Exercise 6.1: Algebraic Fractions
Express in Lowest Terms (1–10):
1.$\frac{a^2b}{a^3c}$
Solution:$\frac{a^2b}{a^3c}$
$= \frac{a^2 \cdot b}{a^2 \cdot a \cdot c}$
$= \frac{b}{ac}$
Answer:$\frac{b}{ac}$
2.$\frac{a^2bc}{ab^2c}$
Solution:$\frac{a^2bc}{ab^2c}$
$= \frac{a \cdot (abc)}{b \cdot (abc)}$
$= \frac{a}{b}$
Answer:$\frac{a}{b}$
3.$\frac{x^3y^3z^3}{x^2y^2z^2}$
Solution:$\frac{x^3y^3z^3}{x^2y^2z^2}$
$= \frac{(x^2y^2z^2) \cdot xyz}{x^2y^2z^2}$
$= xyz$
Answer:$xyz$
4.$\frac{x^2 + x}{xy + y}$
Solution:$\frac{x^2 + x}{xy + y}$
$= \frac{x(x + 1)}{y(x + 1)}$
$= \frac{x}{y}$
Answer:$\frac{x}{y}$
5.$\frac{4a^2b}{6a^3b}$
Solution:$\frac{4a^2b}{6a^3b}$
$= \frac{2 \cdot 2 \cdot a^2 \cdot b}{2 \cdot 3 \cdot a^2 \cdot a \cdot b}$
$= \frac{2}{3a}$
Answer:$\frac{2}{3a}$
6.$\frac{2a – 4ab}{1 – 4b^2}$
Solution:$\frac{2a – 4ab}{1 – 4b^2}$
$= \frac{2a(1 – 2b)}{1^2 – (2b)^2}$
$= \frac{2a(1 – 2b)}{(1 + 2b)(1 – 2b)}$
$= \frac{2a}{1 + 2b}$
Answer:$\frac{2a}{1 + 2b}$
7.$\frac{2a + 3b}{4a^2 – 9b^2}$
Solution:$\frac{2a + 3b}{4a^2 – 9b^2}$
$= \frac{2a + 3b}{(2a)^2 – (3b)^2}$
$= \frac{2a + 3b}{(2a + 3b)(2a – 3b)}$
$= \frac{1}{2a – 3b}$
Answer:$\frac{1}{2a – 3b}$
8.$\frac{a^2 + 4a + 4}{a^2 – 4}$
Solution:
Numerator $= a^2 + 4a + 4$
$= a^2 + 2 \cdot a \cdot 2 + 2^2$
$= (a + 2)^2$
$= (a + 2)(a + 2)$
Denominator $= a^2 – 4$
$= a^2 – 2^2$
$= (a + 2)(a – 2)$
Therefore, $\frac{a^2 + 4a + 4}{a^2 – 4}$
$= \frac{(a + 2)(a + 2)}{(a + 2)(a – 2)}$
$= \frac{a + 2}{a – 2}$
Answer:$\frac{a + 2}{a – 2}$
9.$\frac{x^2 – y^2}{(x + y)^2}$
Solution:$\frac{x^2 – y^2}{(x + y)^2}$
$= \frac{(x + y)(x – y)}{(x + y)(x + y)}$
$= \frac{x – y}{x + y}$
Answer:$\frac{x – y}{x + y}$
10.$\frac{x^2 + 2x – 15}{x^2 + 9x + 20}$
Solution:
Numerator $= x^2 + 2x – 15$
$= x^2 + 5x – 3x – 15$
$= x(x + 5) – 3(x + 5)$
$= (x + 5)(x – 3)$
Denominator $= x^2 + 9x + 20$
$= x^2 + 5x + 4x + 20$
$= x(x + 5) + 4(x + 5)$
$= (x + 5)(x + 4)$
Therefore, $\frac{x^2 + 2x – 15}{x^2 + 9x + 20}$
$= \frac{(x + 5)(x – 3)}{(x + 5)(x + 4)}$
$= \frac{x – 3}{x + 4}$
Answer:$\frac{x – 3}{x + 4}$
Express the fractions with common denominators (11–20):
11.$\frac{a}{bc}, \frac{a}{ac}$
Solution:
LCM of the denominators $bc$ and $ac = abc$
For the 1st fraction, $abc \div bc = a$
1st fraction $= \frac{a}{bc} = \frac{a \times a}{bc \times a} = \frac{a^2}{abc}$
For the 2nd fraction, $abc \div ac = b$
2nd fraction $= \frac{a}{ac} = \frac{a \times b}{ac \times b} = \frac{ab}{abc}$
Answer:$\frac{a^2}{abc}, \frac{ab}{abc}$
12.$\frac{x}{pq}, \frac{y}{pr}$
Solution:
LCM of the denominators $pq$ and $pr = pqr$
