ML Aggarwal Class 8 Maths Solutions Chapter 10 Algebraic Expressions and Identities

Exercise 10.1

Question 1. Identify the terms, their numerical as well as literal coefficients in each of the following expressions:
(i) \( 12x^2yz - 4xy^2 \)
(ii) \( 8 + mn + nl - lm \)
(iii) \( \frac{x^2}{3} + \frac{y}{6} - xy^2 \)
(iv) \( -4p + 2.3q + 1.7r \)
Answer:
For each expression, the terms, their numerical coefficients, and literal coefficients are identified in the table below:

Exam Tip: For any term, pull out the number at the front to get the numerical coefficient. Whatever variables remain form the literal coefficient.

Question 2. Identify monomials, binomials, and trinomials from the following algebraic expressions:
(i) \( 5p \times q \times r^2 \)
(ii) \( 3x^2 + y \div 2z \)
(iii) \( -3 + 7x^2 \)
(iv) \( \frac{5a^2 - 3b^2 + c}{2} \)
(v) \( 7x^5 - \frac{3x}{y} \)
(vi) \( 5p \div 3q - 3p^2 \times q^2 \)
Answer:
(i) \( 5p \times q \times r^2 = 5pqr^2 \) - This has only one term, so it is a monomial.
(ii) \( 3x^2 + y \div 2z = \frac{3x^2}{2z} + \frac{y}{2z} \) - This has two terms, so it is a binomial.
(iii) \( -3 + 7x^2 \) - This has two terms, so it is a binomial.
(iv) \( \frac{5a^2 - 3b^2 + c}{2} \) - This has three terms, so it is a trinomial.
(v) \( 7x^5 - \frac{3x}{y} \) - This has two terms, so it is a binomial.
(vi) \( 5p \div 3q - 3p^2 \times q^2 = \frac{5p}{3q} - 3p^2q^2 \) - This has two terms, so it is a binomial.
In simple words: Count how many terms are in the expression. One term is a monomial, two terms make a binomial, and three terms make a trinomial.

Exam Tip: Simplify the expression first by performing all operations, then count the final number of terms that remain.

Question 3. Identify which of the following expressions are polynomials. If so, write their degrees.
(i) \( \frac{2}{5}x^4 - \sqrt{3}x^2 + 5x - 1 \)
(ii) \( 7x^3 - \frac{3}{x^2} + \sqrt{5} \)
(iii) \( 4a^3b^2 - 3ab^4 + 5ab + \frac{2}{3} \)
(iv) \( 2x^2y - \frac{3}{xy} + 5y^3 + \sqrt{3} \)
Answer:
(i) This is a polynomial. The degree is 4.
(ii) This is not a polynomial because it has a variable in the denominator (the term \( -\frac{3}{x^2} \) gives it a negative exponent).
(iii) This is a polynomial. The degree is 5 (from the term \( 4a^3b^2 \), where the sum of exponents is 3 - 2 = 5).
(iv) This is not a polynomial because it has variables in the denominator (the term \( -\frac{3}{xy} \)).
In simple words: A polynomial can only have variables raised to whole number powers that are not negative. Division by a variable makes it not a polynomial.

Exam Tip: Check if any variable appears in a denominator or with a negative power. If yes, it is not a polynomial. The degree is the highest power of the variables combined in any single term.

Question 4. Add the following expressions:
(i) \( ab - bc, bc - ca, ca - ab \)
(ii) \( 5p^2q^2 + 4pq + 7, 3 + 9pq - 2p^2q^2 \)
(iii) \( l^2 + m^2 + n^2, lm + mn, mn + nl, nl + lm \)
(iv) \( 4x^3 - 7x^2 + 9, 3x^2 - 5x + 4, 7x^3 - 11x + 1, 6x^2 - 13x \)
Answer:
(i) When adding \( ab - bc + bc - ca + ca - ab \), all terms cancel out to give \( 0 \).
(ii) Combining like terms: \( 5p^2q^2 - 2p^2q^2 + 4pq + 9pq + 7 + 3 = 3p^2q^2 + 13pq + 10 \)
(iii) Combining all terms: \( l^2 + m^2 + n^2 + 2lm + 2mn + 2nl \)
(iv) Grouping by degree: \( (4x^3 + 7x^3) + (-7x^2 + 3x^2 + 6x^2) + (-5x - 11x - 13x) + (9 + 4 + 1) = 11x^3 - 2x^2 - 29x + 14 \)
In simple words: Collect like terms together - those with the same variables raised to the same powers. Combine their coefficients by adding.

Exam Tip: Write out all terms clearly, group matching types together, then add or subtract the numbers in front of each group.

Question 5. Subtract:
(i) \( 8a + 3ab - 2b + 7 \) from \( 14a - 5ab + 7b - 5 \)
(ii) \( 8xy + 4yz + 5zx \) from \( 12xy - 3yz - 4zx + 5xyz \)
(iii) \( 4p^2q - 3pq + 5pq^2 - 8p + 7q - 10 \) from \( 18 - 3p - 11q + 5pq - 2pq^2 + 5p^2q \)
Answer:
(i) \( (14a - 5ab + 7b - 5) - (8a + 3ab - 2b + 7) = 14a - 8a - 5ab - 3ab + 7b + 2b - 5 - 7 = 6a - 8ab + 9b - 12 \)
(ii) \( (12xy - 3yz - 4zx + 5xyz) - (8xy + 4yz + 5zx) = 12xy - 8xy - 3yz - 4yz - 4zx - 5zx + 5xyz = 4xy - 7yz - 9zx + 5xyz \)
(iii) \( (18 - 3p - 11q + 5pq - 2pq^2 + 5p^2q) - (4p^2q - 3pq + 5pq^2 - 8p + 7q - 10) = 28 + 5p - 18q + 8pq - 7pq^2 + p^2q \)
In simple words: Reverse the sign of every term being subtracted, then add everything together as one sum.

Exam Tip: Put parentheses around the expression being subtracted and distribute the minus sign carefully to each term before combining.

Question 6. Subtract the sum of \( 3x^2 + 5xy + 7y^2 + 3 \) and \( 2x^2 - 4xy - 3y^2 + 7 \) from \( 9x^2 - 8xy + 11y^2 \)
Answer:
First, add the two expressions: \( 3x^2 + 5xy + 7y^2 + 3 + 2x^2 - 4xy - 3y^2 + 7 = 5x^2 + xy + 4y^2 + 10 \)
Then, subtract this sum from \( 9x^2 - 8xy + 11y^2 \):
\( (9x^2 - 8xy + 11y^2) - (5x^2 + xy + 4y^2 + 10) = 9x^2 - 5x^2 - 8xy - xy + 11y^2 - 4y^2 - 10 = 4x^2 - 9xy + 7y^2 - 10 \)
In simple words: Add the first two expressions together to get one new expression, then subtract that result from the third expression.

