GSEB Class 7 Maths Solutions Chapter 12 Algebraic Expressions InText Questions

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Question 1. Describe expressions are obtained:
(i) 7xy + 5
(ii) x²y
(iii) 4x² - 5x
Answer:
(i) \( 7xy + 5 \)
To get this expression, we first multiply two variables \( x \) and \( y \), which gives \( xy \). Then we multiply their product by 7 to make \( 7xy \). After that, we add 5 to \( 7xy \) to finally get \( 7xy + 5 \).
(ii) \( x^2y \)
First, \( x \) is multiplied by itself, meaning \( x \times x = x^2 \). Then we multiply \( x^2 \) by \( y \) to obtain the necessary expression, so \( x^2 \times y = x^2y \).
(iii) \( 4x^2 - 5x \)
Here, the variable \( x \) is multiplied by itself, so \( x \times x = x^2 \). Then \( x^2 \) is multiplied by 4, which makes \( 4x^2 \). Next, we multiply the variable \( x \) by 5, which results in \( 5x \). Finally, we subtract \( 5x \) from \( 4x^2 \) to get the expression \( 4x^2 - 5x \).
In simple words: To describe how an expression is made, break it down into the steps of multiplying variables, applying exponents, multiplying by numbers, and then adding or subtracting terms.

Exam Tip: When describing expression formation, clearly list the order of operations: multiplication of variables, exponentiation, multiplication by constants, and then addition/subtraction.

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Question 1. What are the terms in the following expressions? Show how the terms are formed. Draw a tree diagram for each expression:
(i) 8y + 3x²
(ii) 7mn - 4
(iii) 2x²y
Answer:
(i) \( 8y + 3x^2 \)
The terms in \( 8y + 3x^2 \) are \( 8y \) and \( 3x^2 \).
The term \( 8y \) is created by multiplying the variable \( y \) by 8.
The term \( 3x^2 \) is created by first multiplying the variable \( x \) by itself to get \( x \times x = x^2 \), and then multiplying \( x^2 \) by 3.
Tree diagram:
Expression: \( 8y + 3x^2 \)
          /          \
Terms:   \( 8y \)       \( 3x^2 \)
         / \       / \ \
Factors: \( 8 \)      \( y \)       \( 3 \)     \( x \)    \( x \)
(ii) \( 7mn - 4 \)
The terms of \( 7mn - 4 \) are \( 7mn \) and \( -4 \).
The term \( 7mn \) is formed by first multiplying variables \( m \) and \( n \) to get \( m \times n = mn \). Then, multiplying \( mn \) by the constant 7 gives \( 7 \times mn = 7mn \). The term \( -4 \) is a constant.
Tree diagram:
Expression: \( 7mn - 4 \) (or \( 7mn + (-4) \))
          /          \
Terms:   \( 7mn \)      \( (-4) \)
         / |\        |\
Factors: \( 7 \)    \( m \)    \( n \)       \( (-4) \)
(iii) \( 2x^2y \)
This expression has only one term, which is \( 2x^2y \). To create this term, we first multiply the variable \( x \) by itself to get \( x \times x = x^2 \). Then the variable \( y \) is multiplied by \( x^2 \) to get \( y \times x^2 = x^2y \). Next, the product \( x^2y \) is multiplied by 2 to achieve \( 2 \times x^2y = 2x^2y \).
Tree diagram:
Expression: \( 2x^2y \)
             |\
Terms:     \( 2x^2y \)
             / |\ \
Factors:   \( 2 \)    \( x \)    \( x \)    \( y \)
In simple words: Identify the individual parts (terms) separated by plus or minus signs. Then, for each term, break it down into its basic multiplying components (factors), which can be numbers or variables.

Exam Tip: A tree diagram helps visualize how an expression is built from its terms and factors. Remember that constants are also factors.

 

Question 2. Write tree expressions each having 4 terms.
(i) \( 2x^3 – 4x^2 + 9xy + 8 \)
(ii) \( 6x^3 + 9y^2 – 3xy^2 – 12 \)
(iii) \( 9x^2 – 3x + 12xy – 1 \)
Answer:
(i) Expression: \( 2x^3 - 4x^2 + 9xy + 8 \)
Terms: \( 2x^3 \), \( -4x^2 \), \( 9xy \), \( 8 \)

• For term \( 2x^3 \): Factors are \( 2, x, x, x \)
• For term \( -4x^2 \): Factors are \( -4, x, x \)
• For term \( 9xy \): Factors are \( 9, x, y \)
• For term \( 8 \): Factor is \( 8 \) (constant)

(ii) Expression: \( 6x^3 + 9y^2 - 3xy^2 - 12 \)
Terms: \( 6x^3 \), \( 9y^2 \), \( -3xy^2 \), \( -12 \)

• For term \( 6x^3 \): Factors are \( 6, x, x, x \)
• For term \( 9y^2 \): Factors are \( 9, y, y \)
• For term \( -3xy^2 \): Factors are \( -3, x, y, y \)
• For term \( -12 \): Factor is \( -12 \) (constant)

(iii) Expression: \( 9x^2 - 3x + 12xy - 1 \)
Terms: \( 9x^2 \), \( -3x \), \( 12xy \), \( -1 \)

• For term \( 9x^2 \): Factors are \( 9, x, x \)
• For term \( -3x \): Factors are \( -3, x \)
• For term \( 12xy \): Factors are \( 12, x, y \)
• For term \( -1 \): Factor is \( -1 \) (constant)

In simple words: For each expression, clearly list all the separate terms. Then, for every term, show what numbers and variables multiply together to make that term. This helps to see the structure of the expression.

