 \).

🎯 Exam Tip: Remember to apply the exponent outside the parentheses to *all* numerical and variable factors inside before performing division.

Question. (v) Divide \( 64m^4(n^2)^3 \) by \( 4mn \)
Answer:
First, simplify the numerator \( 64m^4(n^2)^3 \).
\( 64m^4(n^2)^3 = 64m^4n^{2 \times 3} = 64m^4n^6 \)
Now, perform the division:
\( \frac{64m^4n^6}{4mn} \)
Divide the numbers: \( \frac{64}{4} = 16 \)
Divide the \( m \) terms: \( \frac{m^4}{m^1} = m^{4-1} = m^3 \)
Divide the \( n \) terms: \( \frac{n^6}{n^1} = n^{6-1} = n^5 \)
Combine the results:
\( = 16m^3n^5 \)
The result of the division is \( 16m^3n^5 \). It is important to simplify the exponents correctly.
In simple words: First, fix the power of \( n \) in the top part. Then divide the numbers and subtract the powers of the same letters.

🎯 Exam Tip: When an exponent is applied to a term already having an exponent (e.g., \( (n^2)^3 \)), multiply the exponents: \( n^{2 \times 3} = n^6 \).

Question. (vi) Divide \( (8x^2y^2)^3 \) by \( (8x^2y^2)^2 \)
Answer:
When dividing terms with the same base, we subtract the exponent of the denominator from the exponent of the numerator. Here, the base is \( (8x^2y^2) \).
\( \frac{(8x^2y^2)^3}{(8x^2y^2)^2} \)
Using the rule \( \frac{a^m}{a^n} = a^{m-n} \):
\( = (8x^2y^2)^{3-2} \)
\( = (8x^2y^2)^1 \)
\( = 8x^2y^2 \)
The result of the division is \( 8x^2y^2 \). This applies a basic exponent rule directly.
Alternatively, we could expand the powers first:
\( (8x^2y^2)^3 = 8^3 (x^2)^3 (y^2)^3 = 512x^6y^6 \)
\( (8x^2y^2)^2 = 8^2 (x^2)^2 (y^2)^2 = 64x^4y^4 \)
Then divide:
\( \frac{512x^6y^6}{64x^4y^4} = (\frac{512}{64})x^{6-4}y^{6-4} = 8x^2y^2 \)
Both methods lead to the same correct answer.
In simple words: Since the bracketed terms are exactly the same, you can just subtract the small power numbers outside the brackets. \( (3 - 2 = 1) \), so the answer is simply the term inside the bracket once.

🎯 Exam Tip: When the entire base is the same, use the exponent rule \( \frac{a^m}{a^n} = a^{m-n} \) directly for a quicker solution, rather than expanding each term completely.

Question. (vii) Divide \( 81p^2q^4 \) by \( \sqrt{81p^2q^4} \)
Answer:
First, simplify the square root term in the denominator.
\( \sqrt{81p^2q^4} = \sqrt{81} \times \sqrt{p^2} \times \sqrt{q^4} \)
\( = 9 \times p^{2/2} \times q^{4/2} \)
\( = 9pq^2 \)
Now, perform the division:
\( \frac{81p^2q^4}{9pq^2} \)
Divide the numbers: \( \frac{81}{9} = 9 \)
Divide the \( p \) terms: \( \frac{p^2}{p^1} = p^{2-1} = p^1 \)
Divide the \( q \) terms: \( \frac{q^4}{q^2} = q^{4-2} = q^2 \)
Combine the results:
\( = 9pq^2 \)
The result of the division is \( 9pq^2 \). Simplifying square roots of variables requires dividing their exponents by 2.
In simple words: First, find the square root of the bottom part. Then, divide the numbers and subtract the powers of the same letters.

🎯 Exam Tip: Remember that \( \sqrt{x^n} = x^{n/2} \). Also, carefully simplify the numerical and variable parts of the square root separately.

Question. (viii) Divide \( (4x^2y^3)^0 \) by \( \frac{(x^3)^2}{x^6} \)
Answer:
First, simplify the numerator \( (4x^2y^3)^0 \). Any non-zero expression raised to the power of \( 0 \) is \( 1 \).
\( (4x^2y^3)^0 = 1 \)
Next, simplify the denominator \( \frac{(x^3)^2}{x^6} \).
\( (x^3)^2 = x^{3 \times 2} = x^6 \)
So, the denominator becomes \( \frac{x^6}{x^6} \).
When an expression is divided by itself, the result is \( 1 \).
\( \frac{x^6}{x^6} = x^{6-6} = x^0 = 1 \)
Now, perform the overall division:
\( \frac{(4x^2y^3)^0}{\frac{(x^3)^2}{x^6}} = \frac{1}{1} \)
\( = 1 \)
The result of the division is \( 1 \). This problem combines several exponent rules.
In simple words: The top part is \( 1 \) because anything to the power of zero is \( 1 \). The bottom part is also \( 1 \) because \( (x^3)^2 \) is \( x^6 \), and \( x^6 \) divided by \( x^6 \) is \( 1 \). So, \( 1 \) divided by \( 1 \) is \( 1 \).

🎯 Exam Tip: Master the rule \( a^0 = 1 \) (for \( a \neq 0 \)) and \( (a^m)^n = a^{mn} \). These are frequently used in simplifying complex expressions.

Think (Text Book Page No. 84)

Question. Are the following divisions correct?
(i) \( \frac{4y+3}{4} = y + 3 \)
Answer:
No, the given statement \( \frac{4y+3}{4} = y + 3 \) is not correct.
When dividing an expression like \( (4y+3) \) by \( 4 \), the denominator \( 4 \) applies to *each* term in the numerator. We cannot simply cancel \( 4 \) with \( 4y \) and leave \( 3 \) untouched.
The correct way to divide is:
\( \frac{4y+3}{4} = \frac{4y}{4} + \frac{3}{4} \)
\( = y + \frac{3}{4} \)
The result \( y + \frac{3}{4} \) is not equal to \( y + 3 \). This is a common mistake when distributing division.
In simple words: No, it's wrong. You can't just cancel the \( 4 \) with the \( 4y \). The \( 4 \) in the bottom must divide both the \( 4y \) and the \( 3 \) in the top. So the answer should be \( y + \frac{3}{4} \).

🎯 Exam Tip: Remember that division (or multiplication) must be distributed to *all* terms in the numerator (or polynomial) when dividing by a single term.

Question. (ii) Are the following divisions correct? \( \frac{5m^2+9}{9} = 5m^2 \)
Answer:
No, the given statement \( \frac{5m^2+9}{9} = 5m^2 \) is not correct.
Similar to the previous problem, the denominator \( 9 \) must divide both terms in the numerator.
The correct way to divide is:
\( \frac{5m^2+9}{9} = \frac{5m^2}{9} + \frac{9}{9} \)
\( = \frac{5}{9}m^2 + 1 \)
The result \( \frac{5}{9}m^2 + 1 \) is not equal to \( 5m^2 \). You cannot simply cancel the \( 9 \)s and leave \( 5m^2 \) without the fraction part.
In simple words: No, it's wrong. The \( 9 \) at the bottom must divide both the \( 5m^2 \) and the \( 9 \) at the top. So it should be \( \frac{5}{9}m^2 + 1 \), not just \( 5m^2 \).

🎯 Exam Tip: Avoid the mistake of only canceling terms partially. When a sum or difference is divided by a number, each term in the sum/difference must be divided by that number.

Question. (iii) Are the following divisions correct? \( \frac{2x^2+8}{4} = 2x^2 + 2 \). If not, correct it.
Answer:
No, the given statement \( \frac{2x^2+8}{4} = 2x^2 + 2 \) is not correct.
Again, the denominator \( 4 \) must divide both terms in the numerator.
The correct way to divide is:
\( \frac{2x^2+8}{4} = \frac{2x^2}{4} + \frac{8}{4} \)
Simplify each fraction:
\( = \frac{1}{2}x^2 + 2 \)
The result \( \frac{1}{2}x^2 + 2 \) is not equal to \( 2x^2 + 2 \). The \( x^2 \) term is incorrect in the original statement. This shows the importance of applying division to all parts of the expression.
In simple words: No, it's wrong. The \( 4 \) at the bottom divides both parts at the top. So \( 2x^2 \) divided by \( 4 \) is \( \frac{1}{2}x^2 \), and \( 8 \) divided by \( 4 \) is \( 2 \). The correct answer is \( \frac{1}{2}x^2 + 2 \).