For the 1st fraction, $pqr \div pq = r$
1st fraction $= \frac{x}{pq} = \frac{x \times r}{pq \times r} = \frac{xr}{pqr}$
For the 2nd fraction, $pqr \div pr = q$
2nd fraction $= \frac{y}{pr} = \frac{y \times q}{pr \times q} = \frac{yq}{pqr}$
Answer:$\frac{xr}{pqr}, \frac{yq}{pqr}$
13.$\frac{2x}{3m}, \frac{3y}{2n}$
Solution:
LCM of the denominators $3m$ and $2n = 6mn$
For the 1st fraction, $6mn \div 3m = 2n$
1st fraction $= \frac{2x}{3m} = \frac{2x \times 2n}{3m \times 2n} = \frac{4xn}{6mn}$
For the 2nd fraction, $6mn \div 2n = 3m$
2nd fraction $= \frac{3y}{2n} = \frac{3y \times 3m}{2n \times 3m} = \frac{9ym}{6mn}$
Answer:$\frac{4xn}{6mn}, \frac{9ym}{6mn}$
14.$\frac{a}{a – b}, \frac{b}{a + b}$
Solution:
LCM of the denominators $(a – b)$ and $(a + b) = (a – b)(a + b)$
For the 1st fraction, $(a – b)(a + b) \div (a – b) = a + b$
1st fraction $= \frac{a}{a – b} = \frac{a(a + b)}{(a – b)(a + b)}$
For the 2nd fraction, $(a – b)(a + b) \div (a + b) = a – b$
2nd fraction $= \frac{b}{a + b} = \frac{b(a – b)}{(a – b)(a + b)}$
Answer:$\frac{a(a + b)}{(a – b)(a + b)}, \frac{b(a – b)}{(a – b)(a + b)}$
15.$\frac{x^2}{a^2 – 2ab}, \frac{y^2}{a + 2b}$
Solution:
Denominator of 1st fraction $= a^2 – 2ab = a(a – 2b)$
Denominator of 2nd fraction $= a + 2b$
LCM of the denominators $= a(a – 2b)(a + 2b)$
For the 1st fraction, $a(a – 2b)(a + 2b) \div a(a – 2b) = a + 2b$
1st fraction $= \frac{x^2}{a^2 – 2ab} = \frac{x^2}{a(a – 2b)} = \frac{x^2(a + 2b)}{a(a – 2b)(a + 2b)}$
For the 2nd fraction, $a(a – 2b)(a + 2b) \div (a + 2b) = a(a – 2b)$
2nd fraction $= \frac{y^2}{a + 2b} = \frac{y^2 \cdot a(a – 2b)}{(a + 2b) \cdot a(a – 2b)} = \frac{ay^2(a – 2b)}{a(a – 2b)(a + 2b)}$
Answer:$\frac{x^2(a + 2b)}{a(a – 2b)(a + 2b)}, \frac{ay^2(a – 2b)}{a(a – 2b)(a + 2b)}$
16.$\frac{3}{a^2 – 4}, \frac{2}{a(a + 2)}$
Solution:
Denominator of 1st fraction $= a^2 – 4 = a^2 – 2^2 = (a + 2)(a – 2)$
Denominator of 2nd fraction $= a(a + 2)$
LCM of the denominators $= a(a + 2)(a – 2)$
For the 1st fraction, $a(a + 2)(a – 2) \div (a + 2)(a – 2) = a$
1st fraction $= \frac{3}{a^2 – 4} = \frac{3}{(a + 2)(a – 2)} = \frac{3 \times a}{a(a + 2)(a – 2)} = \frac{3a}{a(a + 2)(a – 2)}$
For the 2nd fraction, $a(a + 2)(a – 2) \div a(a + 2) = a – 2$
2nd fraction $= \frac{2}{a(a + 2)} = \frac{2(a – 2)}{a(a + 2)(a – 2)}$
Answer:$\frac{3a}{a(a + 2)(a – 2)}, \frac{2(a – 2)}{a(a + 2)(a – 2)}$
17.$\frac{a}{a^2 – 9}, \frac{b}{a + 3}$
Solution:
Denominator of 1st fraction $= a^2 – 9 = a^2 – 3^2 = (a + 3)(a – 3)$
Denominator of 2nd fraction $= a + 3$
LCM of the denominators $= (a + 3)(a – 3)$