Exam Tip: Always perform addition first before subtraction when the question asks to "subtract the sum of". Group like terms at each step to avoid errors.

Question 7. What must be subtracted from \( 3a^2 - 5ab - 2b^2 - 3 \) to get \( 5a^2 - 7ab - 3b^2 + 3a \)?
Answer:
Let the expression to be subtracted be \( x \). Then:
\( 3a^2 - 5ab - 2b^2 - 3 - x = 5a^2 - 7ab - 3b^2 + 3a \)
\( x = (3a^2 - 5ab - 2b^2 - 3) - (5a^2 - 7ab - 3b^2 + 3a) \)
\( x = 3a^2 - 5ab - 2b^2 - 3 - 5a^2 + 7ab + 3b^2 - 3a = -2a^2 + 2ab + b^2 - 3a - 3 \)
In simple words: Subtract the target result from the starting expression to find what was removed.

Exam Tip: If asked "what must be subtracted to get [result]?", reverse the subtraction: starting expression minus result equals what was taken away.

Question 8. The perimeter of a triangle is \( 7p^2 - 5p + 11 \) and two of its sides are \( p^2 + 2p - 1 \) and \( 3p^2 - 6p + 3 \). Find the third side of the triangle.
Answer:
The perimeter of a triangle equals the sum of all three sides. Therefore:
\( 7p^2 - 5p + 11 = (p^2 + 2p - 1) + (3p^2 - 6p + 3) + \text{(third side)} \)
First, add the two known sides: \( p^2 + 2p - 1 + 3p^2 - 6p + 3 = 4p^2 - 4p + 2 \)
Now, the third side is:\( (7p^2 - 5p + 11) - (4p^2 - 4p + 2) = 7p^2 - 4p^2 - 5p + 4p + 11 - 2 = 3p^2 - p + 9 \)
In simple words: Add the two known sides together. Then subtract this total from the perimeter to find the missing side.

Exam Tip: Remember that perimeter means the total distance around a shape. When two sides are known, finding the third means subtracting their sum from the total perimeter.

Exercise 10.2

Question 1. Find the product of:
(i) \( 4x^3 \) and \( -3xy \)
(ii) \( 2xyz \) and \( 0 \)
(iii) \( -\frac{2}{3}p^2q \), \( \frac{3}{4}pq^2 \) and \( 5pqr \)
(iv) \( -7ab \), \( -3a^3 \) and \( -\frac{2}{7}ab^2 \)
(v) \( -\frac{1}{2}x^2 \), \( -\frac{3}{5}xy \), \( \frac{2}{3}yz \) and \( \frac{5}{7}xyz \)
Answer:
(i) \( 4x^3 \times (-3xy) = -12x^4y \)
(ii) \( 2xyz \times 0 = 0 \)
(iii) \( -\frac{2}{3}p^2q \times \frac{3}{4}pq^2 \times 5pqr = -\frac{2}{3} \times \frac{3}{4} \times 5 \times p^{2+1+1} \times q^{1+2+1} \times r = -\frac{5}{2}p^4q^4r \)
(iv) \( (-7ab) \times (-3a^3) \times (-\frac{2}{7}ab^2) = (-7) \times (-3) \times (-\frac{2}{7}) \times ab \times a^3 \times ab^2 = -6a^5b^3 \)
(v) \( (-\frac{1}{2}x^2) \times (-\frac{3}{5}xy) \times \frac{2}{3}yz \times \frac{5}{7}xyz = (-\frac{1}{2}) \times (-\frac{3}{5}) \times \frac{2}{3} \times \frac{5}{7} \times x^4 \times y^3 \times z^2 = \frac{1}{7}x^4y^3z^2 \)
In simple words: Multiply all the numbers together, then add up the powers of each variable that repeats.

Exam Tip: When multiplying terms, keep track of signs carefully. A negative times a negative gives a positive result. Add exponents of the same variable when multiplying.

Question 2. Multiply:
(i) \( (3x - 5y + 7z) \) by \( -3xyz \)
(ii) \( (2p^2 - 3pq + 5q^2 + 5) \) by \( -2pq \)
(iii) \( (\frac{2}{3}a^2b - \frac{4}{5}ab^2 + \frac{2}{7}ab + 3) \) by \( 35ab \)
(iv) \( (4x^2 - 10xy + 7y^2 - 8x + 4y + 3) \) by \( 3xy \)
Answer:
(i) \( -3xyz(3x - 5y + 7z) = -9x^2yz + 15xy^2z - 21xyz^2 \)
(ii) \( -2pq(2p^2 - 3pq + 5q^2 + 5) = -4p^3q + 6p^2q^2 - 10pq^3 - 10pq \)
(iii) \( 35ab(\frac{2}{3}a^2b - \frac{4}{5}ab^2 + \frac{2}{7}ab + 3) = \frac{70}{3}a^3b^2 - 28a^2b^3 + 10a^2b^2 + 105ab \)
(iv) \( 3xy(4x^2 - 10xy + 7y^2 - 8x + 4y + 3) = 12x^3y - 30x^2y^2 + 21xy^3 - 24x^2y + 12xy^2 + 9xy \)
In simple words: Multiply the outside term by each part inside the brackets one at a time. Add all the results together.

Exam Tip: Distribute the multiplier to every single term inside the parentheses. Keep the signs correct by tracking positive and negative carefully.

Question 3. Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively:
(i) \( (p^2q, pq^2) \)
(ii) \( (5xy, 7xy^2) \)
Answer:
(i) Area = \( p^2q \times pq^2 = p^3q^3 \)
(ii) Area = \( 5xy \times 7xy^2 = 35x^2y^3 \)
In simple words: The area of a rectangle is length times breadth. Multiply the two monomials together and combine the powers.

Exam Tip: When multiplying monomials to find area, multiply the coefficients together and add exponents of matching variables.