🎯 Exam Tip: When simplifying fractions with polynomial numerators, remember to reduce both the numerical coefficients and the constant terms in the numerator by the denominator.

Try These (Text Book Page No. 84)

Question. (i) Divide \( (16y^5 - 8y^2) \div 4y \)
Answer:
To divide \( (16y^5 - 8y^2) \) by \( 4y \), we divide each term in the numerator by \( 4y \).
\( \frac{16y^5 - 8y^2}{4y} = \frac{16y^5}{4y} - \frac{8y^2}{4y} \)
For the first term:
\( \frac{16y^5}{4y} = (\frac{16}{4})y^{5-1} = 4y^4 \)
For the second term:
\( \frac{8y^2}{4y} = (\frac{8}{4})y^{2-1} = 2y^1 = 2y \)
Combine the simplified terms:
\( = 4y^4 - 2y \)
The result of the division is \( 4y^4 - 2y \). This process simplifies the polynomial expression.
In simple words: Divide the first part of the top by the bottom, then divide the second part of the top by the bottom. Remember to subtract powers of the letters.

🎯 Exam Tip: Ensure that the divisor is applied to *each* term in the numerator separately. Also, correctly apply the exponent rule for division \( \frac{a^m}{a^n} = a^{m-n} \).

Question. (ii) Divide \( (p^5q^2 + 24p^3q - 128q^3) \div 6q \)
Answer:
To divide \( (p^5q^2 + 24p^3q - 128q^3) \) by \( 6q \), we divide each term in the numerator by \( 6q \).
\( \frac{p^5q^2 + 24p^3q - 128q^3}{6q} = \frac{p^5q^2}{6q} + \frac{24p^3q}{6q} - \frac{128q^3}{6q} \)
For the first term:
\( \frac{p^5q^2}{6q} = \frac{1}{6}p^5q^{2-1} = \frac{1}{6}p^5q \)
For the second term:
\( \frac{24p^3q}{6q} = (\frac{24}{6})p^3q^{1-1} = 4p^3q^0 = 4p^3(1) = 4p^3 \)
For the third term:
\( \frac{128q^3}{6q} = (\frac{128}{6})q^{3-1} = \frac{64}{3}q^2 \)
Combine the simplified terms:
\( = \frac{1}{6}p^5q + 4p^3 - \frac{64}{3}q^2 \)
This is the result of the division. Remember that any variable to the power of zero is one.
In simple words: Divide each part of the top by \( 6q \). Simplify the numbers and subtract the powers of \( p \) and \( q \). If \( q \) disappears, it means its power became \( 0 \), so it becomes \( 1 \).

🎯 Exam Tip: Remember that \( q/q = q^0 = 1 \). Also, simplify numerical fractions like \( \frac{128}{6} \) to their lowest terms. You can also write \( \frac{1}{6}p^5q \) as \( \frac{p^5q}{6} \).

Question. (iii) Divide \( (4m^2n + 9n^2m + 3mn) \div 4mn \)
Answer:
To divide \( (4m^2n + 9n^2m + 3mn) \) by \( 4mn \), we divide each term in the numerator by \( 4mn \).
\( \frac{4m^2n + 9n^2m + 3mn}{4mn} = \frac{4m^2n}{4mn} + \frac{9n^2m}{4mn} + \frac{3mn}{4mn} \)
For the first term:
\( \frac{4m^2n}{4mn} = (\frac{4}{4})m^{2-1}n^{1-1} = 1m^1n^0 = m \)
For the second term:
\( \frac{9n^2m}{4mn} = \frac{9}{4} m^{1-1}n^{2-1} = \frac{9}{4}m^0n^1 = \frac{9}{4}n \)
For the third term:
\( \frac{3mn}{4mn} = (\frac{3}{4})m^{1-1}n^{1-1} = \frac{3}{4}m^0n^0 = \frac{3}{4} \)
Combine the simplified terms:
\( = m + \frac{9}{4}n + \frac{3}{4} \)
The result of the division is \( m + \frac{9}{4}n + \frac{3}{4} \). It is helpful to simplify numerical coefficients and exponents step-by-step.
In simple words: Divide each part of the top by \( 4mn \). Simplify the numbers and subtract the powers of \( m \) and \( n \). If a power becomes \( 0 \), that letter becomes \( 1 \).

🎯 Exam Tip: When all variables in a term cancel out (e.g., \( \frac{3mn}{4mn} \)), the result is just the numerical coefficient, not zero. Remember \( m^0 = 1 \) and \( n^0 = 1 \).

Try These (Text Book Page No. 86)

Question 1. Expand the following: \( (p + 2)^2 = \)
Answer:
To expand \( (p + 2)^2 \), we can use the algebraic identity \( (a + b)^2 = a^2 + 2ab + b^2 \).
Here, \( a = p \) and \( b = 2 \).
Substitute these values into the identity:
\( (p + 2)^2 = p^2 + 2(p)(2) + 2^2 \)
\( = p^2 + 4p + 4 \)
Alternatively, we can write \( (p + 2)^2 \) as \( (p + 2)(p + 2) \) and multiply using the distributive property:
\( (p + 2)(p + 2) = p(p + 2) + 2(p + 2) \)
\( = (p \times p) + (p \times 2) + (2 \times p) + (2 \times 2) \)
\( = p^2 + 2p + 2p + 4 \)
\( = p^2 + 4p + 4 \)
Both methods give the same expanded form. Using the identity is often quicker.
In simple words: You can multiply \( (p + 2) \) by \( (p + 2) \). Or, use the rule: first term squared, plus two times (first term times second term), plus second term squared.

🎯 Exam Tip: Recognize that \( (a+b)^2 \) is not just \( a^2+b^2 \); the middle term \( 2ab \) is essential. Memorizing this identity saves time.

Question 2. (3 – a)² = .............................
Answer:
\( (3 - a)^2 = 3^2 - 2(3)(a) + a^2 \)
\( = 9 - 6a + a^2 \) Here, we apply the algebraic identity \( (x - y)^2 = x^2 - 2xy + y^2 \) to expand the given expression.
In simple words: To expand \( (3 - a)^2 \), you multiply (3 - a) by itself. It follows a special pattern: square the first term, subtract twice the product of the two terms, then add the square of the second term.

🎯 Exam Tip: Remember that \( (a - b)^2 \) is not the same as \( a^2 - b^2 \). Always use the correct identity \( (a - b)^2 = a^2 - 2ab + b^2 \) for expansion.

Question 3. (6² – x²) = .............................
Answer:
\( (6^2 - x^2) = (6 + x)(6 - x) \) This expression can be easily factored using the difference of squares identity, which helps simplify it quickly.
In simple words: When you have one square number minus another square number, you can break it down into two brackets: one with a plus sign and one with a minus sign between the square roots.

🎯 Exam Tip: The identity \( a^2 - b^2 = (a + b)(a - b) \) is very useful for factoring expressions where two perfect squares are subtracted from each other.

Question 4. (a + b)² – (a – b)² = .............................
Answer:
\( (a + b)^2 - (a - b)^2 = (a^2 + 2ab + b^2) - (a^2 - 2ab + b^2) \)
\( = a^2 + 2ab + b^2 - a^2 + 2ab - b^2 \)
\( = (1 - 1)a^2 + (2 + 2)ab + (+1 - 1)b^2 = 4ab \) This identity is a shortcut for finding the product of \( (a+b)(a-b) \) when subtracted.
In simple words: If you subtract the square of (a minus b) from the square of (a plus b), you always get 4 times 'a' multiplied by 'b'.

🎯 Exam Tip: This is a common identity, \( (a + b)^2 - (a - b)^2 = 4ab \). Knowing it saves time in simplifying algebraic expressions. It helps avoid expanding each square separately.

Question 5. (a + b)² = (a + b) \( \times \) .............................
Answer:
\( (a + b)^2 = (a + b) \times (a + b) \) Squaring an expression means multiplying it by itself.
In simple words: When you see something like "squared," it means you multiply that thing by itself. So, (a + b) squared means (a + b) multiplied by (a + b).

