We think you are located in United States. Is this correct?

# 1.8 Simplification of fractions

## 1.8 Simplification of fractions (EMAQ)

We have studied procedures for working with fractions in earlier grades.

1. $$\dfrac{a}{b}\times \dfrac{c}{d}=\dfrac{ac}{bd} \qquad \left(b\ne 0; d\ne 0\right)$$

2. $$\dfrac{a}{b}+\dfrac{c}{b}=\dfrac{a+c}{b} \qquad \left(b\ne 0\right)$$

3. $$\dfrac{a}{b}\div\dfrac{c}{d}=\dfrac{a}{b}\times \dfrac{d}{c}=\dfrac{ad}{bc} \qquad \left(b\ne 0; c\ne 0; d\ne 0\right)$$

Note: dividing by a fraction is the same as multiplying by the reciprocal of the fraction.

In some cases of simplifying an algebraic expression, the expression will be a fraction. For example,

$\frac{x^{2} + 3x}{x + 3}$

has a quadratic binomial in the numerator and a linear binomial in the denominator. We have to apply the different factorisation methods in order to factorise the numerator and the denominator before we can simplify the expression.

\begin{align*} \frac{{x}^{2} + 3x}{x + 3} & = \frac{x\left(x + 3\right)}{x + 3}\\ & =x \qquad \qquad \left(x\ne -3\right) \end{align*}

If $$x=-3$$ then the denominator, $$x + 3 = 0$$ and the fraction is undefined.

This video shows some examples of simplifying fractions.

Video: 2DNV

## Worked example 18: Simplifying fractions

Simplify: $\frac{ax-b+x-ab}{a{x}^{2}-abx}, \quad \left(x\ne 0;x\ne b\right)$

### Use grouping to factorise the numerator and take out the common factor $$ax$$ in the denominator

$\frac{\left(ax - ab\right) + \left(x - b\right)}{a{x}^{2} - abx} = \frac{a\left(x - b\right) + \left(x - b\right)}{ax\left(x - b\right)}$

### Take out common factor $$\left(x-b\right)$$ in the numerator

$=\frac{\left(x - b\right)\left(a + 1\right)}{ax\left(x - b\right)}$

### Cancel the common factor in the numerator and the denominator to give the final answer

$= \frac{a + 1}{ax}$
temp text

## Worked example 19: Simplifying fractions

Simplify: $\frac{{x}^{2}-x-2}{{x}^{2}-4}\div\frac{{x}^{2}+x}{{x}^{2}+2x}, \quad \left(x\ne 0;x\ne ±2\right)$

### Factorise the numerator and denominator

$= \frac{\left(x + 1\right)\left(x - 2\right)}{\left(x + 2\right)\left(x - 2\right)} \div \frac{x\left(x + 1\right)}{x\left(x + 2\right)}$

### Change the division sign and multiply by the reciprocal

$= \frac{\left(x + 1\right)\left(x - 2\right)}{\left(x + 2\right)\left(x - 2\right)}\times \frac{x\left(x + 2\right)}{x\left(x + 1\right)}$

$=1$
temp text

## Worked example 20: Simplifying fractions

Simplify: $\frac{x - 2}{{x}^{2} - 4} + \frac{{x}^{2}}{x - 2} - \frac{{x}^{3} + x - 4}{{x}^{2} - 4}, \quad \left(x\ne ±2\right)$

### Factorise the denominators

$\frac{x - 2}{\left(x + 2\right)\left(x - 2\right)} + \frac{{x}^{2}}{x - 2} - \frac{{x}^{3} + x - 4}{\left(x + 2\right)\left(x - 2\right)}$

### Make all denominators the same so that we can add or subtract the fractions

The lowest common denominator is $$\left(x-2\right)\left(x+2\right)$$.

