Exercise 6.2: Addition, Subtraction, and Simplification of Algebraic Fractions
1. Which one of the following pairs expresses fractions $\frac{2}{3a}$ and $\frac{3}{5ab}$ in equal denominators?
Explanation:
LCM of denominators $3a$ and $5ab = 15ab$
For the 1st fraction: $15ab \div 3a = 5b$
1st fraction $= \frac{2}{3a} = \frac{2 \times 5b}{3a \times 5b} = \frac{10b}{15ab}$
For the 2nd fraction: $15ab \div 5ab = 3$
2nd fraction $= \frac{3}{5ab} = \frac{3 \times 3}{5ab \times 3} = \frac{9}{15ab}$
Answer: (a) $\frac{10b}{15ab}, \frac{9}{15ab}$
2. Which one of the following pairs expresses fractions $\frac{x}{yz}$ and $\frac{y}{zx}$ with common denominator?
Explanation:
LCM of denominators $yz$ and $zx = xyz$
For the 1st fraction: $xyz \div yz = x$
1st fraction $= \frac{x}{yz} = \frac{x \times x}{yz \times x} = \frac{x^2}{xyz}$
For the 2nd fraction: $xyz \div zx = y$
2nd fraction $= \frac{y}{zx} = \frac{y \times y}{zx \times y} = \frac{y^2}{xyz}$
Answer: (d) $\frac{x^2}{xyz}, \frac{y^2}{xyz}$
3. If $\frac{x}{2} + 1 = 3$, what is the value of $x$?
Explanation: $\frac{x}{2} + 1 = 3$
or, $\frac{x}{2} = 3 – 1$
or, $\frac{x}{2} = 2$
or, $x = 2 \times 2$
or, $x = 4$
Answer: (b) $4$
4. Which of the following is an equivalent fraction of $\frac{a}{b}$?
Explanation: Multiplying the numerator and denominator of a fraction by the same expression gives an equivalent fraction.
Here, multiplying the numerator and denominator of $\frac{a}{b}$ by $c$, we get $\frac{ac}{bc}$.
Answer: (d) $\frac{ac}{bc}$
5. Which of the following is the lowest term of $\frac{4a^2b – 9b^3}{4a^2b + 6ab^2}$?
Explanation:
Numerator $= 4a^2b – 9b^3 = b(4a^2 – 9b^2) = b\{(2a)^2 – (3b)^2\} = b(2a + 3b)(2a – 3b)$
Denominator $= 4a^2b + 6ab^2 = 2ab(2a + 3b)$
Given expression $= \frac{4a^2b – 9b^3}{4a^2b + 6ab^2} = \frac{b(2a + 3b)(2a – 3b)}{2ab(2a + 3b)} = \frac{2a – 3b}{2a}$