Question 4. Find the volume of rectangular boxes with the following length, breadth and height respectively:
(i) \( 5ab, 3a^2b, 7a^4b^2 \)
(ii) \( 2pq, 4q^2, 8rp \)
Answer:
(i) Volume = \( 5ab \times 3a^2b \times 7a^4b^2 = 5 \times 3 \times 7 \times a^{1+2+4} \times b^{1+1+2} = 105a^7b^4 \)
(ii) Volume = \( 2pq \times 4q^2 \times 8rp = 2 \times 4 \times 8 \times p^{1+1} \times q^{1+2} \times r = 64p^2q^3r \)
In simple words: Volume is length times breadth times height. Multiply all three values and add the powers of each variable.

Exam Tip: For volume, multiply three dimensions together. Keep track of all variables and add their exponents when the same variable appears in multiple factors.

Question 5. Simplify the following expressions and evaluate them as directed:
(i) \( x^2(3 - 2x + x^2) \) for \( x = 1; x = -1; x = \frac{2}{3} \) and \( x = -\frac{1}{2} \)
(ii) \( 5xy(3x + 4y - 7) - 3y(xy - x^2 + 9) - 8 \) for \( x = 2, y = -1 \)
Answer:
(i) First simplify: \( x^2(3 - 2x + x^2) = 3x^2 - 2x^3 + x^4 \)
For \( x = 1 \): \( 3(1)^2 - 2(1)^3 + (1)^4 = 3 - 2 + 1 = 2 \)
For \( x = -1 \): \( 3(-1)^2 - 2(-1)^3 + (-1)^4 = 3 + 2 + 1 = 6 \)
For \( x = \frac{2}{3} \): \( 3(\frac{4}{9}) - 2(\frac{8}{27}) + \frac{16}{81} = \frac{108 - 48 + 16}{81} = \frac{76}{81} \)
For \( x = -\frac{1}{2} \): \( 3(\frac{1}{4}) - 2(-\frac{1}{8}) + \frac{1}{16} = \frac{3}{4} + \frac{1}{4} + \frac{1}{16} = \frac{17}{16} \)
(ii) First expand: \( 5xy(3x + 4y - 7) - 3y(xy - x^2 + 9) - 8 = 15x^2y + 20xy^2 - 35xy - 3xy^2 + 3x^2y - 27y - 8 = 18x^2y + 17xy^2 - 35xy - 27y - 8 \)
For \( x = 2, y = -1 \): \( 18(4)(-1) + 17(2)(1) - 35(2)(-1) - 27(-1) - 8 = -72 + 34 + 70 + 27 - 8 = 51 \)
In simple words: Expand everything using distribution, combine like terms, then put in the given values for the variables.

Exam Tip: Always expand and simplify fully before substituting numerical values. This reduces the chance of making arithmetic errors with the actual numbers.

Question 6. Add the following:
(i) \( 4p(2 - p^2) \) and \( 8p^3 - 3p \)
(ii) \( 7xy(8x + 2y - 3) \) and \( 4xy^2(3y - 7x + 8) \)
Answer:
(i) Expand: \( 4p(2 - p^2) = 8p - 4p^3 \). Adding: \( 8p - 4p^3 + 8p^3 - 3p = 4p^3 + 5p \)
(ii) Expand: \( 7xy(8x + 2y - 3) = 56x^2y + 14xy^2 - 21xy \) and \( 4xy^2(3y - 7x + 8) = 12xy^3 - 28x^2y^2 + 32xy^2 \). Adding: \( 12xy^3 - 28x^2y^2 + 56x^2y + 46xy^2 - 21xy \)
In simple words: Expand each expression fully by distributing, then combine all like terms into one final expression.

Exam Tip: Expand brackets before adding. Grouping like terms (same powers of variables) at the end makes combining easier and reduces errors.

Question 7. Subtract:
(i) \( 6x(x - y + z) - 3y(x + y - z) \) from \( 2z(-x + y + z) \)
(ii) \( 7xy(x^2 - 2xy + 3y^2) - 8x(x^2y - 4xy + 7xy^2) \) from \( 3y(4x^2y - 5xy + 8xy^2) \)
Answer:
(i) Expand each expression:
\( 6x(x - y + z) - 3y(x + y - z) = 6x^2 - 6xy + 6xz - 3xy - 3y^2 + 3yz \)
\( 2z(-x + y + z) = -2xz + 2yz + 2z^2 \)
Subtracting: \( (-2xz + 2yz + 2z^2) - (6x^2 - 6xy + 6xz - 3xy - 3y^2 + 3yz) = -6x^2 + 9xy + 3y^2 - 8xz - yz + 2z^2 \)
(ii) Expand each expression:
\( 7xy(x^2 - 2xy + 3y^2) = 7x^3y - 14x^2y^2 + 21xy^3 \)
\( 8x(x^2y - 4xy + 7xy^2) = 8x^3y - 32x^2y + 56x^2y^2 \)
\( 3y(4x^2y - 5xy + 8xy^2) = 12x^2y^2 - 15xy^2 + 24xy^3 \)
Therefore: \( (12x^2y^2 - 15xy^2 + 24xy^3) - (7x^3y - 14x^2y^2 + 21xy^3 - 8x^3y + 32x^2y - 56x^2y^2) = 82x^2y^2 + 3xy^3 + x^3y - 15xy^2 - 32x^2y \)
In simple words: Expand all terms in both expressions. Reverse the signs of what you subtract, then combine like terms.

Exam Tip: Write out expansions clearly on separate lines. Arrange the result in descending order of exponents to avoid missing or duplicating terms.

Exercise 10.3

Question 1. Multiply:
(i) \( (5x - 2) \) by \( (3x + 4) \)
(ii) \( (ax + b) \) by \( (cx + d) \)
(iii) \( (4p - 7) \) by \( (2 - 3p) \)
(iv) \( (2x^2 + 3) \) by \( (3x - 5) \)
(v) \( (1.5a - 2.5b) \) by \( (1.5a + 2.5b) \)
(vi) \( (\frac{3}{7}p^2 + 4q^2) \) by \( (p^2 - \frac{3}{4}q^2) \)
Answer:
(i) \( (5x - 2)(3x + 4) = 5x(3x + 4) - 2(3x + 4) = 15x^2 + 20x - 6x - 8 = 15x^2 + 14x - 8 \)
(ii) \( (ax + b)(cx + d) = ax(cx + d) + b(cx + d) = acx^2 + adx + bcx + bd \)
(iii) \( (4p - 7)(2 - 3p) = 4p(2 - 3p) - 7(2 - 3p) = 8p - 12p^2 - 14 + 21p = -12p^2 + 29p - 14 \)
(iv) \( (2x^2 + 3)(3x - 5) = 2x^2(3x - 5) + 3(3x - 5) = 6x^3 - 10x^2 + 9x - 15 \)
(v) \( (1.5a - 2.5b)(1.5a + 2.5b) = (1.5a)^2 - (2.5b)^2 = 2.25a^2 - 6.25b^2 \)
(vi) \( (\frac{3}{7}p^2 + 4q^2)(p^2 - \frac{3}{4}q^2) = 7 \times [\frac{3}{7}p^2(p^2 - \frac{3}{4}q^2) + 4q^2(p^2 - \frac{3}{4}q^2)] = 3p^4 - \frac{9}{4}p^2q^2 + 28p^2q^2 - 21q^4 = 3p^4 + \frac{103}{4}p^2q^2 - 21q^4 \)
In simple words: Multiply each term from the first bracket by each term from the second bracket. Then combine like terms together.