🎯 Exam Tip: Always remember that squaring means multiplying a base by itself. It does not mean multiplying the base by 2.

Question 6. (m + n)(.........) = m² – n²
Answer:
\( (m + n)(m - n) = m^2 - n^2 \) This is the difference of squares identity, which is useful for quickly factoring or expanding expressions.
In simple words: When you multiply (m + n) by (m - n), the answer is always the square of 'm' minus the square of 'n'.

🎯 Exam Tip: The identity \( (a + b)(a - b) = a^2 - b^2 \) is very important for factoring algebraic expressions, especially when dealing with polynomials. Always check for this pattern.

Question 7. (m + ......)² = m² + 14m + 49
Answer:
\( (m + 7)^2 = m^2 + 14m + 49 \) The middle term, 14m, is \( 2 \times m \times 7 \), and 49 is \( 7^2 \), which perfectly fits the square of a sum identity.
In simple words: To get \( m^2 + 14m + 49 \), you need to square (m + 7). This is because (m + 7) multiplied by itself gives that answer.

🎯 Exam Tip: When you see an expression like \( a^2 + 2ab + b^2 \), immediately recognize it as \( (a + b)^2 \). Here, \( a=m \) and \( b=7 \).

Question 8. (k² – 49) = (k + ...)(k – ...)
Answer:
\( k^2 - 49 = k^2 - 7^2 = (k + 7)(k - 7) \) This is an application of the difference of squares identity to factor the given expression.
In simple words: The number 49 is the same as 7 squared. So, if you have k squared minus 7 squared, it can be written as (k plus 7) multiplied by (k minus 7).

🎯 Exam Tip: Always look for perfect squares when you see a subtraction. Factoring a difference of squares \( a^2 - b^2 \) into \( (a + b)(a - b) \) is a common and important technique.

Question 9. m² – 6m + 9 = .............................
Answer:
\( m^2 - 6m + 9 = (m - 3)^2 \) This expression is a perfect square trinomial because 9 is \( 3^2 \) and -6m is \( -2 \times m \times 3 \).
In simple words: This expression, \( m^2 - 6m + 9 \), is a special kind of polynomial called a perfect square. It is the same as (m - 3) multiplied by itself.

🎯 Exam Tip: Recognize perfect square trinomials: \( a^2 - 2ab + b^2 = (a - b)^2 \). Here, \( a=m \) and \( b=3 \), making \( 2ab = 2 \times m \times 3 = 6m \).

Question 10. (m – 10)(m + 5) = .............................
Answer:
\( (m - 10)(m + 5) = m^2 + (-10 + 5)m + (-10)(5) = m^2 - 5m - 50 \) This is an application of the identity \( (x + a)(x + b) = x^2 + (a + b)x + ab \), which helps multiply two binomials easily.
In simple words: To multiply these two terms, you multiply 'm' by 'm', then add (-10 + 5) times 'm', and finally add (-10 times 5).

🎯 Exam Tip: Use the FOIL method (First, Outer, Inner, Last) to multiply two binomials like \( (x + a)(x + b) \) to ensure all terms are correctly multiplied.

Think (Text Book Page No. 87)

Question. Which is correct? (3a)² is equal to
(i) 3a²
(ii) 3²a
Answer:
Hint:
\( (3a)^2 = 3^2 a^2 = 9a^2 \)
(iv) 9a² The correct answer is 9a² because both the number and the variable inside the bracket need to be squared.
In simple words: When you square (3a), it means you square the 3 and you square the 'a'. So, it becomes 9a squared.

🎯 Exam Tip: Remember that \( (xy)^n = x^n y^n \). So, \( (3a)^2 = 3^2 a^2 = 9a^2 \). Do not just square the variable or the number alone.

Try These (Text Book Page No. 88)

Question 1. Expand using appropriate identities. (3p + 2q)²
Answer:
Comparing \( (3p + 2q)^2 \) with \( (a + b)^2 \), we get \( a = 3p \) and \( b = 2q \).
\( (a + b)^2 = a^2 + 2ab + b^2 \)
\( (3p + 2q)^2 = (3p)^2 + 2(3p)(2q) + (2q)^2 \)
\( = 9p^2 + 12pq + 4q^2 \) This expansion uses the algebraic identity for the square of a sum.
In simple words: To expand \( (3p + 2q)^2 \), you square the first part (3p), then add two times (3p) multiplied by (2q), and finally add the square of the second part (2q).

🎯 Exam Tip: Always correctly identify 'a' and 'b' in the identity \( (a + b)^2 = a^2 + 2ab + b^2 \). Ensure the entire term (number and variable) is squared, e.g., \( (3p)^2 = 9p^2 \), not \( 3p^2 \).

Question 2. Expand using appropriate identities. (105)²
Answer:
\( (105)^2 = (100 + 5)^2 \)
Comparing \( (100 + 5)^2 \) with \( (a + b)^2 \), we get \( a = 100 \) and \( b = 5 \).
\( (a + b)^2 = a^2 + 2ab + b^2 \)
\( (100 + 5)^2 = (100)^2 + 2(100)(5) + 5^2 \)
\( = 10000 + 1000 + 25 \)
\( 105^2 = 11,025 \) Breaking down 105 into \( (100 + 5) \) allows us to use an identity for easier calculation.
In simple words: To find the square of 105, you can write it as (100 + 5) and use the rule for squaring a sum. This means squaring 100, adding two times 100 multiplied by 5, and then adding the square of 5.

🎯 Exam Tip: When squaring numbers close to multiples of 10, 100, or 1000, use identities like \( (a+b)^2 \) or \( (a-b)^2 \) to simplify calculations without a calculator.

Question 3. Expand using appropriate identities. (2x - 5d)²
Answer:
Comparing with \( (a - b)^2 \), we get \( a = 2x, b = 5d \).
\( (a - b)^2 = a^2 - 2ab + b^2 \)
\( (2x - 5d)^2 = (2x)^2 - 2(2x)(5d) + (5d)^2 \)
\( = 4x^2 - 20xd + 25d^2 \) This expansion uses the algebraic identity for the square of a difference.
In simple words: To expand \( (2x - 5d)^2 \), you square the first part (2x), then subtract two times (2x) multiplied by (5d), and finally add the square of the second part (5d).

🎯 Exam Tip: Pay close attention to the minus sign in \( (a - b)^2 \). The middle term is \( -2ab \), not \( +2ab \). Ensure all terms in 'a' and 'b' are correctly squared.

Question 4. Expand using appropriate identities. (98)²
Answer:
\( (98)^2 = (100 - 2)^2 \)
Comparing \( (100 - 2)^2 \) with \( (a - b)^2 \) we get \( a = 100, b = 2 \).
\( (a - b)^2 = a^2 - 2ab + b^2 \)
\( (100 - 2)^2 = (100)^2 - 2(100)(2) + 2^2 \)
\( = 10000 - 400 + 4 \)
\( = 9600 + 4 \)
\( = 9604 \) Using this identity makes the calculation easier and faster.
In simple words: To find the square of 98, you can write it as (100 - 2) and use the rule for squaring a difference. This means squaring 100, subtracting two times 100 multiplied by 2, and then adding the square of 2.

🎯 Exam Tip: For numbers just below a multiple of 10, 100, etc., use the identity \( (a - b)^2 \) to simplify calculations. It's often easier than direct multiplication.

Question 5. Expand using appropriate identities. (y – 5)(y + 5)
Answer:
Comparing \( (y - 5)(y + 5) \) with \( (a - b)(a + b) \) we get \( a = y; b = 5 \).
\( (a - b)(a + b) = a^2 - b^2 \)
\( (y - 5)(y + 5) = y^2 - 5^2 \)
\( = y^2 - 25 \) This identity provides a quick way to multiply two binomials with the same terms but opposite signs.
In simple words: When you multiply (y minus 5) by (y plus 5), the answer is always y squared minus 5 squared.

🎯 Exam Tip: The difference of squares identity \( (a - b)(a + b) = a^2 - b^2 \) is very important. Recognize this pattern to quickly expand or factor expressions.