Answer: (c) $\frac{2a – 3b}{2a}$
6. What is the value of $\frac{a}{x} + \frac{b}{x} – \frac{c}{x}$?
Explanation:
$\frac{a}{x} + \frac{b}{x} – \frac{c}{x} = \frac{a + b – c}{x}$
Answer: (b) $\frac{a + b – c}{x}$
Find the sum (7–12):
7.$\frac{3a}{5} + \frac{2b}{5}$
Solution:$\frac{3a}{5} + \frac{2b}{5} = \frac{3a + 2b}{5}$
Answer:$\frac{3a + 2b}{5}$
8.$\frac{1}{5x} + \frac{2}{5x}$
Solution:$\frac{1}{5x} + \frac{2}{5x} = \frac{1 + 2}{5x} = \frac{3}{5x}$
Answer:$\frac{3}{5x}$
9.$\frac{x}{2a} + \frac{y}{3b}$
Solution:$\frac{x}{2a} + \frac{y}{3b} = \frac{3b \cdot x + 2a \cdot y}{6ab}$ [LCM of denominators $2a$ and $3b = 6ab$]
$= \frac{3bx + 2ay}{6ab}$
Answer:$\frac{3bx + 2ay}{6ab}$
10.$\frac{2a}{x + 1} + \frac{2a}{x – 2}$
Solution:$\frac{2a}{x + 1} + \frac{2a}{x – 2} = \frac{2a(x – 2) + 2a(x + 1)}{(x + 1)(x – 2)}$ [LCM of denominators $= (x + 1)(x – 2)$]
$= \frac{2ax – 4a + 2ax + 2a}{(x + 1)(x – 2)} = \frac{4ax – 2a}{(x + 1)(x – 2)} = \frac{2a(2x – 1)}{(x + 1)(x – 2)}$
Answer:$\frac{2a(2x – 1)}{(x + 1)(x – 2)}$
11.$\frac{a}{a + 2} + \frac{2}{a – 2}$
Solution:$\frac{a}{a + 2} + \frac{2}{a – 2} = \frac{a(a – 2) + 2(a + 2)}{(a + 2)(a – 2)}$ [LCM of denominators $= (a + 2)(a – 2)$]
$= \frac{a^2 – 2a + 2a + 4}{a^2 – 4} = \frac{a^2 + 4}{a^2 – 4}$
Answer:$\frac{a^2 + 4}{a^2 – 4}$
12.$\frac{3}{x^2 – 4x – 5} + \frac{4}{x + 1}$
Solution:
Denominator of 1st fraction $= x^2 – 4x – 5 = x^2 – 5x + x – 5 = x(x – 5) + 1(x – 5) = (x – 5)(x + 1)$
Given expression $= \frac{3}{(x – 5)(x + 1)} + \frac{4}{x + 1} = \frac{3 + 4(x – 5)}{(x – 5)(x + 1)} = \frac{3 + 4x – 20}{(x – 5)(x + 1)} = \frac{4x – 17}{(x – 5)(x + 1)}$
Answer:$\frac{4x – 17}{(x – 5)(x + 1)}$
Subtract (13–18):
13.$\frac{2a}{7} – \frac{4b}{7}$
Solution:$\frac{2a}{7} – \frac{4b}{7} = \frac{2a – 4b}{7}$
Answer:$\frac{2a – 4b}{7}$
14.$\frac{2x}{5a} – \frac{4y}{5a}$
Solution:$\frac{2x}{5a} – \frac{4y}{5a} = \frac{2x – 4y}{5a}$
Answer:$\frac{2x – 4y}{5a}$
15.$\frac{a}{8x} – \frac{b}{4y}$
Solution: LCM of denominators $8x$ and $4y = 8xy$
$\frac{a}{8x} – \frac{b}{4y} = \frac{ay – 2bx}{8xy}$
Answer:$\frac{ay – 2bx}{8xy}$
16.$\frac{3}{x + 3} – \frac{2}{x + 2}$
Solution:$\frac{3}{x + 3} – \frac{2}{x + 2} = \frac{3(x + 2) – 2(x + 3)}{(x + 3)(x + 2)}$ [LCM of denominators $= (x + 3)(x + 2)$]
$= \frac{3x + 6 – 2x – 6}{(x + 3)(x + 2)} = \frac{x}{(x + 3)(x + 2)}$
Answer:$\frac{x}{(x + 3)(x + 2)}$
17.$\frac{p + q}{pq} – \frac{q + r}{qr}$
Solution:$\frac{p + q}{pq} – \frac{q + r}{qr} = \frac{r(p + q) – p(q + r)}{pqr}$ [LCM of denominators $pq$ and $qr = pqr$]
$= \frac{pr + qr – pq – pr}{pqr} = \frac{qr – pq}{pqr} = \frac{q(r – p)}{pqr} = \frac{r – p}{pr}$