Exam Tip: Use the distributive method: multiply the first term of the first bracket by both terms in the second, then the second term by both terms in the second. Watch out for sign changes.

Question 2. Multiply:
(i) \( (x - 2y + 3) \) by \( (x + 2y) \)
(ii) \( (3 - 5x + 2x^2) \) by \( (4x - 5) \)
Answer:
(i) \( (x - 2y + 3)(x + 2y) = x(x + 2y) - 2y(x + 2y) + 3(x + 2y) = x^2 + 2xy - 2xy - 4y^2 + 3x + 6y = x^2 - 4y^2 + 3x + 6y \)
(ii) \( (3 - 5x + 2x^2)(4x - 5) = 3(4x - 5) - 5x(4x - 5) + 2x^2(4x - 5) = 12x - 15 - 20x^2 + 25x + 8x^3 - 10x^2 = 8x^3 - 30x^2 + 37x - 15 \)
In simple words: With three or more terms in the first bracket, multiply each of them by every term in the second bracket. Combine all like terms at the end.

Exam Tip: Organize by writing each product on a separate line first, then collecting and combining. This prevents losing or forgetting any terms.

Question 3. Multiply:
(i) \( (3x^2 - 2x - 1) \) by \( (2x^2 + x - 5) \)
(ii) \( (2 - 3y - 5y^2) \) by \( (2y - 1 + 3y^2) \)
Answer:
(i) \( (3x^2 - 2x - 1)(2x^2 + x - 5) = 3x^2(2x^2 + x - 5) - 2x(2x^2 + x - 5) - 1(2x^2 + x - 5) = 6x^4 + 3x^3 - 15x^2 - 4x^3 - 2x^2 + 10x - 2x^2 - x + 5 = 6x^4 - x^3 - 19x^2 + 9x + 5 \)
(ii) \( (2 - 3y - 5y^2)(2y - 1 + 3y^2) = 2(2y - 1 + 3y^2) - 3y(2y - 1 + 3y^2) - 5y^2(2y - 1 + 3y^2) = 4y - 2 + 6y^2 - 6y^2 + 3y - 9y^3 - 10y^3 + 5y^2 - 15y^4 = -15y^4 - 19y^3 + 5y^2 + 7y - 2 \)
In simple words: When both expressions have many terms, systematically multiply each term from the first by every term from the second. Arrange the final answer in order of highest to lowest power.

Exam Tip: Use a table or column method if needed to keep track of all products. Arrange results by power level to spot and combine like terms efficiently.

Question 4. Simplify:
(i) \( (x^2 + 3)(x - 3) + 9 \)
(ii) \( (x + 3)(x - 3)(x + 4)(x - 4) \)
(iii) \( (x + 5)(x + 6)(x + 7) \)
(iv) \( (p + q - 2r)(2p - q + r) - 4qr \)
(v) \( (p + q)(r + s) + (p - q)(r - s) - 2(pr + qs) \)
(vi) \( (x + y + z)(x - y + z) + (x + y - z)(-x + y + z) - 4zx \)
Answer:
(i) \( (x^2 + 3)(x - 3) + 9 = x^3 - 3x^2 + 3x - 9 + 9 = x^3 - 3x^2 + 3x \)
(ii) Using difference of squares: \( [(x + 3)(x - 3)][(x + 4)(x - 4)] = (x^2 - 9)(x^2 - 16) = x^4 - 16x^2 - 9x^2 + 144 = x^4 - 25x^2 + 144 \)
(iii) First multiply the first two: \( (x + 5)(x + 6) = x^2 + 11x + 30 \). Then: \( (x^2 + 11x + 30)(x + 7) = x^3 + 11x^2 + 30x + 7x^2 + 77x + 210 = x^3 + 18x^2 + 107x + 210 \)
(iv) \( (p + q - 2r)(2p - q + r) - 4qr = 2p^2 - pq + pr + 2pq - q^2 + qr - 4pr + 2qr - 2r^2 - 4qr = 2p^2 - q^2 - 2r^2 + pq - 3pr - 2qr \)
(v) \( (p + q)(r + s) + (p - q)(r - s) - 2(pr + qs) = pr + ps + qr + qs + pr - ps - qr + qs - 2pr - 2qs = 0 \)
(vi) \( (x + y + z)(x - y + z) + (x + y - z)(-x + y + z) - 4zx = x^2 + 2xz + z^2 - y^2 - x^2 + 2xy + y^2 + z^2 - 4zx = 0 \)
In simple words: Expand all products fully, combine like terms, and simplify to reach the final result. Some expressions may cancel completely to give zero.

Exam Tip: Look for patterns like difference of squares which simplify quickly. Always collect like terms at the end, and double-check your arithmetic when terms cancel.

Question 5. If two adjacent sides of a rectangle are \( 5x^2 + 25xy + 4y^2 \) and \( 2x^2 - 2xy + 3y^2 \), find its area.
Answer:
Area of rectangle = Product of adjacent sides
\( = (5x^2 + 25xy + 4y^2)(2x^2 - 2xy + 3y^2) \)
\( = 5x^2(2x^2 - 2xy + 3y^2) + 25xy(2x^2 - 2xy + 3y^2) + 4y^2(2x^2 - 2xy + 3y^2) \)
\( = 10x^4 - 10x^3y + 15x^2y^2 + 50x^3y - 50x^2y^2 + 75xy^3 + 8x^2y^2 - 8xy^3 + 12y^4 \)
\( = 10x^4 + 40x^3y - 27x^2y^2 + 67xy^3 + 12y^4 \)
In simple words: Multiply the two side lengths together using the distributive method. Combine all like terms to get the final area expression.

Exam Tip: When multiplying two trinomials, write all nine products from multiplying each term of the first by each of the second. Then group and combine terms with matching powers.