Question 6. Expand using appropriate identities. (3x)² - 5²
Answer:
Comparing \( (3x)^2 - 5^2 \) with \( a^2 - b^2 \) we have \( a = 3x; b = 5 \).
\( (a^2 - b^2) = (a + b)(a - b) \)
\( (3x)^2 - 5^2 = (3x + 5)(3x - 5) \) This allows us to factor the expression into two binomials.
In simple words: When you have (3x) squared minus 5 squared, you can write it as (3x plus 5) multiplied by (3x minus 5).

🎯 Exam Tip: Always make sure to consider the entire term, including its coefficient, as 'a' or 'b' in the difference of squares identity. Here, 'a' is \( 3x \), not just 'x'.

Question 7. Expand using appropriate identities. (2m + n)(2m + p)
Answer:
Comparing \( (2m + n)(2m + p) \) with \( (x + a)(x + b) \) we have \( x = 2m; a = n; b = p \).
\( (x + a)(x + b) = x^2 + (a + b)x + ab \)
\( (2m + n)(2m + p) = (2m)^2 + (n + p)(2m) + (n)(p) \)
\( = 4m^2 + 2mn + 2mp + np \) This identity helps simplify the multiplication of two binomials that share a common term.
In simple words: When multiplying two brackets where the first part is the same (like 2m), you square the common part, then add the sum of the other parts multiplied by the common part, and finally add the product of the other parts.

🎯 Exam Tip: For binomials of the form \( (x + a)(x + b) \), the expansion is \( x^2 + (a + b)x + ab \). Be careful to correctly substitute the common term for 'x' and the different terms for 'a' and 'b'.

Question 8. Expand using appropriate identities. 203 \( \times \) 197
Answer:
\( 203 \times 197 = (200 + 3)(200 - 3) \)
Comparing \( (a + b)(a - b) \) we have \( a = 200, b = 3 \).
\( (a + b)(a - b) = a^2 - b^2 \)
\( (200 + 3)(200 - 3) = 200^2 - 3^2 \)
\( = 40000 - 9 \)
\( 203 \times 197 = 39991 \) This method simplifies multiplication by using algebraic identities, making calculations faster and less prone to errors.
In simple words: To multiply 203 by 197, you can think of it as (200 + 3) multiplied by (200 - 3). This is a special pattern that equals 200 squared minus 3 squared, which is easy to calculate.

🎯 Exam Tip: For products of numbers just above and below a multiple of 100 (or 10, or 1000), use the identity \( (a + b)(a - b) = a^2 - b^2 \). This provides a quick mental math shortcut.

Question 9. Find the area of the square whose side is (x – 2) units.
Answer:
Side of a square \( = x - 2 \)
Area \( = \text{Side} \times \text{Side} \)
\( = (x - 2)(x - 2) = x(x - 2) - 2(x - 2) \)
\( = x(x) + x(-2) - 2(x) - 2(-2) \)
\( = x^2 - 2x - 2x + 4 \)
\( = x^2 - 4x + 4 \) units square. The area of a square is found by squaring its side length.
In simple words: The area of a square is its side length multiplied by itself. So, for a side of (x - 2), the area is (x - 2) times (x - 2), which works out to \( x^2 - 4x + 4 \).

🎯 Exam Tip: Remember the formula for the area of a square: Area = \( \text{side}^2 \). When the side is an algebraic expression, use the square of a binomial identity \( (a-b)^2 = a^2 - 2ab + b^2 \) for expansion.

Question 10. Find the area of the rectangle whose length and breadth are (y + 4) units and (y – 3) units.
Answer:
Length of the rectangle \( = y + 4 \)
Breadth of the rectangle \( = y - 3 \)
Area of the rectangle \( = \text{length} \times \text{breadth} \)
\( = (y + 4)(y - 3) = y^2 + (4 - 3)y + (4)(-3) \)
\( = y^2 + y - 12 \) units square. The area of a rectangle is calculated by multiplying its length by its breadth.
In simple words: To find the area of a rectangle, you multiply its length by its width. Here, multiply (y + 4) by (y - 3), which gives \( y^2 + y - 12 \).

🎯 Exam Tip: For area problems involving algebraic expressions, write down the formula first, then substitute the given expressions and expand carefully, using identities like \( (x+a)(x+b) = x^2 + (a+b)x + ab \) or the FOIL method.

Try These (Text Book Page No. 91)

Question. Expand: (i) (x + 5)³
Answer:
Comparing \( (x + 5)^3 \) with \( (a + b)^3 \), we have \( a = x \) and \( b = 5 \).
\( (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \)
\( (x + 5)^3 = x^3 + 3x^2(5) + 3(x)(5)^2 + 5^3 \)
\( = x^3 + 15x^2 + 75x + 125 \) This uses the binomial expansion for a cube.
In simple words: To expand \( (x + 5)^3 \), you cube the first term (x), add three times the square of the first term multiplied by the second (5), add three times the first term multiplied by the square of the second (5), and finally cube the second term (5).

🎯 Exam Tip: Remember the identity \( (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \). Be careful with the coefficients and powers for each term in the expansion.

Question. Expand: (ii) (y – 2)³
Answer:
Comparing \( (y - 2)^3 \) with \( (a - b)^3 \) we have \( a = y \) and \( b = 2 \).
\( (a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 \)
\( (y - 2)^3 = y^3 - 3y^2(2) + 3y(2)^2 - 2^3 \)
\( = y^3 - 6y^2 + 12y - 8 \) This uses the binomial expansion for a cube with a negative term.
In simple words: To expand \( (y - 2)^3 \), you cube 'y', then subtract three times 'y' squared multiplied by 2, add three times 'y' multiplied by 2 squared, and finally subtract 2 cubed.

🎯 Exam Tip: For \( (a-b)^3 \), the signs alternate: \( a^3 - 3a^2b + 3ab^2 - b^3 \). Make sure to include the correct signs for each term when expanding.

Question. Expand: (iii) (x + 1)(x + 4)(x + 6)
Answer:
Comparing \( (x + 1)(x + 4)(x + 6) \) with \( (x + a)(x + b)(x + c) \) we have \( a = 1, b = 4 \) and \( c = 6 \).
\( (x + a)(x + b)(x + c) = x^3 + (a + b + c)x^2 + (ab + bc + ca)x + abc \)
\( = x^3 + (1 + 4 + 6)x^2 + (1)(4) + (4)(6) + (6)(1)x + (1)(4)(6) \)
\( = x^3 + 11x^2 + (4 + 24 + 6)x + 24 \)
\( = x^3 + 11x^2 + 34x + 24 \) This identity is useful for expanding three binomial factors that share a common variable.
In simple words: When multiplying three brackets like these, you get x cubed, plus (the sum of the numbers) times x squared, plus (the sum of products of numbers taken two at a time) times x, plus (the product of all the numbers).

🎯 Exam Tip: Remember the expansion for \( (x+a)(x+b)(x+c) \). This identity helps avoid lengthy step-by-step multiplication. Carefully sum and multiply the constants for each term.

Find the factors

🎯 Exam Tip: To find factors for a given product and sum, list all pairs of numbers that multiply to the product, then check which pair also adds up to the sum. Pay close attention to positive and negative signs.

Think (Text Book Page No. 94)

Question. x² - 4(x – 2) = (x² – 4)(x – 2) Is this correct? If not, correct it.
Answer:
No, this statement is not correct.
Corrected version:
\( x^2 - 4(x - 2) = x^2 - 4x + 8 \)
The statement is incorrect because the distributive property was not applied correctly on the left side, and the right side uses a different factorization approach. The term \( -4(x-2) \) expands to \( -4x+8 \), not \( (x^2-4)(x-2) \).
In simple words: The first statement is wrong. When you have -4 multiplied by (x - 2), it should become -4x + 8. You cannot just change the problem to multiply (x squared - 4) by (x - 2).

🎯 Exam Tip: Always remember the distributive property: \( a(b - c) = ab - ac \). Apply it carefully to avoid common errors when multiplying a single term by a binomial.