Answer:$\frac{r – p}{pr}$
18.$\frac{2x}{x^2 – 4y^2} – \frac{x}{xy + 2y^2}$
Solution:
Denominator of 1st fraction $= x^2 – 4y^2 = (x + 2y)(x – 2y)$
Denominator of 2nd fraction $= xy + 2y^2 = y(x + 2y)$
LCM of denominators $= y(x + 2y)(x – 2y)$
Given expression $= \frac{2x}{(x + 2y)(x – 2y)} – \frac{x}{y(x + 2y)} = \frac{2x \cdot y – x(x – 2y)}{y(x + 2y)(x – 2y)} = \frac{2xy – x^2 + 2xy}{y(x^2 – 4y^2)} = \frac{4xy – x^2}{y(x^2 – 4y^2)} = \frac{x(4y – x)}{y(x^2 – 4y^2)}$
Answer:$\frac{x(4y – x)}{y(x^2 – 4y^2)}$
Simplify (19–24):
19.$\frac{5}{a^2 – 6a + 5} + \frac{1}{a – 1}$
Solution:
Denominator of 1st fraction $= a^2 – 6a + 5 = a^2 – 5a – a + 5 = a(a – 5) – 1(a – 5) = (a – 5)(a – 1)$
Given expression $= \frac{5}{(a – 5)(a – 1)} + \frac{1}{a – 1} = \frac{5 + 1(a – 5)}{(a – 5)(a – 1)} = \frac{5 + a – 5}{(a – 5)(a – 1)} = \frac{a}{(a – 5)(a – 1)}$
Answer:$\frac{a}{(a – 5)(a – 1)}$
20.$\frac{1}{x + 2} – \frac{1}{x^2 – 4}$
Solution: Denominator of 2nd fraction $= x^2 – 4 = (x + 2)(x – 2)$
Given expression $= \frac{1}{x + 2} – \frac{1}{(x + 2)(x – 2)} = \frac{1(x – 2) – 1}{(x + 2)(x – 2)} = \frac{x – 2 – 1}{(x + 2)(x – 2)} = \frac{x – 3}{x^2 – 4}$
Answer:$\frac{x – 3}{x^2 – 4}$
21.$\frac{a}{3} + \frac{a}{6} – \frac{3a}{8}$
Solution:$\frac{a}{3} + \frac{a}{6} – \frac{3a}{8} = \frac{8a + 4a – 3(3a)}{24}$ [LCM of denominators $3, 6, 8 = 24$]
$= \frac{8a + 4a – 9a}{24} = \frac{12a – 9a}{24} = \frac{3a}{24} = \frac{a}{8}$
Answer:$\frac{a}{8}$
22.$\frac{a}{b} – \frac{3a}{2b} + \frac{2a}{3b}$
Solution:$\frac{a}{b} – \frac{3a}{2b} + \frac{2a}{3b} = \frac{6a – 3(3a) + 2(2a)}{6b}$ [LCM of denominators $b, 2b, 3b = 6b$]
$= \frac{6a – 9a + 4a}{6b} = \frac{10a – 9a}{6b} = \frac{a}{6b}$
Answer:$\frac{a}{6b}$
23.$\frac{x}{yz} – \frac{y}{zx} + \frac{z}{xy}$
Solution:$\frac{x}{yz} – \frac{y}{zx} + \frac{z}{xy} = \frac{x \cdot x – y \cdot y + z \cdot z}{xyz}$ [LCM of denominators $yz, zx, xy = xyz$]
$= \frac{x^2 – y^2 + z^2}{xyz}$
Answer:$\frac{x^2 – y^2 + z^2}{xyz}$
24.$\frac{x – y}{xy} + \frac{y – z}{yz} + \frac{z – x}{zx}$
Solution:$\frac{x – y}{xy} + \frac{y – z}{yz} + \frac{z – x}{zx} = \frac{z(x – y) + x(y – z) + y(z – x)}{xyz}$ [LCM of denominators $xy, yz, zx = xyz$]
$= \frac{xz – yz + xy – xz + yz – xy}{xyz} = \frac{0}{xyz} = 0$
Answer:$0$
25. Three algebraic fractions are: $\frac{x}{x + y}, \frac{x}{x – 4y}, \frac{y}{x^2 – 3xy – 4y^2}$
(a) Factorize: $a^2 – 3a + 2$
Solution:$a^2 – 3a + 2 = a^2 – 2a – a + 2 = a(a – 2) – 1(a – 2) = (a – 2)(a – 1)$
Answer:$(a – 2)(a – 1)$
(b) Express the first and second fractions with equal denominators.
