Multiplying and Dividing Algebraic Fractions

Solution

Write it as one fraction: $\dfrac{3a \times 5ab}{4b^2 \times 3}$.

Write out the factors: $\dfrac{3 \times a \times 5 \times a \times b}{4 \times b \times b \times 3}\text{.}$

Cancel the $3$ on top with the $3$ below, and one $b$ on top with one $b$ below: $\dfrac{5 \times a \times a}{4 \times b} = \dfrac{5a^2}{4b}\text{.}$

Check with $a = 2$, $b = 2$: $\dfrac{6}{16} \times \dfrac{20}{3} = \dfrac{120}{48} = \dfrac{5}{2}\text{,}$ and $\dfrac{5 \times 4}{4 \times 2} = \dfrac{5}{2}$.

Solution

(a) $\dfrac{a \times b \times 4 \times c}{c \times 6 \times a \times a \times b}\text{.}$ Cancel $a$, $b$ and $c$ once each, and $\dfrac{4}{6} = \dfrac{2}{3}$. One $a$ is left below: $\dfrac{2}{3a}$.

(b) $\dfrac{2 \times x \times 9 \times y \times y}{3 \times y \times 4 \times x \times x \times x}\text{.}$ Numbers: $\dfrac{18}{12} = \dfrac{3}{2}$. One $x$ cancels, leaving $x^2$ below; one $y$ cancels, leaving $y$ on top: $\dfrac{3y}{2x^2}$.

Check (b) with $x = 1$, $y = 2$: $\dfrac{2}{6} \times \dfrac{36}{4} = \dfrac{1}{3} \times 9 = 3\text{,}$ and $\dfrac{6}{2} = 3$.

Solution

Factorise: $4x - 16 = 4{(x - 4)}$ and $5x + 5y = 5{(x + y)}$.

$\dfrac{4(x - 4)}{x + y} \times \dfrac{5(x + y)}{8} = \dfrac{4 \times 5 \times (x - 4)\cancel{(x + y)}}{\cancel{(x + y)} \times 8}\text{.}$

The ${(x + y)}$ on top of the second fraction cancels with the ${(x + y)}$ below the first. Then $\dfrac{20}{8} = \dfrac{5}{2}$.

So the product is $\dfrac{5(x - 4)}{2}$.

Check with $x = 5$, $y = 1$: $\dfrac{4}{6} \times \dfrac{30}{8} = \dfrac{120}{48} = \dfrac{5}{2}\text{,}$ and $\dfrac{5 \times 1}{2} = \dfrac{5}{2}$.

Solution

Factorise: $h^2 - 2h + 1 = {(h - 1)}^2\text{,}$ a perfect square.

$\dfrac{h}{(h - 1)(h - 1)} \times \dfrac{(h - 1)(h - 1)(h - 1)}{2h + 1}\text{.}$

Two of the three factors ${(h - 1)}$ on top cancel with the two below, leaving one on top: $\dfrac{h(h - 1)}{2h + 1}$.

Check with $h = 2$: $\dfrac{2}{1} \times \dfrac{1}{5} = \dfrac{2}{5}$, and $\dfrac{2 \times 1}{5} = \dfrac{2}{5}$.

Solution

Factorise $9 - 6x$ so that it shows the bracket ${(2x - 3)}$: $9 - 6x = -3{(2x - 3)}\text{.}$ Check by expanding: $-3 \times 2x = -6x$ and $-3 \times {(-3)} = 9$.

$\dfrac{(2x - 3)(2x - 3)}{a} \times \dfrac{a \times a}{-3(2x - 3)}\text{.}$

Cancel one ${(2x - 3)}$ and one $a$: $\dfrac{(2x - 3) \times a}{-3} = -\dfrac{a(2x - 3)}{3}\text{.}$

This can also be written $\dfrac{a(3 - 2x)}{3}$, since $-{(2x - 3)} = 3 - 2x$.

Check with $x = 2$, $a = 1$: $\dfrac{1}{1} \times \dfrac{1}{-3} = -\dfrac{1}{3}\text{,}$ and $\dfrac{1 \times (3 - 4)}{3} = -\dfrac{1}{3}\text{.}$

Solution

(a) $\dfrac{3a}{4} \times \dfrac{10}{9a^2} = \dfrac{3 \times a \times 10}{4 \times 9 \times a \times a}\text{.}$

Numbers: $\dfrac{30}{36} = \dfrac{5}{6}$. One $a$ cancels, leaving $a$ below: $\dfrac{5}{6a}$.

(b) $\dfrac{8x}{3y} \times \dfrac{9y}{2x^2} = \dfrac{72xy}{6x^2y}\text{.}$ Numbers: $\dfrac{72}{6} = 12$. One $x$ and the $y$ cancel, leaving $x$ below: $\dfrac{12}{x}$.