Try These (Text Book Page No. 95)

Question 1. 3y + 6
Answer:
\( 3y + 6 = 3 \times y + 2 \times 3 \)
Taking out the common factor 3 from each term we get \( 3(y + 2) \).
So, \( 3y + 6 = 3(y + 2) \) Factoring helps simplify expressions by writing them as a product of their common parts.
In simple words: In the expression 3y + 6, both parts can be divided by 3. So, we can take 3 outside the bracket, leaving (y + 2) inside.

🎯 Exam Tip: Always look for the greatest common factor (GCF) among all terms in an expression. Factoring out the GCF is the first step in most factorization problems.

Question 2. 10x² + 15y²
Answer:
\( 10x^2 + 15y^2 = (2 \times 5 \times x \times x) + (3 \times 5 \times y \times y) \)
Taking out the common factor 5 we have
\( 10x^2 + 15y^2 = 5(2x^2 + 3y^2) \) This shows that 5 is the common factor for both terms.
In simple words: Both 10x squared and 15y squared can be divided by 5. So, you can pull the 5 out, and what's left inside the bracket is (2x squared + 3y squared).

🎯 Exam Tip: When factoring out the GCF, ensure you consider both the numerical coefficients and any common variables. Here, only the number 5 is common.

Question 3. 7m(m - 5) + 1(5 – m)
Answer:
\( 7m(m - 5) + 1(5 - m) \)
We can rewrite \( (5 - m) \) as \( -1(m - 5) \).
So, \( 7m(m - 5) + 1(5 - m) = 7m(m - 5) + (-1)(m - 5) \)
\( = 7m(m - 5) - 1(m - 5) \)
Taking out the common binomial factor \( (m - 5) \), we get \( (m - 5)(7m - 1) \) This technique is useful when terms appear to be similar but have opposite signs, allowing for further factorization.
In simple words: Notice that (5 - m) is just the negative of (m - 5). So, we change 1(5 - m) to -1(m - 5). Then, both parts have (m - 5) in common, which we can take out.

🎯 Exam Tip: If you see terms like \( (a - b) \) and \( (b - a) \), remember that \( (b - a) = -(a - b) \). This trick is essential for creating common factors in more complex expressions.

Question 4. 64 - x²
Answer:
\( 64 - x^2 = 8^2 - x^2 \)
This is of the form \( a^2 - b^2 \).
Comparing with \( a^2 - b^2 \) we have \( a = 8, b = x \).
\( a^2 - b^2 = (a + b)(a - b) \)
\( 64 - x^2 = (8 + x)(8 - x) \) This is a classic example of factoring the difference of two perfect squares.
In simple words: The number 64 is 8 squared. So, if you have 8 squared minus x squared, you can write it as (8 plus x) multiplied by (8 minus x).

🎯 Exam Tip: Always recognize the pattern \( a^2 - b^2 \). This identity is one of the most frequently used factoring formulas and often leads to simpler expressions.

Question 5. x² - 3x + 2
Answer:
We need two numbers that multiply to 2 and add up to -3. These numbers are -2 and -1.
\( x^2 - 3x + 2 = x^2 - 2x - x + 2 \)
\( = x(x - 2) - 1(x - 2) \)
\( = (x - 2)(x - 1) \) This method involves splitting the middle term to find common factors, which is a key technique for factoring quadratic trinomials.
In simple words: To factor \( x^2 - 3x + 2 \), find two numbers that multiply to 2 and add to -3. These are -2 and -1. Then, rewrite the middle term, group the terms, and factor out the common parts.

🎯 Exam Tip: For trinomials like \( ax^2 + bx + c \), look for two numbers that multiply to 'ac' and add up to 'b'. Then, rewrite the middle term and factor by grouping.

Question 6. y² – 4y – 32
Answer:
We need two numbers that multiply to -32 and add up to -4. These numbers are -8 and 4.
\( y^2 - 4y - 32 = y^2 - 8y + 4y - 32 \)
\( = y(y - 8) + 4(y - 8) \)
\( = (y - 8)(y + 4) \) This method of factoring quadratic trinomials is called splitting the middle term.
In simple words: To factor \( y^2 - 4y - 32 \), find two numbers that multiply to -32 and add to -4. These are -8 and 4. Then, split the middle term, group, and factor.

🎯 Exam Tip: When the last term is negative, the two numbers you are looking for will have opposite signs. The sign of the middle term will tell you which number (the larger or smaller absolute value) carries the negative sign.

Question 7. p² + 2p – 15
Answer:
We need two numbers that multiply to -15 and add up to 2. These numbers are 5 and -3.
\( p^2 + 2p - 15 = p^2 + 5p - 3p - 15 \)
\( = p(p + 5) - 3(p + 5) \)
\( = (p + 5)(p - 3) \) This demonstrates the process of factoring a quadratic expression by grouping terms after splitting the middle term.
In simple words: To factor \( p^2 + 2p - 15 \), find two numbers that multiply to -15 and add to 2. These are 5 and -3. Rewrite the middle part, then group and factor.

🎯 Exam Tip: If the last term is negative and the middle term is positive, the larger of the two numbers you find (in absolute value) should be positive, and the smaller should be negative.

Question 8. m² + 14m + 48
Answer:
We need two numbers that multiply to 48 and add up to 14. These numbers are 6 and 8.
\( m^2 + 14m + 48 = m^2 + 8m + 6m + 48 \)
\( = m(m + 8) + 6(m + 8) \)
\( = (m + 6)(m + 8) \) This illustrates how to factor a quadratic trinomial by finding two numbers whose product is the constant term and whose sum is the coefficient of the middle term.
In simple words: To factor \( m^2 + 14m + 48 \), look for two numbers that multiply to 48 and add to 14. These are 6 and 8. Then, rewrite the middle term, group the terms, and factor out the common parts.

🎯 Exam Tip: When both the middle and last terms are positive, the two numbers you find will both be positive. This simplifies the search for factors.

Question 9. x² - x - 90
Answer:
We need two numbers that multiply to -90 and add up to -1. These numbers are -10 and 9.
\( x^2 - x - 90 = x^2 - 10x + 9x - 90 \)
\( = x(x - 10) + 9(x - 10) \)
\( = (x + 9)(x - 10) \) This demonstrates the factoring of a quadratic expression by splitting the middle term and then grouping.
In simple words: To factor \( x^2 - x - 90 \), find two numbers that multiply to -90 and add to -1. These are -10 and 9. Then, split the middle term, group, and factor.

🎯 Exam Tip: For large constant terms, systematically list pairs of factors and check their sums to quickly find the correct combination. Remember that a negative constant term implies one positive and one negative factor.

Question 10. 9x² - 6x – 8
Answer:
We need two numbers that multiply to \( 9 \times -8 = -72 \) and add up to -6. These numbers are -12 and 6.
\( 9x^2 - 6x - 8 = 9x^2 - 12x + 6x - 8 \)
\( = 3x(3x - 4) + 2(3x - 4) \)
\( = (3x + 2)(3x - 4) \) This method involves multiplying 'a' and 'c' (the coefficient of \( x^2 \) and the constant term), then finding factors for that product that sum to 'b' (the coefficient of x).
In simple words: For \( 9x^2 - 6x - 8 \), find two numbers that multiply to \( 9 \times -8 = -72 \) and add to -6. These are -12 and 6. Split the middle term, then group the terms and factor them out.

🎯 Exam Tip: When the coefficient of \( x^2 \) is not 1, multiply 'a' by 'c' first. Find two numbers that multiply to 'ac' and add to 'b', then use these numbers to split the middle term and factor by grouping.

Try These (Text Book Page No. 99)

Question. Identify which among the following are linear equations. (i) 2 + x = 19
Answer:
\( 2 + x = 19 \)
This is a linear equation as the degree of the variable x is 1. A linear equation has its variables raised to the power of one, making a straight line when graphed.
In simple words: Yes, this is a linear equation because the power of 'x' is just 1.

🎯 Exam Tip: A linear equation is one where the highest power of any variable is 1. There are no terms with \( x^2 \), \( y^3 \), \( xy \), etc.

Question. Identify which among the following are linear equations. (ii) 7x² – 5 = 3
Answer:
\( 7x^2 - 5 = 3 \)
This is not a linear equation as the highest degree of x is 2. Because the variable is squared, it forms a parabolic curve, not a straight line.
In simple words: No, this is not a linear equation because 'x' is squared, meaning its highest power is 2.