$\frac{x - 2}{\left(x + 2\right)\left(x - 2\right)} + \frac{\left({x}^{2}\right)\left(x + 2\right)}{\left(x + 2\right)\left(x - 2\right)} - \frac{{x}^{3} + x - 4}{\left(x + 2\right)\left(x - 2\right)}$

### Write as one fraction

$\frac{x - 2 + \left({x}^{2}\right)\left(x + 2\right) - \left({x}^{3} + x - 4\right)}{\left(x + 2\right)\left(x - 2\right)}$

### Simplify

$\frac{x - 2 + {x}^{3} + 2{x}^{2} - {x}^{3} - x + 4}{\left(x + 2\right)\left(x - 2\right)} = \frac{2{x}^{2} + 2}{\left(x + 2\right)\left(x - 2\right)}$

### Take out the common factor and write the final answer

$\frac{2\left({x}^{2} + 1\right)}{\left(x + 2\right)\left(x - 2\right)}$
temp text

## Worked example 21: Simplifying fractions

Simplify: $\frac{2}{{x}^{2} - x} + \frac{{x}^{2} + x + 1}{{x}^{3} - 1} - \frac{x}{{x}^{2} - 1}, \quad \left(x\ne 0;x\ne ±1\right)$

### Factorise the numerator and denominator

$\frac{2}{x\left(x - 1\right)} + \frac{\left({x}^{2} + x + 1\right)}{\left(x - 1\right)\left({x}^{2} + x + 1\right)} - \frac{x}{\left(x - 1\right)\left(x + 1\right)}$

### Simplify and find the common denominator

$\frac{2\left(x + 1\right) + x\left(x + 1\right) - {x}^{2}}{x\left(x - 1\right)\left(x + 1\right)}$

$\frac{2x + 2 + {x}^{2} + x - {x}^{2}}{x\left(x - 1\right)\left(x + 1\right)} = \frac{3x + 2}{x\left(x - 1\right)\left(x + 1\right)}$
Textbook Exercise 1.10

Simplify (assume all denominators are non-zero):

$$\dfrac{3a}{15}$$

$\frac{3a}{15} = \frac{a}{5}$

$$\dfrac{2a + 10}{4}$$

\begin{align*} \frac{2a + 10}{4}& = \frac{2(a+5)}{4}\\ &=\frac{a + 5}{2} \end{align*}

$$\dfrac{5a + 20}{a + 4}$$

\begin{align*} \frac{5a + 20}{a + 4}& = \frac{5(a + 4)}{a + 4}\\ &=5 \end{align*}

$$\dfrac{a^{2} - 4a}{a - 4}$$

\begin{align*} \frac{a^{2} - 4a}{a - 4}& = \frac{a(a - 4)}{a - 4}\\ &=a \end{align*}

$$\dfrac{3a^{2} - 9a}{2a - 6}$$

\begin{align*} \frac{3a^{2} - 9a}{2a - 6}& = \frac{3a(a - 3)}{2(a - 3)}\\ &=\frac{3a}{2} \end{align*}

$$\dfrac{9a + 27}{9a + 18}$$

\begin{align*} \frac{9a + 27}{9a + 18}& = \frac{9(a + 3)}{9(a + 2)}\\ &=\frac{a + 3}{a + 2} \end{align*}

Note restriction: $$a \ne -2$$.

$$\dfrac{6ab + 2a}{2b}$$

\begin{align*} \frac{6ab + 2a}{2b}& = \frac{2a(3b + 1)}{2b}\\ &=\frac{a(3b + 1)}{b} \end{align*}

Note restriction: $$b \ne 0$$.

$$\dfrac{16x^{2}y - 8xy}{12x - 6}$$

\begin{align*} \frac{16x^{2}y - 8xy}{12x - 6}& = \frac{8xy(2x - 1)}{6(2x - 1)}\\ & = \frac{8xy}{6}\\ &=\frac{4xy}{3} \end{align*}

$$\dfrac{4xyp - 8xp}{12xy}$$

\begin{align*} \frac{4xyp - 8xp}{12xy} & = \frac{4xp(y - 2)}{12xy}\\ &=\frac{p(y - 2)}{3y} \end{align*}

Note restriction: $$y \ne 0$$.