Exercise 10.4

Question 1. Divide:
(i) \( -39pq^2r^5 \) by \( -24p^3q^3r \)
(ii) \( -a^2b^3 \) by \( a^3b^2 \)
Answer:
(i) \( \frac{-39pq^2r^5}{-24p^3q^3r} = \frac{39}{24} \times \frac{p}{p^3} \times \frac{q^2}{q^3} \times \frac{r^5}{r} = \frac{13}{8} \times \frac{r^4}{p^2q} = \frac{13r^4}{8p^2q} \)
(ii) \( \frac{-a^2b^3}{a^3b^2} = -\frac{a^2}{a^3} \times \frac{b^3}{b^2} = -\frac{1}{a} \times b = -\frac{b}{a} \)
In simple words: Divide the numbers. For each variable, subtract the power in the denominator from the power in the numerator.

Exam Tip: When dividing like variables, subtract exponents. If an exponent becomes negative, move that variable to the denominator.

Question 2. Divide:
(i) \( 9x^4 - 8x^3 - 12x + 3 \) by \( 3x \)
(ii) \( 14p^2q^3 - 32p^3q^2 + 15pq^2 - 22p + 18q \) by \( -2p^2q \)
Answer:
(i) \( \frac{9x^4 - 8x^3 - 12x + 3}{3x} = \frac{9x^4}{3x} - \frac{8x^3}{3x} - \frac{12x}{3x} + \frac{3}{3x} = 3x^3 - \frac{8}{3}x^2 - 4 + \frac{1}{x} \)
(ii) \( \frac{14p^2q^3 - 32p^3q^2 + 15pq^2 - 22p + 18q}{-2p^2q} = -7q^2 + 16pq - \frac{15q}{2p} + \frac{11}{pq} - \frac{9}{p^2} \)
In simple words: Divide each term in the numerator by the divisor separately. Combine the results.

Exam Tip: Break the division into separate fractions, one for each term. Simplify each fraction individually, then write the answer as a sum or difference.

Question 3. Divide:
(i) \( 6x^2 + 13x + 5 \) by \( 2x + 1 \)
(ii) \( 1 + y^3 \) by \( 1 + y \)
(iii) \( 5 + x - 2x^2 \) by \( x + 1 \)
(iv) \( x^3 - 6x^2 + 12x - 8 \) by \( x - 2 \)
Answer:
(i) Using long division: \( 6x^2 + 13x + 5 = (2x + 1)(3x + 5) \), so quotient is \( 3x + 5 \) and remainder is \( 0 \).
(ii) Using long division: \( 1 + y^3 = (1 + y)(1 - y + y^2) \), so quotient is \( 1 - y + y^2 \) and remainder is \( 0 \).
(iii) Using long division: \( 5 + x - 2x^2 = (x + 1)(-2x + 3) - (-2) \), so quotient is \( -2x + 3 \) and remainder is \( -2 \).
(iv) Using long division: \( x^3 - 6x^2 + 12x - 8 = (x - 2)(x^2 - 4x + 4) \), so quotient is \( x^2 - 4x + 4 \) and remainder is \( 0 \).
In simple words: Use long division. Divide the first term of the dividend by the first term of the divisor. Multiply the divisor by this result and subtract from the dividend. Repeat until complete.

Exam Tip: Arrange terms in order from highest to lowest power before starting long division. Write down each step clearly to avoid arithmetic mistakes. Check your answer by multiplying the quotient by the divisor and adding the remainder.

Question 1. Using suitable identities, find the following products:
(i) (3x + 5) (3x + 5)
(ii) (9y - 5) (9y - 5)
(iii) (4x + 11y) (4x - 11y)
(iv) (3m/2 + 2n/3) (3m/2 - 2n/3)
(v) (2/a + 5/b) (2a + 5/b)
(vi) (p²/2 + 2/q²) (p²/2 - 2/q²)
Answer: (i) We recognize that (3x + 5)(3x + 5) is the same as (3x + 5)². Using the identity (a + b)² = a² + 2ab + b², we get (3x)² + 2(3x)(5) + 5² = 9x² + 30x + 25.
(ii) Similarly, (9y - 5)(9y - 5) = (9y - 5)². Applying (a - b)² = a² - 2ab + b², we obtain (9y)² - 2(9y)(5) + 5² = 81y² - 90y + 25.
(iii) For (4x + 11y)(4x - 11y), this matches (a + b)(a - b) = a² - b². So we get (4x)² - (11y)² = 16x² - 121y².
(iv) Using the same difference of squares pattern, (3m/2 + 2n/3)(3m/2 - 2n/3) gives (3m/2)² - (2n/3)² = 9m²/4 - 4n²/9.
(v) The expression (2/a + 5/b)(2a + 5/b) can be rewritten as (2/a + 5/b)². By the perfect square identity, this equals (2/a)² + 2(2/a)(5/b) + (5/b)² = 4/a² + 20a/b + 25/b².
(vi) For (p²/2 + 2/q²)(p²/2 - 2/q²), we apply the difference formula: (p²/2)² - (2/q²)² = p⁴/4 - 4/q⁴.
In simple words: When two expressions have the same terms but different signs, or are repeated exactly, use the perfect square and difference of squares identities to find the answer quickly without expanding everything.

Exam Tip: Recognize pattern matching first - spot whether you have (a + b)², (a - b)², or (a + b)(a - b) before expanding. This saves time and reduces errors on the exam.

Question 2. Using the identities, evaluate the following:
(i) 81²
(ii) 97²
(iii) 105²
(iv) 997²
(v) 6.1²
(vi) 496 × 504
(vii) 20.5 × 19.5
(viii) 9.6²
Answer: (i) Express 81 as 80 + 1. Then (80 + 1)² = 80² + 2(80)(1) + 1² = 6400 + 160 + 1 = 6561.
(ii) Write 97 as 100 - 3. So (100 - 3)² = 100² - 2(100)(3) + 3² = 10000 - 600 + 9 = 9409.
(iii) Breaking 105 into 100 + 5: (100 + 5)² = 100² + 2(100)(5) + 5² = 10000 + 1000 + 25 = 11025.
(iv) Since 997 = 1000 - 3, we have (1000 - 3)² = 1000² - 2(1000)(3) + 3² = 1000000 - 6000 + 9 = 994009.
(v) For 6.1, write it as 6 + 0.1. Then (6 + 0.1)² = 36 + 2(6)(0.1) + 0.01 = 36 + 1.2 + 0.01 = 37.21.
(vi) Notice 496 × 504 = (500 - 4)(500 + 4). Using (a - b)(a + b) = a² - b², we get 500² - 4² = 250000 - 16 = 249984.
(vii) Similarly, 20.5 × 19.5 = (20 + 0.5)(20 - 0.5) = 20² - 0.5² = 400 - 0.25 = 399.75.
(viii) Express 9.6 as 10 - 0.4. Then (10 - 0.4)² = 100 - 2(10)(0.4) + 0.16 = 100 - 8 + 0.16 = 92.16.
In simple words: Always break the number into a nearby round number plus or minus a small amount, then use the squaring formulas. This makes the math much faster than multiplying directly.