🎯 Exam Tip: To check if an equation is linear, look at the highest power of any variable. If it's anything other than 1 (like 2, 3, etc.), it's not a linear equation.

Question. Identify which among the following are linear equations. (iii) 4p³ = 12
Answer:
\( 4p^3 = 12 \)
This is not a linear equation as the highest degree of p is 3. An equation is linear only if all variables have an exponent of 1.
In simple words: No, this is not a linear equation because 'p' is cubed, meaning its highest power is 3.

🎯 Exam Tip: The degree of a variable is its highest exponent. For an equation to be linear, all variables must have a degree of 1.

Question. Identify which among the following are linear equations. (iv) 6m + 2
Answer:
\( 6m + 2 \)
This is linear, but not an equation. An equation must contain an equality sign \( (=) \) to show that two expressions are equal. This is an algebraic expression.
In simple words: This is a linear expression because 'm' has a power of 1. But it's not an equation because there is no equals sign.

🎯 Exam Tip: An expression is a combination of terms. An equation sets two expressions equal to each other. Both are important in algebra but serve different purposes.

Question. Identify which among the following are linear equations. (v) n = 10
Answer:
\( n = 10 \)
This is a linear equation as the degree of n is 1. This is a very simple form of a linear equation, showing a direct value for the variable.
In simple words: Yes, this is a linear equation because 'n' has a power of 1, and it has an equals sign.

🎯 Exam Tip: A single variable set equal to a constant (e.g., \( x=5 \)) is a linear equation because the variable's highest power is 1.

Question. Identify which among the following are linear equations. (vi) 7k – 12 = 0
Answer:
\( 7k - 12 = 0 \)
This is a linear equation as the degree of k is 1. This is a common form of a linear equation in one variable.
In simple words: Yes, this is a linear equation because 'k' has a power of 1, and it has an equals sign.

🎯 Exam Tip: Standard form for a linear equation in one variable is \( ax + b = 0 \), where 'a' is not zero. This example fits that form.

Question. Identify which among the following are linear equations. (vii) \( \frac{6 x}{8} + y = 1 \)
Answer:
\( \frac{6x}{8} + y = 1 \)
This is a linear equation as the degree of x and y is 1. Even with two variables, if their powers are both one, the equation is linear.
In simple words: Yes, this is a linear equation because both 'x' and 'y' have a power of 1, and it has an equals sign.

🎯 Exam Tip: Linear equations can have one or more variables. As long as the highest power of each variable is 1, it remains a linear equation.

Question. Identify which among the following are linear equations. (viii) 5 + y = 3x
Answer:
\( 5 + y = 3x \)
This is a linear equation as the degree of y and x is 1. This can be rearranged into a standard linear form, such as \( 3x - y - 5 = 0 \).
In simple words: Yes, this is a linear equation because both 'y' and 'x' have a power of 1, and it has an equals sign.

🎯 Exam Tip: An equation is linear if rearranging it results in a form where all variable terms have an exponent of 1, and there are no products of variables.

Question. Identify which among the following are linear equations. (ix) 10p + 2q = 3
Answer:
\( 10p + 2q = 3 \)
This is a linear equation as the degree of p and q is 1. It represents a straight line in a 2D coordinate system.
In simple words: Yes, this is a linear equation because both 'p' and 'q' have a power of 1, and it has an equals sign.

🎯 Exam Tip: In a linear equation with multiple variables, the power of each individual variable must be 1. Products like 'pq' would make it non-linear.

Question. Identify which among the following are linear equations. (x) x² – 2x – 4
Answer:
\( x^2 - 2x - 4 \)
This is not a linear equation as the highest degree of x is 2. Also, it is an expression, not an equation, as it lacks an equality sign.
In simple words: No, this is not a linear equation because 'x' is squared. It's also an expression, not an equation, as it has no equals sign.

🎯 Exam Tip: Carefully check both the degree of variables and the presence of an equality sign to correctly identify linear equations versus other types of expressions.

Think (Text Book Page No. 99)

Question. (i) Is t(t – 5) = 10 a linear equation? Why?
Answer:
\( t(t - 5) = 10 \)
\( t \times t - 5 \times t = 10 \)
\( t^2 - 5t = 10 \)
This is not a linear equation as the highest degree of the variable 't' is 2. When expanded, the \( t^2 \) term shows it's a quadratic equation.
In simple words: No, this is not a linear equation. When you multiply 't' by (t - 5), you get \( t^2 \). Since 't' is squared, it's not linear.

🎯 Exam Tip: Always expand or simplify equations first before determining if they are linear. The highest power of a variable might not be obvious in its initial form.

Question. (ii) Is x² = 2x, a linear equation? Why?
Answer:
\( x^2 = 2x \)
\( x^2 - 2x = 0 \)
This is not a linear equation as the highest degree of the variable 'x' is 2. Even though it can be factored, it is still a quadratic equation.
In simple words: No, this is not a linear equation. If you move 2x to the other side, you still have \( x^2 \). Because 'x' is squared, it is not linear.

🎯 Exam Tip: An equation like \( x^2 = 2x \) has two solutions (\( x=0 \) and \( x=2 \)), characteristic of a quadratic equation, not a linear one which typically has only one solution.

Try These (Text Book Page No. 100)

Question. Convert the following statements into linear equations: On subtracting 8 from the product of 5 and a number, I get 32.
Answer:
Let the number be 'x'.
The product of 5 and the number is \( 5 \times x = 5x \).
Subtracting 8 from this product means \( 5x - 8 \).
The result is 32, so the equation is: \( 5x - 8 = 32 \) This process translates a word problem into a mathematical equation, which can then be solved.
In simple words: If you take a number (let's call it x), multiply it by 5, then take away 8, you get 32. So the equation is \( 5x - 8 = 32 \).

🎯 Exam Tip: Carefully read word problems to identify the unknown quantity (assign a variable like 'x'), and then translate each phrase into its corresponding mathematical operation (product, difference, sum, etc.).

Question 2. The sum of three consecutive integers is 78.
Answer:
Let the first integer be 'x'.
Then the next consecutive integer is \( x + 1 \).
The third consecutive integer is \( x + 2 \).
The sum of three consecutive integers is 78, so:
\( x + (x + 1) + (x + 2) = 78 \)
\( x + x + 1 + x + 2 = 78 \)
\( 3x + 3 = 78 \) This forms a linear equation by defining consecutive integers algebraically.
In simple words: If you take three numbers that come one after another (like x, x+1, and x+2) and add them up, the total is 78. This makes the equation \( 3x + 3 = 78 \).

🎯 Exam Tip: For consecutive integers, use \( x, x+1, x+2, \dots \). For consecutive even or odd integers, use \( x, x+2, x+4, \dots \).

Question 3. Peter had a Two hundred rupee note. After buying 7 copies of a book he was left with 60.
Answer:
Let the cost of one book be 'x' Rs.
The cost of 7 copies of the book is \( 7 \times x = 7x \) Rs.
Peter started with 200 Rs. After buying books, he was left with 60 Rs.
So, \( 200 - 7x = 60 \) This equation represents the remaining money after a purchase.
In simple words: Peter had 200 rupees. He bought 7 books, each costing 'x' rupees. He had 60 rupees left. So, 200 minus the cost of 7 books equals 60.

🎯 Exam Tip: In problems involving money, clearly define what each variable represents. Distinguish between the initial amount, the amount spent, and the amount remaining.

Question 4. The base angles of an isosceles triangle are equal and the vertex angle measures 80°. Applying triangle property, sum of all angles is 180°.
Answer:
Let each base angle be 'x'.
The vertex angle is 80°.
The sum of angles in a triangle is 180°.
So, \( x + x + 80 = 180 \)
\( 2x + 80 = 180 \) This linear equation uses the fundamental property of triangles to relate its angles.
In simple words: In an isosceles triangle, the two base angles are the same. If the top angle is 80 degrees, and all angles add up to 180 degrees, then x (for each base angle) plus x plus 80 must equal 180.

🎯 Exam Tip: Always remember the properties of different types of triangles (e.g., isosceles, equilateral) and the fundamental rule that the sum of interior angles in any triangle is 180°.

