Sec 2 Maths: Algebraic Fractions & Formulae, practice questions & worked solutions

(a)

Multiply by the reciprocal of the second fraction, then cancel:

$$\begin{aligned}\frac{5t^2}{v^3}\div\frac{25t}{v^2}&=\frac{5t^2}{v^3}\times\frac{v^2}{25t}\\ &=\frac{5t^2v^2}{25tv^3}\\ &=\frac{t}{5v}\end{aligned}$$

(b)

The common denominator is $(2x-3)(x+2)$:

$$\begin{aligned}\frac{6}{2x-3}-\frac{1}{x+2}&=\frac{6(x+2)-(2x-3)}{(2x-3)(x+2)}\\ &=\frac{6x+12-2x+3}{(2x-3)(x+2)}\\ &=\frac{4x+15}{(2x-3)(x+2)}\end{aligned}$$

(a)

Cancel $7$, one $p$ and one $q$ from top and bottom:

$$\begin{aligned}\frac{7p^2qr}{14pq^3}&=\frac{pr}{2q^2}\end{aligned}$$

(b)

Since $2-x=-(x-2)$, write $\dfrac{1}{2-x}=-\dfrac{1}{x-2}$. The common denominator is then $(x-2)^2$:

$$\begin{aligned}\frac{5}{(x-2)^2}+\frac{1}{2-x}&=\frac{5}{(x-2)^2}-\frac{1}{x-2}\\ &=\frac{5}{(x-2)^2}-\frac{x-2}{(x-2)^2}\\ &=\frac{5-x+2}{(x-2)^2}\\ &=\frac{7-x}{(x-2)^2}\end{aligned}$$

(a)

$$\begin{aligned}\frac{12a^3b}{7}\div\frac{ac^2}{b^2}&=\frac{12a^3b}{7}\times\frac{b^2}{ac^2}\\ &=\frac{12a^3b^3}{7ac^2}\\ &=\frac{12a^2b^3}{7c^2}\end{aligned}$$

(b)

The numerator is a difference of two squares. The denominator factorises as $(3a+1)(a+5)$, since $3a^2+15a+a+5=3a^2+16a+5$:

$$\begin{aligned}\frac{9a^2-1}{3a^2+16a+5}&=\frac{(3a+1)(3a-1)}{(3a+1)(a+5)}\\ &=\frac{3a-1}{a+5}\end{aligned}$$

(a)

$$\begin{aligned}\frac{9p^2y^2}{7a^4}\div\frac{3p^3}{2a^8y}&=\frac{9p^2y^2}{7a^4}\times\frac{2a^8y}{3p^3}\\ &=\frac{18a^8p^2y^3}{21a^4p^3}\\ &=\frac{6a^4y^3}{7p}\end{aligned}$$

(b)

Write $\dfrac{1}{5-x}=-\dfrac{1}{x-5}$, so the common denominator is $(x-5)(x+1)$:

$$\begin{aligned}\frac{3x}{(x-5)(x+1)}+\frac{1}{5-x}&=\frac{3x}{(x-5)(x+1)}-\frac{1}{x-5}\\ &=\frac{3x-(x+1)}{(x-5)(x+1)}\\ &=\frac{2x-1}{(x-5)(x+1)}\end{aligned}$$

(a)

$$\begin{aligned}\frac{a(a-b)}{2b}\div\frac{(a-b)^2}{2a}&=\frac{a(a-b)}{2b}\times\frac{2a}{(a-b)^2}\\ &=\frac{a^2}{b(a-b)}\end{aligned}$$

(b)

Factorise $x^2-9=(x-3)(x+3)$ and write $\dfrac{1}{3-x}=-\dfrac{1}{x-3}$:

$$\begin{aligned}\frac{3}{x^2-9}+\frac{1}{3-x}&=\frac{3}{(x-3)(x+3)}-\frac{1}{x-3}\\ &=\frac{3}{(x-3)(x+3)}-\frac{x+3}{(x-3)(x+3)}\\ &=\frac{3-x-3}{(x-3)(x+3)}\\ &=-\frac{x}{(x-3)(x+3)}\end{aligned}$$