Exam Tip: Choose the decomposition wisely - split into numbers that are easy to square and easy to multiply. Always double-check your arithmetic in the middle steps.

Question 3. Find the following squares, using the identities:
(i) (pq + 5r)²
(ii) (5a/2 - 3b/5)²
(iii) (√2a + √3b)²
(iv) (2x/3y - 3y/2x)²
Answer: (i) Applying (a + b)² = a² + 2ab + b², we have (pq)² + 2(pq)(5r) + (5r)² = p²q² + 10pqr + 25r².
(ii) Using (a - b)² = a² - 2ab + b² on (5a/2 - 3b/5)², we get (5a/2)² - 2(5a/2)(3b/5) + (3b/5)² = 25a²/4 - 3ab + 9b²/25.
(iii) For (√2a + √3b)², the identity gives (√2a)² + 2(√2a)(√3b) + (√3b)² = 2a² + 2√6ab + 3b².
(iv) Squaring (2x/3y - 3y/2x): (2x/3y)² - 2(2x/3y)(3y/2x) + (3y/2x)² = 4x²/9y² - 2 + 9y²/4x².
In simple words: Identify the first term, the second term, and the sign between them. Then apply the right perfect square formula to expand quickly and correctly.

Exam Tip: Show all three parts of the expansion (first squared, twice the product of both, second squared) even if some simplify - examiners expect to see your method clearly.

Question 4. Using the identity, (x + a)(x + b) = x² + (a + b)x + ab, find the following products:
(i) (x + 7)(x + 3)
(ii) (3x + 4)(3x - 5)
(iii) (p² + 2q)(p² - 3q)
(iv) (abc + 3)(abc - 5)
Answer: (i) Here x = x, a = 7, b = 3. So x² + (7 + 3)x + 7(3) = x² + 10x + 21.
(ii) Treat 3x as the common term. Then (3x)² + (4 - 5)(3x) + 4(-5) = 9x² - 3x - 20.
(iii) Let x = p², a = 2q, b = -3q. We get (p²)² + (2q - 3q)p² + 2q(-3q) = p⁴ - p²q - 6q².
(iv) Taking abc as the variable x, we have (abc)² + (3 - 5)(abc) + 3(-5) = a²b²c² - 2abc - 15.
In simple words: Find what repeats in both brackets - that is your "x". Add the other numbers to get the middle term, and multiply them to get the last term.

Exam Tip: Always identify the common repeated part first; treating it as a single variable makes the algebra straightforward and less error-prone.

Question 5. Using the identity, (x + a)(x + b) = x² + (a + b)x + ab, evaluate the following:
(i) 203 × 204
(ii) 8.2 × 8.7
(iii) 107 × 93
Answer: (i) Write 203 × 204 as (200 + 3)(200 + 4). Then 200² + (3 + 4)(200) + 3(4) = 40000 + 1400 + 12 = 41412.
(ii) Express 8.2 × 8.7 as (8 + 0.2)(8 + 0.7). So 8² + (0.2 + 0.7)(8) + 0.2(0.7) = 64 + 7.2 + 0.14 = 71.34.
(iii) For 107 × 93, rewrite as (100 + 7)(100 - 7). This becomes 100² + (7 - 7)(100) + 7(-7) = 10000 + 0 - 49 = 9951.
In simple words: Split each number into a round part and a small extra or deficit. Then multiply using the formula instead of doing the calculation the long way.

Exam Tip: Choose decompositions where numbers are close together or where one is very close to a round number - the formula works fastest in these cases.

Question 6. Using the identity a² - b² = (a + b)(a - b), find
(i) 53² - 47²
(ii) (2.05)² - (0.95)²
(iii) (14.3)² - (5.7)²
Answer: (i) We recognize this as a difference of squares. So 53² - 47² = (53 + 47)(53 - 47) = 100(6) = 600. Note: The calculation should be (50 + 3)² - (50 - 3)² = (50)² - 3² = 2500 - 9 = 2491 if working directly, but the identity approach gives (53 + 47)(53 - 47) = 100(6) = 600. Using the identity correctly: 53² - 47² = (53 + 47)(53 - 47) = 100 × 6 = 600. However, rechecking: this should equal (53)(53) - (47)(47). Let me recalculate: a = 53, b = 47, so (a+b)(a-b) = 100 × 6 = 600. But 53² = 2809, 47² = 2209, difference = 600. Yes, this is correct.
(ii) We have (2.05)² - (0.95)² = (2.05 + 0.95)(2.05 - 0.95) = 3(1.1) = 3.3.
(iii) Similarly, (14.3)² - (5.7)² = (14.3 + 5.7)(14.3 - 5.7) = 20(8.6) = 172.
In simple words: Whenever you see one square minus another square, add and subtract the two numbers, then multiply these two results together. You get the answer without any actual squaring.

Exam Tip: Recognize the "difference of two squares" pattern instantly - it turns a hard calculation into simple addition and subtraction followed by one multiplication.

Question 7. Simplify the following:
(i) (2x + 5y)² + (2x - 5y)²
(ii) (7a/2 - 5b/2)² - (5a/2 - 7b/2)²
(iii) (p² - q²r)² + 2p²q²r
Answer: (i) Expand both squares separately: (2x + 5y)² = 4x² + 20xy + 25y² and (2x - 5y)² = 4x² - 20xy + 25y². When we add them, the 20xy and -20xy cancel, leaving 4x² + 25y² + 4x² + 25y² = 8x² + 50y².
(ii) First square: (7a/2 - 5b/2)² = 49a²/4 - 35ab/2 + 25b²/4. Second square: (5a/2 - 7b/2)² = 25a²/4 - 35ab/2 + 49b²/4. Subtracting the second from the first: (49a²/4 - 25a²/4) + (25b²/4 - 49b²/4) = 24a²/4 - 24b²/4 = 6a² - 6b².
(iii) Expanding (p² - q²r)² gives (p²)² - 2(p²)(q²r) + (q²r)² = p⁴ - 2p²q²r + q⁴r². Adding 2p²q²r removes the middle term: p⁴ - 2p²q²r + q⁴r² + 2p²q²r = p⁴ + q⁴r².
In simple words: Expand carefully, then combine like terms. Watch for terms that cancel out (like opposite signs) - they simplify the final answer greatly.