$$\dfrac{9x^2 - 16}{6x - 8}$$
\begin{align*} \frac{9x^2 - 16}{6x - 8} &= \frac{(3x-4)(3x+4)}{2(3x-4)} \\ &= \frac{3x+4}{2} \end{align*}
$$\dfrac{b^2 - 81a^2}{18a-2b}$$
\begin{align*} \frac{b^2 - 81a^2}{18a-2b} &= \frac{(b-9)(b+9)}{2(9-b)} \\ &= -\frac{b+9}{2} \end{align*}
$$\dfrac{t^2 - s^2}{s^2 - 2st + t^2}$$
\begin{align*} \frac{t^2 - s^2}{s^2 - 2st + t^2} &= \frac{(t-s)(t+s)}{(s-t)^2} \\ &= \frac{t+s}{t-s} \end{align*}

Note restriction: $$s \ne t$$

$$\dfrac{x^2 - 2x - 15}{5x - 25}$$
\begin{align*} \frac{x^2 - 2x - 15}{5x - 25} &= \frac{(x-5)(x+3)}{5(x - 5)} \\ &= \frac{x+3}{5} \end{align*}
$$\dfrac{x^2 + 2x - 15}{x^2 + 8x + 15}$$
\begin{align*} \frac{x^2 + 2x - 15}{x^2 + 8x + 15} &= \frac{(x+5)(x-3)}{(x+3)(x+5)} \\ &= \frac{x-3}{x+3} \end{align*}

Note restriction: $$x \ne -3$$.

$$\dfrac{x^2 - x -6}{x^3 - 27}$$
\begin{align*} \frac{x^2 - x -6}{x^3 - 27} &= \frac{(x-3)(x+2)}{(x-3)(x^2 + 3x + 9)} \\ &= \frac{x+2}{x^2 + 3x + 9} \end{align*}
$$\dfrac{a^2 + 6a - 16}{a^3 - 8}$$
\begin{align*} \frac{a^2 + 6a - 16}{a^3 - 8} &= \frac{(a+8)(a-2)}{(a - 2)(a^2 + 2a + 4)} \\ &= \frac{a+8}{a^2 + 2a + 4} \end{align*}
$$\dfrac{a^2 - 4ab - 12b^2}{a^2 + 4ab + 4b^2}$$
\begin{align*} \frac{a^2 - 4ab - 12b^2}{a^2 + 4ab + 4b^2} &= \frac{(a-6b)(a +2b)}{(a+2b)^2} \\ &= \frac{a-6b}{a+2b} \end{align*}

Note restriction: $$a \ne -2b$$.

$$\dfrac{6a^2 - 7a - 3}{3ab + b}$$
\begin{align*} \frac{6a^2 - 7a - 3}{3ab + b} &= \frac{(2a-3)(3a+1)}{b(3a + 1)} \\ &= \frac{2a-3}{b} \end{align*}

Note restriction: $$b \ne 0$$.

$$\dfrac{2x^2 - x -1}{x^3 - x}$$
\begin{align*} \frac{2x^2 - x -1}{x^3 - x} &= \frac{(2x+1)(x-1)}{x(x-1)(x+1)} \\ &= \frac{2x+1}{x(x+1)} \end{align*}

Note restrictions: $$x \ne -1$$ and $$x \ne 0$$.

$$\dfrac{qz + qr + 16z+16r}{z+r}$$

\begin{align*} \frac{qz + qr + 16z + 16r}{(z + r)} &= \frac{q(z + r) + 16(z + r)}{(z + r)}\\ &= \frac{(z+r)(q+16)}{(z+r)}\\ &= q + 16 \end{align*}

$$\dfrac{pz - pq + 5z-5q}{z-q}$$

\begin{align*} \frac{pz - pq + 5z - 5q}{(z-q)} &= \frac{p(z-q) + 5(z-q)}{(z-q)}\\ &= \frac{(z-q)(p+5)}{(z-q)}\\ &= p+5 \end{align*}

$$\dfrac{hx - hg + 13x-13g}{x-g}$$

\begin{align*} \frac{hx - hg + 13x-13g}{(x-g)} &= \frac{h(x-g) + 13(x-g)}{(x-g)}\\ &= \frac{(x-g)(h+13)}{(x-g)}\\ &= h+13 \end{align*}

$$\dfrac{f^{2}a - fa^{2}}{f - a}$$

\begin{align*} \frac{f^{2}a - fa^{2}}{f - a} & = \frac{af(f - a)}{(f - a)}\\ & = af \end{align*}