(a)(i)

$$\begin{aligned}\frac{3ab}{5yz}\div\frac{18b^3}{25y^2z}&=\frac{3ab}{5yz}\times\frac{25y^2z}{18b^3}\\ &=\frac{75ab\,y^2z}{90b^3yz}\\ &=\frac{5ay}{6b^2}\end{aligned}$$

(a)(ii)

Factorise the numerator by grouping:

$$\begin{aligned}\frac{3ax+3ay-2bx-2by}{3a-2b}&=\frac{3a(x+y)-2b(x+y)}{3a-2b}\\ &=\frac{(3a-2b)(x+y)}{3a-2b}\\ &=x+y\end{aligned}$$

(b)

$x^2-6x+9=(x-3)^2$, so the common denominator is $(x-3)^2$:

$$\begin{aligned}\frac{6}{x-3}-\frac{12}{(x-3)^2}&=\frac{6(x-3)-12}{(x-3)^2}\\ &=\frac{6x-30}{(x-3)^2}\\ &=\frac{6(x-5)}{(x-3)^2}\end{aligned}$$

Factorise the top and the bottom. For the denominator, $(3x+2)(x-4)=3x^2-12x+2x-8=3x^2-10x-8$:

$$\begin{aligned}\frac{3x-12}{3x^2-10x-8}&=\frac{3(x-4)}{(3x+2)(x-4)}\\ &=\frac{3}{3x+2}\end{aligned}$$

(a)

Check: $(3x+5)(2x-3)=6x^2-9x+10x-15=6x^2+x-15$.

$$\begin{aligned}6x^2+x-15&=(3x+5)(2x-3)\end{aligned}$$

(b)

Use (a) for the numerator and the difference of two squares $4x^2-9=(2x+3)(2x-3)$ for the denominator:

$$\begin{aligned}\frac{6x^2+x-15}{7(4x^2-9)}&=\frac{(3x+5)(2x-3)}{7(2x+3)(2x-3)}\\ &=\frac{3x+5}{7(2x+3)}\end{aligned}$$

Factorise $2q-6=2(q-3)$ and $p^2-16=(p+4)(p-4)$, then multiply by the reciprocal:

$$\begin{aligned}\frac{2q-6}{p+4}\div\frac{q-3}{p^2-16}&=\frac{2(q-3)}{p+4}\times\frac{(p+4)(p-4)}{q-3}\\ &=2(p-4)\end{aligned}$$

Factorise each part: $5x-35=5(x-7)$, $x^2-5x=x(x-5)$ and $x^2-14x+49=(x-7)^2$.

$$\begin{aligned}\frac{x}{5x-35}\div\frac{x^2-5x}{x^2-14x+49}&=\frac{x}{5(x-7)}\times\frac{(x-7)^2}{x(x-5)}\\ &=\frac{x-7}{5(x-5)}\end{aligned}$$

(a)

$$\begin{aligned}\frac{3x-24}{x^2-8x}&=\frac{3(x-8)}{x(x-8)}\\ &=\frac{3}{x}\end{aligned}$$

(b)

Combine the fractions in the denominator first: $\dfrac{y}{x}-\dfrac{x}{y}=\dfrac{y^2-x^2}{xy}$.

$$\begin{aligned}\frac{y+x}{\dfrac{y}{x}-\dfrac{x}{y}}&=(y+x)\div\frac{y^2-x^2}{xy}\\ &=(y+x)\times\frac{xy}{(y+x)(y-x)}\\ &=\frac{xy}{y-x}\end{aligned}$$

Factorise $4x^2-12x+9=(2x-3)^2$, so the common denominator is $(2x-3)^2$:

$$\begin{aligned}\frac{2}{2x-3}-\frac{6-4x}{(2x-3)^2}&=\frac{2(2x-3)-(6-4x)}{(2x-3)^2}\\ &=\frac{4x-6-6+4x}{(2x-3)^2}\\ &=\frac{8x-12}{(2x-3)^2}\\ &=\frac{4(2x-3)}{(2x-3)^2}\\ &=\frac{4}{2x-3}\end{aligned}$$

Comparing with $\dfrac{a(x+b)}{(4-x)(x+1)}$: $a=5$ and $b=-1$.