Solution:
Denominator of 1st fraction $= x + y$
Denominator of 2nd fraction $= x – 4y$
LCM of denominators $= (x + y)(x – 4y)$
1st fraction $= \frac{x}{x + y} = \frac{x(x – 4y)}{(x + y)(x – 4y)} = \frac{x^2 – 4xy}{(x + y)(x – 4y)}$
2nd fraction $= \frac{x}{x – 4y} = \frac{x(x + y)}{(x + y)(x – 4y)} = \frac{x^2 + xy}{(x + y)(x – 4y)}$
Answer:$\frac{x(x – 4y)}{(x + y)(x – 4y)}, \frac{x(x + y)}{(x + y)(x – 4y)}$
(c) Find the sum of the given three fractions.
Solution:
Denominator of 3rd fraction $= x^2 – 3xy – 4y^2 = x^2 – 4xy + xy – 4y^2 = x(x – 4y) + y(x – 4y) = (x – 4y)(x + y)$
Sum of the three fractions $= \frac{x}{x + y} + \frac{x}{x – 4y} + \frac{y}{(x + y)(x – 4y)}$
$= \frac{x(x – 4y) + x(x + y) + y}{(x + y)(x – 4y)} = \frac{x^2 – 4xy + x^2 + xy + y}{(x + y)(x – 4y)} = \frac{2x^2 – 3xy + y}{(x + y)(x – 4y)}$
Answer:$\frac{2x^2 – 3xy + y}{(x + y)(x – 4y)}$
26. Three algebraic fractions: $\frac{1}{a^2 + 3a}, \frac{1}{a^2 + 5a + 6}, \frac{1}{a^2 – a – 12}$
(a) Factorize the denominator of the 3rd fraction.
Solution: Denominator of 3rd fraction $= a^2 – a – 12 = a^2 – 4a + 3a – 12 = a(a – 4) + 3(a – 4) = (a – 4)(a + 3)$
Answer:$(a – 4)(a + 3)$
(b) Convert the 1st and 2nd fractions into fractions with equal denominators.
Solution:
Denominator of 1st fraction $= a^2 + 3a = a(a + 3)$
Denominator of 2nd fraction $= a^2 + 5a + 6 = a^2 + 3a + 2a + 6 = a(a + 3) + 2(a + 3) = (a + 3)(a + 2)$
LCM of denominators $= a(a + 3)(a + 2)$
1st fraction $= \frac{1}{a^2 + 3a} = \frac{1}{a(a + 3)} = \frac{1 \times (a + 2)}{a(a + 3)(a + 2)} = \frac{a + 2}{a(a + 3)(a + 2)}$
2nd fraction $= \frac{1}{a^2 + 5a + 6} = \frac{1}{(a + 3)(a + 2)} = \frac{1 \times a}{a(a + 3)(a + 2)} = \frac{a}{a(a + 3)(a + 2)}$
Answer:$\frac{a + 2}{a(a + 3)(a + 2)}, \frac{a}{a(a + 3)(a + 2)}$
(c) Find the sum of the 1st, 2nd, and 3rd fractions.