Simplify (assume all denominators are non-zero):

$$\dfrac{b^2+10b+21}{3(b^2-9)} \div \dfrac{2b^2+14b}{30b^2-90b}$$

\begin{align*} \frac{b^2+10b+21}{3(b^2-9)} \div \frac{2b^2+14b}{30b^2-90b} &= \frac{b^2+10b+21}{3(b^2-9)} \times \frac{30b^2-90b}{2b^2+14b}\\ &= \frac{(b+7)(b+3)}{3(b-3)(b+3)} \times \frac{30b(b-3)}{2b(b+7)}\\ &= \frac{1}{3} \times \frac{30}{2}\\ &= 5 \end{align*}

$$\dfrac{x^2+17x+70}{5(x^2-100)} \div \dfrac{3x^2+21x}{45x^2-450x}$$

\begin{align*} \frac{x^2+17x+70}{5(x^2-100)} \div \frac{3x^2+21x}{45x^2-450x} &= \frac{x^2+17x+70}{5(x^2-100)} \times \frac{45x^2-450x}{3x^2+21x}\\ &= \frac{(x+7)(x+10)}{5(x-10)(x+10)} \times \frac{45x(x-10)}{3x(x+7)}\\ &= \frac{1}{5} \times \frac{45}{3}\\ &= 3 \end{align*}

$$\dfrac{z^2+17z+66}{3(z^2-121)} \div \dfrac{2z^2+12z}{24z^2-264z}$$

\begin{align*} \frac{z^2+17z+66}{3(z^2-121)} \div \frac{2z^2+12z}{24z^2-264z} &= \frac{z^2+17z+66}{3(z^2-121)} \times \frac{24z^2-264z}{2z^2+12z}\\ &= \frac{(z+6)(z+11)}{3(z-11)(z+11)} \times \frac{24z(z-11)}{2z(z+6)}\\ &= \frac{1}{3} \times \frac{24}{2}\\ &= 4 \end{align*}

$$\dfrac{3a + 9}{14} \div \dfrac{7a + 21}{a + 3}$$

\begin{align*} \frac{3a + 9}{14} \div \frac{7a + 21}{a + 3} & = \frac{3(a + 3)}{14} \div \frac{7(a + 3)}{a + 3}\\ & = \frac{3(a + 3)}{14} \div 7\\ & = \frac{3(a + 3)}{14} \times \frac{1}{7}\\ &= \frac{3(a + 3)}{98} \end{align*}

$$\dfrac{a^{2} - 5a}{2a + 10} \times \dfrac{4a}{3a + 15}$$

\begin{align*} \frac{{a}^{2} - 5a}{2a + 10} \times \frac{4a}{3a + 15} & = \frac{a(a - 5)}{2(a + 5)} \times \frac{4a}{3(a + 5)}\\ & = \frac{[a(a - 5)][4a]}{[2(a + 5)][3(a + 5)]} \\ & = \frac{4a^2(a - 5)}{6(a + 5)^2} \end{align*}

Note restriction: $$a \ne -5$$.

$$\dfrac{3xp + 4p}{8p} \div \dfrac{12p^{2}}{3x + 4}$$

\begin{align*} \frac{3xp + 4p}{8p} \div \frac{12p^{2}}{3x + 4} & = \frac{p(3x + 4)}{8p} \div \frac{12p^{2}}{3x + 4}\\ & = \frac{3x + 4}{8} \times \frac{3x + 4}{12p^{2}}\\ & = \frac{[3x + 4][3x + 4]}{[8][12p^{2}]} \\ & = \frac{(3x + 4)^{2}}{96p^2} \end{align*}

Note restriction: $$p \ne 0$$.

$$\dfrac{24a - 8}{12} \div \dfrac{9a - 3}{6}$$

\begin{align*} \frac{24a - 8}{12} \div \frac{9a - 3}{6} & = \frac{8(3a - 1)}{12} \div \frac{3(a - 1)}{6}\\ & = \frac{2(3a - 1)}{3} \times \frac{2}{a - 1}\\ & = \frac{[2(3x - 1)][2]}{[3][a - 1]} \\ & = \frac{4(3a - 1)}{3(a - 1)} \end{align*}