Exam Tip: Write out each squared term fully before combining; rushing this step leads to sign mistakes, which cost marks even if your method is right.

Question 8. Show that:
(i) (4x + 7y)² - (4x - 7y)² = 112xy
(ii) (3p/7 - 7q/6)² + pq = 9p²/49 + 49q²/36
(iii) (p - q)(p + q) + (q - r)(q + r) + (r - p)(r + p) = 0
Answer: (i) Starting with the left side: (4x + 7y)² = 16x² + 56xy + 49y² and (4x - 7y)² = 16x² - 56xy + 49y². Subtracting, the 16x² cancels, 49y² cancels, leaving 56xy - (-56xy) = 112xy, which equals the right side. ✓
(ii) Expand (3p/7 - 7q/6)²: (3p/7)² - 2(3p/7)(7q/6) + (7q/6)² = 9p²/49 - pq + 49q²/36. Adding pq: 9p²/49 - pq + 49q²/36 + pq = 9p²/49 + 49q²/36, which is the right side. ✓
(iii) Using the difference of squares formula (a + b)(a - b) = a² - b² three times: (p - q)(p + q) = p² - q², (q - r)(q + r) = q² - r², (r - p)(r + p) = r² - p². Adding all three: (p² - q²) + (q² - r²) + (r² - p²) = p² + q² + r² - q² - r² - p² = 0. ✓
In simple words: Work through the algebra step by step, expanding each square or product carefully. Keep track of positive and negative signs throughout, and verify that both sides match at the end.

Exam Tip: Always show every step of your working for a "show that" proof - examiners want to see your reasoning clearly, and one small algebraic error will lose marks even if the final answer is written correctly.

Question 9. If x + 1/x = 2, evaluate:
(i) x² + 1/x²
(ii) x⁴ + 1/x⁴
Answer: (i) Square both sides of x + 1/x = 2: (x + 1/x)² = 4. Expanding the left side: x² + 2(x)(1/x) + 1/x² = 4. Since 2(x)(1/x) = 2, we have x² + 2 + 1/x² = 4. Therefore x² + 1/x² = 2.
(ii) Square the result from part (i): (x² + 1/x²)² = 2² = 4. Expanding: x⁴ + 2(x²)(1/x²) + 1/x⁴ = 4. Since 2(x²)(1/x²) = 2, we get x⁴ + 2 + 1/x⁴ = 4. Thus x⁴ + 1/x⁴ = 2.
In simple words: When you square an equation, the cross term (the 2ab part) often gives you something simple like 2. Use that to simplify quickly and find the answer.

Exam Tip: Always identify the middle term when squaring - if the product of the two parts equals 1, that middle term becomes exactly 2, which greatly simplifies your calculation.

Question 10. If x - 1/x = 7, evaluate:
(i) x² + 1/x²
(ii) x⁴ + 1/x⁴
Answer: (i) Square the given equation: (x - 1/x)² = 49. Expanding: x² - 2(x)(1/x) + 1/x² = 49. Since -2(x)(1/x) = -2, we have x² - 2 + 1/x² = 49. Therefore x² + 1/x² = 51.
(ii) Square the result from part (i): (x² + 1/x²)² = 51² = 2601. Expanding: x⁴ + 2(x²)(1/x²) + 1/x⁴ = 2601. Since 2(x²)(1/x²) = 2, we get x⁴ + 2 + 1/x⁴ = 2601. Thus x⁴ + 1/x⁴ = 2599.
In simple words: Follow the same squaring steps as before, but watch carefully for negative signs - when you square a difference (a - b), the middle term is still positive 2ab, not negative.

Exam Tip: Pay close attention to signs throughout. The phrase "x - 1/x" has a minus, but when squared, the middle term becomes +2, not -2. This sign reversal is a common source of errors.

Question 11. If x² + 1/x² = 23, evaluate:
(i) x + 1/x
(ii) x - 1/x
Answer: (i) We use the identity (x + 1/x)² = x² + 2 + 1/x². Substituting x² + 1/x² = 23: (x + 1/x)² = 23 + 2 = 25. Taking square roots: x + 1/x = ±5.
(ii) We use the identity (x - 1/x)² = x² - 2 + 1/x². Substituting x² + 1/x² = 23: (x - 1/x)² = 23 - 2 = 21. Taking square roots: x - 1/x = ±√21.
In simple words: Rearrange the identity so that the part you want is on one side. Then substitute what you know and solve for the unknown by taking square roots.

Exam Tip: Remember to write both the positive and negative square root solutions unless the problem specifies which one to choose. Missing one answer costs marks on multi-part questions.

Question 12. If a + b = 9 and ab = 10, find the value of a² + b².
Answer: Square the equation a + b = 9: (a + b)² = 81. Expanding: a² + 2ab + b² = 81. Substitute ab = 10: a² + 2(10) + b² = 81. Simplify: a² + b² + 20 = 81. Therefore a² + b² = 61.
In simple words: When you know the sum and product of two numbers, square the sum first. Then use the product to replace the middle term and solve for the sum of squares.

Exam Tip: This is a standard technique - always square the sum (or difference) to create a connection between a² + b², the sum/difference, and the product ab.

Question 13. If a - b = 6 and a² + b² = 42, find the value of ab.
Answer: Square the equation a - b = 6: (a - b)² = 36. Expanding: a² - 2ab + b² = 36. Rearrange: a² + b² - 2ab = 36. Substitute a² + b² = 42: 42 - 2ab = 36. Solve for ab: 2ab = 42 - 36 = 6, so ab = 3.
In simple words: Square the difference to break it into the sum of squares and the product. Then substitute what you already know and solve for what you need.

Exam Tip: Watch the order of operations - you expand (a - b)² first, then rearrange to isolate the 2ab term, then substitute and solve.

Question 14. If a² + b² = 41 and ab = 4, find the values of
(i) a + b
(ii) a - b
Answer: (i) Use the identity (a + b)² = a² + 2ab + b². Substitute a² + b² = 41 and ab = 4: (a + b)² = 41 + 2(4) = 41 + 8 = 49. Taking square roots: a + b = ±7.
(ii) Use the identity (a - b)² = a² - 2ab + b². Substitute a² + b² = 41 and ab = 4: (a - b)² = 41 - 2(4) = 41 - 8 = 33. Taking square roots: a - b = ±√33.
In simple words: Write down the identity you need, plug in the numbers you know, then simplify inside the square root. Finally, take the square root to get both positive and negative answers.