Solution:
From (a) and (b), we get:
Denominator of 1st fraction $= a(a + 3)$
Denominator of 2nd fraction $= (a + 3)(a + 2)$
Denominator of 3rd fraction $= (a – 4)(a + 3)$
LCM of the denominators $= a(a + 3)(a + 2)(a – 4)$
Sum of the three fractions $= \frac{1}{a(a + 3)} + \frac{1}{(a + 3)(a + 2)} + \frac{1}{(a – 4)(a + 3)}$
$= \frac{(a + 2)(a – 4) + a(a – 4) + a(a + 2)}{a(a + 3)(a + 2)(a – 4)} = \frac{(a^2 – 4a + 2a – 8) + (a^2 – 4a) + (a^2 + 2a)}{a(a + 3)(a + 2)(a – 4)}$
$= \frac{a^2 – 2a – 8 + a^2 – 4a + a^2 + 2a}{a(a + 3)(a + 2)(a – 4)} = \frac{3a^2 – 4a – 8}{a(a + 3)(a + 2)(a – 4)}$
Answer:$\frac{3a^2 – 4a – 8}{a(a + 3)(a + 2)(a – 4)}$
Sample Questions (Multiple Choice)
1. What is the value of $\frac{1}{a + b} + \frac{1}{a – b}$?
Explanation: LCM of denominators $(a + b)$ and $(a – b) = (a + b)(a – b) = a^2 – b^2$
Given expression $= \frac{1}{a + b} + \frac{1}{a – b} = \frac{(a – b) + (a + b)}{(a + b)(a – b)} = \frac{a – b + a + b}{a^2 – b^2} = \frac{2a}{a^2 – b^2}$
Answer: (c) $\frac{2a}{a^2 – b^2}$
2. For the fraction $\frac{x^2}{xy}$—
i. Lowest form is $\frac{x}{y}$
ii. Equivalent fraction is $\frac{x^2 + 1}{xy + 1}$
iii. A fraction with the same denominator is $\frac{x^2 + 3}{xy}$
Which of the following is correct?
(a) i and ii
(b) i and iii
(c) ii and iii
(d) i, ii, and iii
Explanation:
i. True: $\frac{x^2}{xy} = \frac{x \cdot x}{x \cdot y} = \frac{x}{y}$
ii. False: Adding or subtracting a number from the numerator and denominator does not yield an equivalent fraction (it is only obtained by multiplying or dividing by the same non-zero quantity).
iii. True: Both fractions $\frac{x^2}{xy}$ and $\frac{x^2 + 3}{xy}$ have the same denominator ($xy$), making them fractions with common denominators.
Therefore, i and iii are correct.
Answer: (b) i and iii
Based on the information below, answer to questions no. 3 & 4.
$\frac{x^2 + 2x}{(x + 2)(x – 2)}$ and $\frac{x^2 + 4x – 4}{x^2 – 4}$ two algebraic fractions.
3. What is the simplest form of the first fraction?
Solution:
Numerator of 1st fraction $= x^2 + 2x = x(x + 2)$
Denominator $= (x + 2)(x – 2)$
Therefore, 1st fraction $= \frac{x(x + 2)}{(x + 2)(x – 2)} = \frac{x}{x – 2}$
Answer: (a) $\frac{x}{x – 2}$
4. What is the result when the first fraction is subtracted from the second?
Solution:
2nd fraction $= \frac{x^2 + 4x – 4}{x^2 – 4} = \frac{x^2 + 4x – 4}{(x + 2)(x – 2)}$
1st fraction $= \frac{x}{x – 2}$
Difference $= \frac{x^2 + 4x – 4}{(x + 2)(x – 2)} – \frac{x}{x – 2} = \frac{(x^2 + 4x – 4) – x(x + 2)}{(x + 2)(x – 2)} = \frac{x^2 + 4x – 4 – x^2 – 2x}{(x + 2)(x – 2)} = \frac{2x – 4}{(x + 2)(x – 2)} = \frac{2(x – 2)}{(x + 2)(x – 2)} = \frac{2}{x + 2}$
Answer: (b) $\frac{2}{x + 2}$
Creative Question
5. Consider the four fractions: $\frac{3}{x^2 – x – 12}$, $\frac{4}{x^2 – 9}$, $\frac{x – 1}{x^2 + 3x}$, and $\frac{3x + 6}{x(x^2 + 5x + 6)}$.
(a) Express $\frac{a^2 – 2a}{3(a – 2)}$ in the simplest form.