Question 5. In a triangle ABC, \( \angle A \) is 100 more than \( \angle B \). Also \( \angle C \) is three times \( \angle A \). Express the equation in terms of angle B.
Answer: Let angle B be \( b \).
Angle A is \( 10^\circ \) more than angle B, so \( \angle A = 10^\circ + b \).
Angle C is three times angle A, so \( \angle C = 3 \times (10^\circ + b) = 30^\circ + 3b \).
The sum of angles in a triangle is \( 180^\circ \).
So, \( \angle A + \angle B + \angle C = 180^\circ \).
Substituting the expressions: \( (10^\circ + b) + b + (30^\circ + 3b) = 180^\circ \).
Combining like terms, we get \( 5b + 40^\circ = 180^\circ \).
This is the equation in terms of angle B.
In simple words: First, write down what each angle is in terms of 'b'. Then, add all three angles together and set them equal to 180 degrees, because all angles in a triangle always add up to 180 degrees.

🎯 Exam Tip: Remember that the sum of angles in any triangle is always \( 180^\circ \). This fundamental property is crucial for solving problems involving triangle angles.

Question. Can you get more than one solution for a linear equation?
Answer: Yes, a linear equation with two or more variables typically has many solutions. For example, consider the equation \( x + y = 5 \).
If \( x = 1 \), then \( y = 4 \).
If \( x = 2 \), then \( y = 3 \).
If \( x = 3 \), then \( y = 2 \).
If \( x = 4 \), then \( y = 1 \).
There are many more pairs of \( x \) and \( y \) that satisfy this equation, such as \( x=0, y=5 \) or \( x=5, y=0 \), or even fractional values like \( x=2.5, y=2.5 \). Each pair is a solution.
In simple words: Yes, for a line equation with two different letters (like x and y), there are usually endless answers. You can pick many pairs of numbers for x and y that still make the equation true.

🎯 Exam Tip: A single linear equation with two variables always has infinitely many solutions, forming a straight line when plotted on a graph.

Try These (Text Book Page No. 101)

Question. Identify which among the following are linear equations.
(i) \( 2 + x = 10 \)
Answer: \( 2 + x = 10 \)
\( \implies x = 10 - 2 \)
\( \implies x = 8 \)
This is a linear equation because the highest power of the variable \( x \) is 1.
In simple words: This is a linear equation because the letter 'x' has a power of 1, which means it's a straight line when you draw it.

🎯 Exam Tip: A linear equation is an equation where the highest exponent of the variable (or variables) is 1. There are no terms with \( x^2, y^3 \), or square roots of variables.

Question. Identify which among the following are linear equations.
(ii) \( 3 + x = 5 \)
Answer: \( 3 + x = 5 \)
\( \implies x = 5 - 3 \)
\( \implies x = 2 \)
This is a linear equation because the highest power of the variable \( x \) is 1.
In simple words: Yes, this is a linear equation. The letter 'x' is raised to the power of 1, so it is a simple, straight-line equation.

🎯 Exam Tip: Linear equations involve only basic arithmetic operations (addition, subtraction, multiplication, division) on variables, not exponents higher than one or other complex functions.

Question. Identify which among the following are linear equations.
(iii) \( x - 6 = 10 \)
Answer: \( x - 6 = 10 \)
\( \implies x = 10 + 6 \)
\( \implies x = 16 \)
This is a linear equation because the highest power of the variable \( x \) is 1.
In simple words: This equation is linear because 'x' has a power of 1. It forms a straight line.

🎯 Exam Tip: Remember that the highest power (or degree) of the variable determines if an equation is linear. For linear equations, this degree must always be 1.

Question. Identify which among the following are linear equations.
(iv) \( 3x + 5 = 2 \)
Answer: \( 3x + 5 = 2 \)
\( \implies 3x = 2 - 5 \)
\( \implies 3x = -3 \)
\( \implies x = -1 \)
This is a linear equation because the highest power of the variable \( x \) is 1.
In simple words: This is a linear equation. The variable 'x' has an exponent of 1, making it a simple, straight-line relationship.

🎯 Exam Tip: Linear equations are foundational in algebra. Recognize them by the 'power of 1' rule for all variables involved.

Question. Identify which among the following are linear equations.
(v) \( \frac{2 x}{7} = 3 \)
Answer: \( \frac{2 x}{7} = 3 \)
\( \implies 2x = 3 \times 7 \)
\( \implies 2x = 21 \)
\( \implies x = \frac{21}{2} \)
This is a linear equation because the highest power of the variable \( x \) is 1.
In simple words: Yes, this is a linear equation because the letter 'x' is only raised to the power of 1. It can be solved directly.

🎯 Exam Tip: Fractions with variables in the numerator (like \( \frac{2x}{7} \)) are still linear as long as the variable itself doesn't have a power other than 1.

Question. Identify which among the following are linear equations.
(vi) \( -2 = 4m - 6 \)
Answer: \( -2 = 4m - 6 \)
\( \implies -2 + 6 = 4m \)
\( \implies 4 = 4m \)
\( \implies m = \frac{4}{4} \)
\( \implies m = 1 \)
This is a linear equation because the highest power of the variable \( m \) is 1.
In simple words: This is a linear equation because 'm' is only raised to the power of 1. It is simple and forms a straight line.

🎯 Exam Tip: When an equation involves only one variable, it is linear if that variable has an exponent of 1, even if numbers are on both sides of the equal sign.

Question. Identify which among the following are linear equations.
(vii) \( 4(3x - 1) = 80 \)
Answer: \( 4(3x - 1) = 80 \)
\( \implies 12x - 4 = 80 \)
\( \implies 12x = 80 + 4 \)
\( \implies 12x = 84 \)
\( \implies x = \frac{84}{12} \)
\( \implies x = 7 \)
This is a linear equation because the highest power of the variable \( x \) is 1.
In simple words: This is a linear equation because when you multiply out the bracket, the letter 'x' still has a power of 1.

🎯 Exam Tip: Distribute terms carefully when brackets are present. Even after distribution, if the variable's highest power remains 1, the equation is linear.

Question. Identify which among the following are linear equations.
(viii) \( 3x - 8 = 7 - 2x \)
Answer: \( 3x - 8 = 7 - 2x \)
\( \implies 3x + 2x = 7 + 8 \)
\( \implies 5x = 15 \)
\( \implies x = \frac{15}{5} \)
\( \implies x = 3 \)
This is a linear equation because the highest power of the variable \( x \) is 1.
In simple words: This is a linear equation because the highest power of 'x' is 1. You can move terms around to solve it easily.

🎯 Exam Tip: To solve linear equations with variables on both sides, gather all variable terms on one side and all constant terms on the other, then simplify.

Question. Identify which among the following are linear equations.
(ix) \( 7 - y = 3(5 - y) \)
Answer: \( 7 - y = 3(5 - y) \)
\( \implies 7 - y = 15 - 3y \)
\( \implies 3y - y = 15 - 7 \)
\( \implies 2y = 8 \)
\( \implies y = \frac{8}{2} \)
\( \implies y = 4 \)
This is a linear equation because the highest power of the variable \( y \) is 1.
In simple words: This is a linear equation because when you expand the right side, the letter 'y' is still only to the power of 1.

🎯 Exam Tip: Always expand brackets first when solving equations, then collect like terms to simplify and solve for the variable.

Question. Identify which among the following are linear equations.
(x) \( 4(1 - 2y) - 2(3 - y) = 0 \)
Answer: \( 4(1 - 2y) - 2(3 - y) = 0 \)
\( \implies 4 - 8y - 6 + 2y = 0 \)
\( \implies -2 - 6y = 0 \)
\( \implies 6y = -2 \)
\( \implies y = \frac{-2}{6} \)
\( \implies y = \frac{-1}{3} \)
This is a linear equation because the highest power of the variable \( y \) is 1.
In simple words: Even with multiple brackets, if the variable 'y' never gets a power higher than 1 when you expand everything, the equation is linear.

🎯 Exam Tip: Be careful with negative signs when distributing, especially when a negative number is outside the bracket (e.g., \( -2(3 - y) = -6 + 2y \)).