Exam Tip: Do not forget the ± symbol when taking square roots - unless told otherwise, both answers are valid. Leaving out one solution is a common mistake that loses marks.

Question 10. Using identities, find the following products:
(i) (3x + 4y) (3x + 4y)
(ii) (5a/2 - b) (5a/2 – b)
(iii) (3.5m – 1.5n) (3.5m + 1.5n)
(iv) (7xy – 2) (7xy + 7)
Answer:
(i) (3x + 4y) (3x + 4y) = (3x + 4y)²
\[ = (3x)^2 + 2 \times 3x \times 4y + (4y)^2 \]
\[ = 9x^2 + 24xy + 16y^2 \]

(ii) (5a/2 - b) (5a/2 – b) = (5a/2 - b)²
\[ = (5a/2)^2 + 2 \times \frac{5a}{2} \times (-b) + (b)^2 \]
\[ = \frac{25a^2}{4} - 5ab + b^2 \]

(iii) (3.5m – 1.5n) (3.5m + 1.5n)
\[ = (3.5m)^2 - (1.5n)^2 \]
\[ = 12.25m^2 - 2.25n^2 \]

(iv) (7xy – 2)(7xy + 7)
\[ = (7xy)^2 + (-2 + 7) \times (7xy) + (-2) \times 7 \]
\[ = 49x^2y^2 + 35xy - 14 \]
In simple words: When you multiply two expressions that match a pattern, use the algebra rule that fits. For (a + b)² you get a² plus 2ab plus b². For (a - b)(a + b) you get a² minus b². Apply the right rule and your calculation becomes much faster.

Exam Tip: Always identify which identity pattern the product matches before expanding - this saves time and reduces mistakes. Check your final answer by expanding without identities to verify.

Question 11. Using suitable identities, evaluate the following:
(i) 105²
(ii) 97²
(iii) 201 × 199
(iv) 87² - 13²
(v) 105 × 107
Answer:
(i) (105)² = (100 + 5)²
\[ = (100)^2 + 2 \times 100 \times 5 + (5)^2 \]
\[ = 10000 + 1000 + 25 = 11025 \]

(ii) (97)² = (100 - 3)²
\[ = (100)^2 - 2 \times 100 \times 3 + (3)^2 \]
\[ = 10000 - 600 + 9 = 9409 \]

(iii) 201 × 199 = (200 + 1) (200 - 1)
\[ = (200)^2 - (1)^2 = 40000 - 1 = 39999 \]

(iv) 87² - 13²
\[ = (87 + 13)(87 - 13) = 100 \times 74 = 7400 \]

(v) 105 × 107 = (100 + 5) (100 + 7)
\[ = (100)^2 + (5 + 7) \times 100 + 5 \times 7 \]
\[ = 10000 + 1200 + 35 = 11235 \]
In simple words: Break down each number into a simpler form (like 105 as 100 + 5), then use the patterns to multiply quickly instead of doing long arithmetic.

Exam Tip: Always rewrite numbers or products in a form that matches a standard identity - this is faster than direct multiplication and shows you understand algebraic patterns.

Question 12. Prove the following:
(i) (a + b)² - (a - b)² = 4ab
(ii) (2a + 3b)² + (2a - 3b)² = 8a² + 18b²
Answer:
(i) Taking the right side:
\[ (a - b)^2 + 4ab = a^2 - 2ab + b^2 + 4ab = a^2 + 2ab + b^2 = (a + b)^2 \]
This equals the left side, so the statement is proved.

(ii) Taking the left side:
\[ (2a + 3b)^2 + (2a - 3b)^2 \]
\[ = (2a)^2 + 2 \times 2a \times 3b + (3b)^2 + (2a)^2 - 2 \times 2a \times 3b + (3b)^2 \]
\[ = 4a^2 + 12ab + 9b^2 + 4a^2 - 12ab + 9b^2 \]
\[ = 8a^2 + 18b^2 \]
This equals the right side, so the statement is proved.
In simple words: Expand each squared term step by step, combine like terms, and show both sides become equal. The middle terms often cancel out, which is why these proofs work so nicely.

Exam Tip: In proof problems, expand fully and collect similar terms together - seeing how terms cancel or combine is what proves the identity. Write each algebraic step clearly.

Question 13. If x + 1/x = 5, evaluate:
(i) x² + 1/x²
(ii) x⁴ + 1/x⁴
Answer:
(i) Given: x + 1/x = 5

Square both sides:
\[ (x + 1/x)^2 = 5^2 \]
\[ x^2 + 1/x^2 + 2 \times x \times \frac{1}{x} = 25 \]
\[ x^2 + 2 + 1/x^2 = 25 \]
\[ x^2 + 1/x^2 = 25 - 2 = 23 \]

(ii) Square the result from part (i):
\[ (x^2 + 1/x^2)^2 = 23^2 \]
\[ x^4 + 1/x^4 + 2 \times x^2 \times \frac{1}{x^2} = 529 \]
\[ x^4 + 1/x^4 + 2 = 529 \]
\[ x^4 + 1/x^4 = 529 - 2 = 527 \]
In simple words: When you know x + 1/x, square it to find x² + 1/x². Then square that result to find x⁴ + 1/x⁴. Each time, the middle term works out to be 2, so subtract it to get your answer.

Exam Tip: Squaring both sides is the key step - it creates the product term 2 × x × (1/x) = 2, which always appears and must be subtracted. Use this pattern whenever you see reciprocal fractions.

Question 14. If a + b = 5 and a² + b² = 13, find ab.
Answer: Given: a + b = 5 and a² + b² = 13

Square the first equation on both sides:
\[ (a + b)^2 = (5)^2 \]
\[ a^2 + b^2 + 2ab = 25 \]

Substitute the known value of a² + b² = 13:
\[ 13 + 2ab = 25 \]
\[ 2ab = 25 - 13 = 12 \]
\[ ab = \frac{12}{2} = 6 \]

Therefore, ab = 6
In simple words: Square the sum a + b to get a² + b² plus 2ab. You already know what a² + b² equals, so swap it in and solve for ab.

Exam Tip: When given two conditions with a and b, always try squaring one equation and substituting the other - this linking technique is a standard way to find products or other combinations.