Solution: Given expression $= \frac{a^2 – 2a}{3(a – 2)} = \frac{a(a – 2)}{3(a – 2)} = \frac{a}{3}$
Answer:$\frac{a}{3}$
(b) Express the first and second fractions with a common denominator.
Solution:
Denominator of 1st fraction $= x^2 – x – 12 = x^2 – 4x + 3x – 12 = x(x – 4) + 3(x – 4) = (x – 4)(x + 3)$
Denominator of 2nd fraction $= x^2 – 9 = x^2 – 3^2 = (x + 3)(x – 3)$
LCM of denominators $= (x + 3)(x – 3)(x – 4)$
1st fraction $= \frac{3}{(x – 4)(x + 3)} = \frac{3(x – 3)}{(x + 3)(x – 3)(x – 4)}$
2nd fraction $= \frac{4}{(x + 3)(x – 3)} = \frac{4(x – 4)}{(x + 3)(x – 3)(x – 4)}$
Answer:$\frac{3(x – 3)}{(x + 3)(x – 3)(x – 4)}, \frac{4(x – 4)}{(x + 3)(x – 3)(x – 4)}$
(c) Find the sum of the third and fourth fractions.
Solution:
3rd fraction $= \frac{x – 1}{x^2 + 3x} = \frac{x – 1}{x(x + 3)}$
4th fraction $= \frac{3x + 6}{x(x^2 + 5x + 6)}$
Here, $x^2 + 5x + 6 = x^2 + 3x + 2x + 6 = x(x + 3) + 2(x + 3) = (x + 3)(x + 2)$
and Numerator $= 3x + 6 = 3(x + 2)$
Therefore, 4th fraction $= \frac{3(x + 2)}{x(x + 3)(x + 2)} = \frac{3}{x(x + 3)}$
Sum $= \frac{x – 1}{x(x + 3)} + \frac{3}{x(x + 3)} = \frac{x – 1 + 3}{x(x + 3)} = \frac{x + 2}{x(x + 3)}$
Answer:$\frac{x + 2}{x(x + 3)}$
Short-Answer Questions
6. (a) Express the fractions $\frac{1}{x – 4}$ and $\frac{x}{x + 2}$ with a common denominator.
Solution: LCM of denominators $= (x – 4)(x + 2)$
1st fraction $= \frac{1}{x – 4} = \frac{1 \cdot (x + 2)}{(x – 4)(x + 2)} = \frac{x + 2}{(x – 4)(x + 2)}$
2nd fraction $= \frac{x}{x + 2} = \frac{x(x – 4)}{(x – 4)(x + 2)} = \frac{x^2 – 4x}{(x – 4)(x + 2)}$
Answer:$\frac{x + 2}{(x – 4)(x + 2)}, \frac{x^2 – 4x}{(x – 4)(x + 2)}$
6. (b) Find the result when $\frac{bx}{b – a}$ is subracted from $\frac{-ax}{a – b}$.
Solution:
$\frac{-ax}{a – b} – \frac{bx}{b – a} = \frac{-ax}{a – b} – \frac{bx}{-(a – b)} = \frac{-ax}{a – b} + \frac{bx}{a – b}$
$= \frac{-ax + bx}{a – b} = \frac{x(b – a)}{a – b} = \frac{-x(a – b)}{a – b} = -x$
Answer:$-x$
6. (c) What fraction should be added to $\frac{a + b}{a – b}$ so that the sum is $1$.
Solution: Let the required fraction be $P$.
According to the question,
$\frac{a + b}{a – b} + P = 1$
or, $P = 1 – \frac{a + b}{a – b}$
or, $P = \frac{(a – b) – (a + b)}{a – b}$
or, $P = \frac{a – b – a – b}{a – b}$
or, $P = \frac{-2b}{a – b}$
or, $P = \frac{2b}{b – a}$
Answer:$\frac{2b}{b – a}$ (or $\frac{-2b}{a – b}$)