Exam Tip: When asked to write "tree expressions", ensure you clearly identify all terms and then all factors for each term, including coefficients and constants. Pay attention to negative signs; they should be attached to the numerical factor.

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Question 1. Identify the coefficients of the terms of following expressions:
(i) 4x – 3y
(ii) a + b + 5
(iii) 2y + 5
(iv) 2xy
Answer:
(i) \( 4x - 3y \)
The number that multiplies \( x \) in \( 4x \) is 4. So, the coefficient of \( x \) is 4.
The number that multiplies \( y \) in \( -3y \) is \( -3 \). So, the coefficient of \( y \) is \( -3 \).
(ii) \( a + b + 5 \)
The number that multiplies \( a \) in \( a \) is 1 (since \( 1 \times a = a \)). So, the coefficient of \( a \) is 1.
The number that multiplies \( b \) in \( b \) is 1. So, the coefficient of \( b \) is 1.
The number 5 is a constant term. Its coefficient is 1 (since \( 1 \times 5 = 5 \)).
(iii) \( 2y + 5 \)
The number that multiplies \( y \) in \( 2y \) is 2. So, the coefficient of \( y \) is 2.
The number 5 is a constant term. Its coefficient is 1.
(iv) \( 2xy \)
The number that multiplies \( xy \) in \( 2xy \) is 2. So, the coefficient of \( xy \) is 2.
The part that multiplies \( y \) in \( 2xy \) is \( 2x \). So, the coefficient of \( y \) is \( 2x \).
The part that multiplies \( x \) in \( 2xy \) is \( 2y \). So, the coefficient of \( x \) is \( 2y \).
In simple words: A coefficient is the numerical part that multiplies a variable or a group of variables. If no number is shown, the coefficient is 1. For a constant term, its coefficient is itself.

Exam Tip: Remember that a coefficient is the numerical factor multiplying a variable. In a term like \( 2xy \), \( 2 \) is the coefficient of \( xy \), but \( 2x \) is the coefficient of \( y \), and \( 2y \) is the coefficient of \( x \).

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Question 1. Group the like terms together from the following: 12x, 12, -25x, -25, -25y, 1, x, 12y, y
Answer:
We have:
(i) Terms with variable \( x \): \( 12x \), \( -25x \), and \( x \) are like terms.
(ii) Terms with variable \( y \): \( -25y \), \( 12y \), and \( y \) are like terms.
(iii) Constant terms: \( 12 \), \( -25 \), and \( 1 \) are like terms.
In simple words: Like terms are those that have exactly the same variable parts, including the same powers. Numbers without any variables are also considered like terms (constants).

Exam Tip: For terms to be "like terms", both the variable(s) and their powers must match exactly. The numerical coefficient can be different.

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Question 1. Classify the following expressions as a monomial, a binomial or a trinomial: a, a + b, ab + a + b, ab + a + b - 5, xy, xy + 5, 5x² - x + 2, 4pq – 3q + 5p, 7, 4m – 7n + 10, 4mn + 7.
Answer:
(i) \( a \) contains 1 term. It is a monomial.
(ii) \( a + b \) contains 2 terms. It is a binomial.
(iii) \( ab + a + b \) contains 3 terms. It is a trinomial.
(iv) \( ab + a + b - 5 \) contains 4 terms. It is a polynomial (any expression with more than 3 terms).
(v) \( xy \) contains only 1 term. It is a monomial.
(vi) \( xy + 5 \) contains 2 terms. It is a binomial.
(vii) \( 5x^2 - x + 2 \) contains 3 terms. It is a trinomial.
(viii) \( 4pq - 3q + 5p \) contains 3 terms. It is a trinomial.
(ix) \( 7 \) contains only 1 term. It is a monomial.
(x) \( 4m - 7n + 10 \) contains 3 terms. It is a trinomial.
(xi) \( 4mn + 7 \) contains 2 terms. It is a binomial.
In simple words: An expression with one term is a monomial. If it has two terms, it's a binomial. If it has three terms, it's a trinomial. More than three terms means it's a polynomial.

Exam Tip: The classification of algebraic expressions (monomial, binomial, trinomial) depends solely on the number of unlike terms they contain, separated by plus or minus signs.