Check (a) with $a = 1$: $\dfrac{3}{4} \div \dfrac{9}{10} = \dfrac{3}{4} \times \dfrac{10}{9} = \dfrac{30}{36} = \dfrac{5}{6}\text{,}$ and $\dfrac{5}{6 \times 1} = \dfrac{5}{6}$.

Solution

Turn the second fraction over: $\dfrac{m^2 - 4}{m^2 - 3m + 2} \times \dfrac{m - 1}{m}\text{.}$

Factorise: $m^2 - 4 = {(m + 2)}{(m - 2)}$ and $m^2 - 3m + 2 = {(m - 1)}{(m - 2)}\text{.}$

$\dfrac{(m + 2)\cancel{(m - 2)}}{\cancel{(m - 1)}\cancel{(m - 2)}} \times \dfrac{\cancel{(m - 1)}}{m} = \dfrac{m + 2}{m}\text{.}$

Undefined values: the first denominator is $0$ at $m = 1$ or $m = 2$; the second denominator at $m = 1$; and the fraction we divide by is $0$ at $m = 0$. So $m \neq 0$, $1$, $2$.

Check with $m = 3$: $\dfrac{5}{2} \div \dfrac{3}{2} = \dfrac{5}{3}$, and $\dfrac{3 + 2}{3} = \dfrac{5}{3}$.

Solution

(a) Dividing by ${(x + 3)}$ is multiplying by its reciprocal $\dfrac{1}{x + 3}$:

$\dfrac{(x + 3)(x - 3)}{2x} \times \dfrac{1}{x + 3} = \dfrac{x - 3}{2x}\text{,}$ for $x \neq 0$ and $x \neq -3$.

(b) The divisor needs factorising: $x^2 - 1 = {(x + 1)}{(x - 1)}\text{.}$ Turn it over and multiply:

$\dfrac{x + 1}{4x} \times \dfrac{2x^2}{(x + 1)(x - 1)}\text{.}$ The bracket ${(x + 1)}$ cancels; $2x^2 = 2x \times x$ on top and $4x = 2x \times 2$ below share the factor $2x$.

This leaves $\dfrac{x}{2(x - 1)}$. The excluded values: $x = 0$ (the denominators $4x$ and $2x^2$), and $x = \pm 1$, where the fraction we divide by is $0$ or has no value.

Check (a) with $x = 1$: $\dfrac{-8}{2} \div 4 = -1$, and $\dfrac{1 - 3}{2} = -1$. Check (b) with $x = 2$: $\dfrac{3}{8} \div \dfrac{3}{8} = 1$, and $\dfrac{2}{2 \times 1} = 1$.

Solution

Turn over only the fraction after $\div$: $\dfrac{p}{q} \times \dfrac{q^2}{p^2r} \times \dfrac{pr^2}{2q}\text{.}$

One fraction: $\dfrac{p \times q^2 \times p \times r^2}{q \times p^2 \times r \times 2q}\text{.}$

On top there are two factors $p$ ($p \times p = p^2$); they cancel with $p^2$ below.

$q^2 = q \times q$ on top cancels with $q \times q$ below.

$r$ below cancels with one $r$ of $r^2$ on top, leaving $r$ on top. Only $2$ is left below.

So the answer is $\dfrac{r}{2}$.

Check with $p = 2$, $q = 1$, $r = 3$: $\dfrac{2}{1} \div \dfrac{12}{1} = \dfrac{1}{6}\text{,}$ then $\dfrac{1}{6} \times \dfrac{2 \times 9}{2} = \dfrac{9}{6} = \dfrac{3}{2}\text{,}$ and $\dfrac{r}{2} = \dfrac{3}{2}$.

Solution

(a) The first fraction was turned over instead of the second. Correct: $\dfrac{2}{x} \times \dfrac{x^2}{4} = \dfrac{2x^2}{4x} = \dfrac{x}{2}\text{.}$

Test with $x = 4$: $\dfrac{2}{4} \div \dfrac{4}{16} = \dfrac{1}{2} \div \dfrac{1}{4} = 2\text{.}$ And $\dfrac{x}{2} = 2$, but the wrong answer $\dfrac{2}{x}$ gives $\dfrac{1}{2}$.

(b) The $x$ was cancelled, but $x$ is a term of $x + 2$ and of $x + 4$, not a factor. Only the numbers cancel: $\dfrac{6}{3} = 2$. Correct: $\dfrac{2(x + 2)}{x + 4}$.

Test with $x = 1$: $\dfrac{3}{3} \times \dfrac{6}{5} = \dfrac{6}{5}$, and $\dfrac{2 \times 3}{5} = \dfrac{6}{5}$, not $1$.