Multiply both sides by $(x-3)$, where $x\neq 3$, to get a quadratic equation:

$$\begin{aligned}-2&=x(x-3)\\ -2&=x^2-3x\\ x^2-3x+2&=0\\ (x-1)(x-2)&=0\end{aligned}$$
$$\begin{aligned}x=1\quad&\text{or}\quad x=2\end{aligned}$$

Neither value makes the denominator zero, so both are solutions.

(a)

$$\begin{aligned}h&=3(2)+\frac12(5)^2\\ &=6+12.5\\ &=18.5\end{aligned}$$

(b)

$$\begin{aligned}h&=3g+\frac12t^2\\ 2h&=6g+t^2\\ t^2&=2h-6g\\ t&=\pm\sqrt{2h-6g}\end{aligned}$$

(a)

$$\begin{aligned}12&=\frac{n}{1-\frac13}\\ 12&=\frac{n}{\frac23}\\ n&=12\times\frac23\\ &=8\end{aligned}$$

(b)

$$\begin{aligned}S&=\frac{n}{1-r}\\ S(1-r)&=n\\ S-Sr&=n\\ Sr&=S-n\\ r&=\frac{S-n}{S}\end{aligned}$$

This can also be written as $r=1-\dfrac{n}{S}$.

Note: the school answer key prints $\pm\sqrt{\dfrac{A}{2b}}+3$, with the $+3$ outside the root. The root must cover the whole of $\dfrac{A}{2b}+3$, as the line $d^2=\dfrac{A}{2b}+3$ shows.

(a)

Use brackets round the negative value: $(-2)^2=4$ and $(-2)^3=-8$.

$$\begin{aligned}p&=\frac{3-(-2)^2+3}{2(3)(-2)^3}\\ &=\frac{3-4+3}{2(3)(-8)}\\ &=\frac{2}{-48}\\ &=-\frac{1}{24}\end{aligned}$$

(b)

Clear the fraction, then collect the terms in $q$ on one side and factorise:

$$\begin{aligned}2pqr^3&=q-r^2+3\\ 2pqr^3-q&=3-r^2\\ q(2pr^3-1)&=3-r^2\\ q&=\frac{3-r^2}{2pr^3-1}\end{aligned}$$

(a)

$$\begin{aligned}a&=\frac{4(-6)+3(-2)}{2-(-6)}\\ &=\frac{-24-6}{8}\\ &=-\frac{30}{8}\\ &=-3.75\end{aligned}$$

(b)

$$\begin{aligned}a(2-b)&=4b+3c\\ 2a-ab&=4b+3c\\ 2a-3c&=4b+ab\\ 2a-3c&=b(4+a)\\ b&=\frac{2a-3c}{4+a}\end{aligned}$$

(a)

$$\begin{aligned}3a&=\frac{5(9)-3(12)}{9+2(12)}\\ 3a&=\frac{9}{33}=\frac{3}{11}\\ a&=\frac{1}{11}\end{aligned}$$

(b)

$$\begin{aligned}3a(b+2c)&=5b-3c\\ 3ab+6ac&=5b-3c\\ 6ac+3c&=5b-3ab\\ 6ac+3c&=b(5-3a)\\ b&=\frac{6ac+3c}{5-3a}\end{aligned}$$

Write the left-hand side as a single fraction, then cross-multiply:

$$\begin{aligned}\frac{2}{n}-5&=\frac{4}{m}\\ \frac{2-5n}{n}&=\frac{4}{m}\\ m(2-5n)&=4n\\ m&=\frac{4n}{2-5n}\end{aligned}$$