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Question 1. Think of at least two situations in which you need to form two algebraic expressions and add or subtract them.
Answer:
**Situation - I: Monthly Earnings**
Here's a scenario where you would form and combine expressions:
Let Vibha's monthly earnings be \( V \) Rs.
Rohan's monthly earnings are twice Vibha's, so Rohan earns \( 2V \) Rs.
The sum of Rohan's and Vibha's earnings is \( V + 2V = 3V \) Rs.
Kavita's monthly earnings are 400 Rs. more than the sum of Rohan's and Vibha's earnings.
So, Kavita's earnings would be \( (3V + 400) \) Rs.
This situation needs us to form two expressions (\( V \) and \( 2V \)) and then add them, and finally add a constant to find Kavita's total earnings.

**Situation - II: Number of Toys**
Here's another situation involving expressions for toy counts:
Let Kanta have \( K \) toys.
Mahesh has twice the number of toys Kanta has, so Mahesh has \( 2K \) toys.
The total number of toys Mahesh and Kanta have together is \( K + 2K = 3K \) toys.
Lata has 5 toys more than twice the number of toys Mahesh and Kanta together have.
So, Lata has \( 2(3K) + 5 = (6K + 5) \) toys.
This problem required forming expressions for Kanta's and Mahesh's toys, adding them, and then using that sum to create an expression for Lata's toys through multiplication and addition.
In simple words: When you have unknown amounts that relate to each other through words like "twice," "more than," or "less than," you use letters for the unknowns. Then you write mathematical statements (expressions) for each amount and add or subtract them to find a total or difference.

Exam Tip: To solve word problems involving algebraic expressions, first assign variables to the unknown quantities. Then, translate the phrases into mathematical expressions using those variables and perform the required operations like addition or subtraction.

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Question 1. Add and subtract:
(i) m - n, m +n
(ii) mn + 5 - 2, mn + 3
Answer:
(i) To add \( (m - n) \) and \( (m + n) \):
\( (m - n) + (m + n) = m - n + m + n \)
Collecting the like terms, we get:
\( (m + m) + (-n + n) = (1 + 1)m + (-1 + 1)n \)
\( = 2m + 0n \)
\( = 2m \)
To subtract \( (m - n) \) from \( (m + n) \):
\( (m + n) - (m - n) = m + n - m + n \)
Collecting the like terms, we get:
\( (m - m) + (n + n) = (1 - 1)m + (1 + 1)n \)
\( = 0m + 2n \)
\( = 2n \)
(ii) To add \( (mn + 5 - 2) \) and \( (mn + 3) \):
First, simplify \( mn + 5 - 2 \) to \( mn + 3 \).
So, we are adding \( (mn + 3) \) and \( (mn + 3) \).
\( (mn + 3) + (mn + 3) = mn + 3 + mn + 3 \)
Collecting the like terms, we get:
\( (mn + mn) + (3 + 3) = (1 + 1)mn + 6 \)
\( = 2mn + 6 \)
To subtract \( (mn + 3) \) from \( (mn + 5 - 2) \):
First, simplify \( mn + 5 - 2 \) to \( mn + 3 \).
So, we are subtracting \( (mn + 3) \) from \( (mn + 3) \).
\( (mn + 3) - (mn + 3) = mn + 3 - mn - 3 \)
Collecting the like terms, we get:
\( (mn - mn) + (3 - 3) = 0mn + 0 \)
\( = 0 \)
In simple words: When adding or subtracting expressions, combine the parts that have the same variables and powers. Remember to change the signs of all terms when subtracting an entire expression.

Exam Tip: Always group like terms together before performing addition or subtraction. Be careful with signs, especially when subtracting an entire expression, as every term within the parentheses changes its sign.

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Question 1. (The number of segments required to make the figure is given to the right. Also, the expression for the number of segments required to make n shapes is also given).
(i) P, (4n+1)
(ii) H, (3n+2)
Answer:
(i) Figure: The letter P
When we make the letter 'P' using segments, a single 'P' requires 5 segments. If we connect multiple 'P's, each additional 'P' shares one vertical segment with the previous one. Therefore, for each new 'P', only 4 additional segments are needed (1 vertical, 1 horizontal, 2 for the curve or other parts).
For \( n \) shapes of 'P':
Number of segments \( = 5 + (n-1) \times 4 \)
\( = 5 + 4n - 4 \)
\( = 4n + 1 \)
This matches the given expression \( 4n + 1 \).
(ii) Figure: The letter H
For the letter 'H', a single 'H' requires 5 segments (2 vertical, 1 horizontal, and 2 more vertical segments). When 'H' shapes are connected in a series, each subsequent 'H' shares two vertical segments with the previous one. So, each additional 'H' adds only 3 new segments (1 horizontal and 2 new vertical ones).
For \( n \) shapes of 'H':
Number of segments \( = 5 + (n-1) \times 3 \)
\( = 5 + 3n - 3 \)
\( = 3n + 2 \)
This matches the given expression \( 3n + 2 \).
In simple words: For pattern questions, count the segments for the first shape, then figure out how many *new* segments are added for each extra shape. Use this to create a formula that predicts the total segments for any number of shapes.

Exam Tip: When analyzing patterns for segment count, always identify the initial number of segments for the first figure. Then, determine the constant increase in segments for each subsequent figure in the series. This difference will be the coefficient of 'n' in your algebraic expression.