Think (Text Book Page No. 102)

Question 1. "An equation is multiplied or divided by a non zero number on either side:' Will there be any change in the solution?
Answer: No, there will be no change in the solution. If you multiply or divide both sides of an equation by the same non-zero number, the equality remains true, and the solution for the variable stays the same. For example, if \( 2x = 4 \), then \( x = 2 \). If you multiply by 3, you get \( 6x = 12 \), which still gives \( x = 2 \).
In simple words: No, the answer stays the same. Doing the same multiplication or division on both sides of an equation keeps it balanced, like a seesaw.

🎯 Exam Tip: Multiplying or dividing both sides of an equation by a non-zero number is a fundamental property that helps isolate the variable without changing the solution.

Question 2. “An equation is multiplied or divided by two different numbers on either side. What will happen to the equation?
Answer: If an equation is multiplied or divided by two *different* numbers on each side, the equation will change, and its solution will also change. This is because the balance of the equation is disturbed. For example, if \( x = 5 \), multiplying the left by 2 and the right by 3 gives \( 2x = 15 \), which means \( x = 7.5 \), a different solution.
In simple words: If you use different numbers to multiply or divide each side, the equation becomes unbalanced. This changes the problem and gives you a different answer.

🎯 Exam Tip: Always perform the same operation (addition, subtraction, multiplication, division) with the same number on *both* sides of an equation to maintain equality and find the correct solution.

Think (Text Book Page No. 104)

Question 1. Suppose a length of 200 cm is cut into two pieces. If the second piece is \( x \) cm, then the first piece is \( (200 - x) \) cm. If the first piece is 40 cm smaller than twice the second piece, how will the steps vary and will the final answer for the lengths be different?
Answer: Let the second piece be \( x \) cm.
Then the first piece is \( (200 - x) \) cm.
According to the problem, the first piece is 40 cm smaller than twice the second piece.
\( \implies 200 - x = 2x - 40 \)
Now, we solve for \( x \).
Add 40 to both sides:
\( \implies 200 + 40 = 2x + x \)
\( \implies 240 = 3x \)
Divide by 3:
\( \implies x = \frac{240}{3} = 80 \)
So, the second piece is \( 80 \) cm.
The first piece is \( 200 - x = 200 - 80 = 120 \) cm.
The steps vary in how the equation is set up, but the final answer for the lengths will not change, as it must accurately reflect the problem's conditions.
In simple words: We can write the problem as an equation: \( 200 - x = 2x - 40 \). Solving this equation gives us the length of the second piece as 80 cm and the first piece as 120 cm. Even if we set it up differently, the real lengths won't change.

🎯 Exam Tip: When solving problems by setting up equations, choosing different variables or expressions for the unknowns can change the intermediate steps, but the final values for the physical quantities (like lengths, in this case) must remain consistent.

Think (Text Book Page No. 109)

Question 1. If instead of (4,3), we write (3,4) and tn to mark it, will it represent 'M' again?
Answer: No, if we mark (3, 4) instead of (4, 3), it will represent a different point. In a coordinate system, the order of numbers in an ordered pair matters. (4, 3) means 4 units along the x-axis and 3 units along the y-axis, while (3, 4) means 3 units along the x-axis and 4 units along the y-axis. These are distinct locations.
In simple words: No, it will be a different point. In math, the order of numbers in a point (like x, y) is very important. Changing the order changes the spot on the graph.

🎯 Exam Tip: Remember that in ordered pairs \( (x, y) \), the first number always refers to the x-coordinate (horizontal position) and the second number refers to the y-coordinate (vertical position). The order is crucial.

Try These (Text Book Page No. 111)

Question 1. Complete the table given below.
S.No | Point | Sign of X-coordinate | Sign of Y-coordinate | Quadrant
---|---|---|---|---
1 | (-7,2) | | |
2 | (10,-2) | | |
3 | (-3,-7) | | |
4. | (3,1) | | |
5. | (7,0) | | |
6. | (0,-4) | | |
Answer:

In simple words: Look at the signs of the x and y coordinates. If both are positive, it's Quadrant I. If x is negative and y is positive, it's Quadrant II. If both are negative, it's Quadrant III. If x is positive and y is negative, it's Quadrant IV. If a coordinate is zero, the point lies on an axis.

🎯 Exam Tip: Remember that points on the x-axis have a y-coordinate of 0, and points on the y-axis have an x-coordinate of 0. These points are not in any quadrant but on the axis itself.

Question 2. Write the coordinates of the points marked in the following figure
Answer: The coordinates of the points marked in the figure are:

B: (5, 2)
C: (5, -3)
D: (-3, -3)
E: (-1, 4)
F: (1, 2)
G: (7, 4)
H: (0, 2)
I: (0, 3)
J: (-3, 0)
K: (5, 0)
L: (-1, 0)
M: (-2, 0)
N: (-2, -1)
O: (0, 0)
P: (-1, -1)
Q: (1, -1)
R: (2, -1)
S: (0, -3)
T: (7, 0)
U: (7, -2)

In simple words: Each point on the graph has two numbers, an x-coordinate for how far left or right it is from the center, and a y-coordinate for how far up or down. The center point is (0,0).

🎯 Exam Tip: Always remember that the x-coordinate comes first and the y-coordinate second in an ordered pair (x, y). Count carefully from the origin (0,0) to find the exact position.

Think (Text Book Page No. 114)

Question 1. Which of the points (5, -10) (0, 5) (5, 20) lie on the straight line x = 5?
Answer: A straight line defined by \( x = 5 \) means that every point on this line must have an x-coordinate of 5, regardless of its y-coordinate. Therefore, we look for points where the first number is 5.
- (5, -10): The x-coordinate is 5. So, this point lies on the line \( x = 5 \).
- (0, 5): The x-coordinate is 0, not 5. So, this point does not lie on the line \( x = 5 \).
- (5, 20): The x-coordinate is 5. So, this point lies on the line \( x = 5 \).
Thus, the points (5, -10) and (5, 20) lie on the line \( x = 5 \).
In simple words: Only points where the first number (the x-coordinate) is 5 will be on the line \( x = 5 \). So, (5, -10) and (5, 20) are on the line, but (0, 5) is not.

🎯 Exam Tip: A vertical line is represented by \( x = c \) (where c is a constant). All points on this line will have the same x-coordinate, which is \( c \). Similarly, a horizontal line is represented by \( y = c \), and all points on it have the same y-coordinate.

Try These (Text Book Page No. 117)

Question 1. Identify and correct the errors
(i)
Answer: The original graph shows incorrect labeling and scaling, particularly on the y-axis where negative values are not sequential. Also, the x-axis has incorrect negative labels shown. The corrected graph below displays proper numbering and scaling for both axes, with an even interval between each number.

In simple words: A coordinate graph must have clear and consistent numbering on both axes. The numbers must increase evenly, and the positive and negative directions should be clearly marked from the origin.

🎯 Exam Tip: Always check that the numbering on both the x-axis and y-axis is uniform and correctly extends into both positive and negative directions from the origin (0,0).

Question 1. Identify and correct the errors
(ii)
Answer: The original graph contains an error in its scale definition. The x-axis labels (10, 20, 30) do not match the declared scale of "X axis 1cm = 1unit". The corrected graph below adjusts the x-axis scale to "X axis 1cm = 10 units" to accurately reflect the numbering on the axis.

In simple words: The scale on a graph tells us what each unit mark means. It's important that the numbers marked on the axes match the scale described, or else the graph will be misleading.

🎯 Exam Tip: Always verify that the scale provided for a graph matches the numbering on the axes. A common mistake is to have a visual scale (e.g., marks every 1 unit) but a written scale (e.g., 1cm = 10 units) that contradicts it.

Question 1. Identify and correct the errors
(iii)
Answer: The original graph contains errors where the marked points (the small circles) do not match their assigned coordinate labels. For example, a label might say (-2,1), but the actual circle is at (-2,2). The corrected graph below shows the points accurately placed at their labeled coordinates.

In simple words: When marking points on a graph, always make sure the dot or marker is exactly at the position indicated by its coordinates. The label should point to or be near the correct spot.

🎯 Exam Tip: Pay close attention to detail when plotting points. A small error in one coordinate can shift the point significantly, leading to an incorrect graphical representation.