For the 1st fraction, $(a + 3)(a – 3) \div (a + 3)(a – 3) = 1$
1st fraction $= \frac{a}{a^2 – 9} = \frac{a}{(a + 3)(a – 3)}$
For the 2nd fraction, $(a + 3)(a – 3) \div (a + 3) = a – 3$
2nd fraction $= \frac{b}{a + 3} = \frac{b(a – 3)}{(a + 3)(a – 3)}$
Answer:$\frac{a}{(a + 3)(a – 3)}, \frac{b(a – 3)}{(a + 3)(a – 3)}$
18.$\frac{a}{a + b}, \frac{b}{a – b}, \frac{c}{a – c}$
Solution:
LCM of the denominators $= (a + b)(a – b)(a – c)$
For the 1st fraction, $(a + b)(a – b)(a – c) \div (a + b) = (a – b)(a – c)$
1st fraction $= \frac{a}{a + b} = \frac{a(a – b)(a – c)}{(a + b)(a – b)(a – c)}$
For the 2nd fraction, $(a + b)(a – b)(a – c) \div (a – b) = (a + b)(a – c)$
2nd fraction $= \frac{b}{a – b} = \frac{b(a + b)(a – c)}{(a + b)(a – b)(a – c)}$
For the 3rd fraction, $(a + b)(a – b)(a – c) \div (a – c) = (a + b)(a – b)$
3rd fraction $= \frac{c}{a – c} = \frac{c(a + b)(a – b)}{(a + b)(a – b)(a – c)}$
Answer:$\frac{a(a – b)(a – c)}{(a + b)(a – b)(a – c)}, \frac{b(a + b)(a – c)}{(a + b)(a – b)(a – c)}, \frac{c(a + b)(a – b)}{(a + b)(a – b)(a – c)}$
19.$\frac{a}{a – b}, \frac{b}{a + b}, \frac{c}{a(a + b)}$
Solution:
LCM of the denominators $= a(a – b)(a + b)$
For the 1st fraction, $a(a – b)(a + b) \div (a – b) = a(a + b)$
1st fraction $= \frac{a}{a – b} = \frac{a \cdot a(a + b)}{a(a – b)(a + b)} = \frac{a^2(a + b)}{a(a – b)(a + b)}$
For the 2nd fraction, $a(a – b)(a + b) \div (a + b) = a(a – b)$
2nd fraction $= \frac{b}{a + b} = \frac{b \cdot a(a – b)}{a(a – b)(a + b)} = \frac{ab(a – b)}{a(a – b)(a + b)}$
For the 3rd fraction, $a(a – b)(a + b) \div a(a + b) = a – b$
3rd fraction $= \frac{c}{a(a + b)} = \frac{c(a – b)}{a(a – b)(a + b)}$
Answer:$\frac{a^2(a + b)}{a(a – b)(a + b)}, \frac{ab(a – b)}{a(a – b)(a + b)}, \frac{c(a – b)}{a(a – b)(a + b)}$
20.$\frac{2}{x^2 – x – 2}, \frac{3}{x^2 + x – 6}$
Solution:
Denominator of 1st fraction $= x^2 – x – 2 = x^2 – 2x + x – 2 = x(x – 2) + 1(x – 2) = (x – 2)(x + 1)$
Denominator of 2nd fraction $= x^2 + x – 6 = x^2 + 3x – 2x – 6 = x(x + 3) – 2(x + 3) = (x + 3)(x – 2)$
LCM of the denominators $= (x – 2)(x + 1)(x + 3)$
For the 1st fraction, $(x – 2)(x + 1)(x + 3) \div (x – 2)(x + 1) = x + 3$
1st fraction $= \frac{2}{x^2 – x – 2} = \frac{2}{(x – 2)(x + 1)} = \frac{2(x + 3)}{(x – 2)(x + 1)(x + 3)}$
For the 2nd fraction, $(x – 2)(x + 1)(x + 3) \div (x + 3)(x – 2) = x + 1$
2nd fraction $= \frac{3}{x^2 + x – 6} = \frac{3}{(x + 3)(x – 2)} = \frac{3(x + 1)}{(x – 2)(x + 1)(x + 3)}$
Answer:$\frac{2(x + 3)}{(x – 2)(x + 1)(x + 3)}, \frac{3(x + 1)}{(x – 2)(x + 1)(x + 3)}$