Note restriction: $$a \ne 1$$.

$$\dfrac{a^{2} + 2a}{5} \div \dfrac{2a + 4}{20}$$

\begin{align*} \frac{a^{2} + 2a}{5} \div \frac{2a + 4}{20} & = \frac{a(a + 2)}{5} \div \frac{2(a + 2)}{20}\\ & = \frac{a(a + 2)}{5} \times \frac{10}{a + 2}\\ & = \frac{[a(a + 2)][10]}{[5][a + 2]} \\ & = \frac{10a}{5} \\ & = 2a \end{align*}

$$\dfrac{p^{2} + pq}{7p} \times \dfrac{21q}{8p + 8q}$$

\begin{align*} \frac{p^{2} + pq}{7p} \times \frac{21q}{8p + 8q} & = \frac{p(p + q)}{7p} \times \frac{21q}{8(p + q)}\\ & = \frac{[p(p + q)][21q]}{[7p][8(p + q)]} \\ & = \frac{21pq}{56p} \\ & = \frac{3q}{8} \end{align*}

$$\dfrac{5ab - 15b}{4a - 12} \div \dfrac{6b^{2}}{a + b}$$

\begin{align*} \frac{5ab - 15b}{4a - 12} \div \frac{6b^{2}}{a + b} & = \frac{5b(a - 3)}{4(a - 3)} \div \frac{6b^{2}}{a + b}\\ & = \frac{5b}{4} \times \frac{a + b}{6b^{2}} \\ & = \frac{[5b][a + b]}{[4][6b^{2}]} \\ & = \frac{30b^{3}}{4(a + b)} \end{align*}

Note restriction: $$a \ne -b$$.

$$\dfrac{16 - x^2}{x^2 - x - 12} \times \dfrac{x+3}{x+4}$$
\begin{align*} \frac{16 - x^2}{x^2 - x - 12} \times \frac{x+3}{x+4} &=\frac{(4-x)(4+x)}{(x-4)(x+3)} \times \frac{x+3}{x+4} \\ &= - 1 \end{align*}
$$\dfrac{a^3 + b^3}{a^3} \times \dfrac{5a + 5b}{a^2 + 2ab + b^2}$$
\begin{align*} \frac{a^3 + b^3}{a^3} \times \frac{5a + 5b}{a^2 + 2ab + b^2} &= \frac{(a+b)(a^2 -ab + b^2)}{a^3} \times \frac{5(a + b)}{(a+b)^2} \\ &= \frac{a^2 -ab + b^2}{a^3} \times 5 \\ &= \frac{5(a^2 -ab + b^2)}{a^3} \end{align*}

Note restrictions: $$a \ne \pm 0$$.

$$\dfrac{a-4}{a + 5a + 4} \times \dfrac{a^2 + 2a + 1}{a^2 - 3a -4}$$
\begin{align*} \frac{a-4}{a + 5a + 4} \times \frac{a^2 + 2a + 1}{a^2 - 3a -4} &= \frac{a-4}{(a+4)(a+1)} \times \frac{(a+1)^2}{(a-4)(a+1)} \\ &= \frac{1}{a+4} \end{align*}

Note restrictions: $$a \ne -4$$.

$$\dfrac{3x+2}{x^2 - 6x + 8} \times \dfrac{x-2}{3x^2 + 8x + 4}$$
\begin{align*} \frac{3x+2}{x^2 - 6x + 8} \times \frac{x-2}{3x^2 + 8x + 4} &= \frac{3x+2}{(x-4)(x-2)} \times \frac{x-2}{(3x+2)(x+2)} \\ &= \frac{1}{(x-4)(x+2)} \end{align*}

Note restrictions: $$x \ne 4$$ and $$x \ne -2$$.

$$\dfrac{a^2 - 2a + 8}{a^2 + 6a + 8} \times \dfrac{a^2 + a - 12}{3} - \dfrac{3}{2}$$
\begin{align*} \frac{a^2 - 2a + 8}{a^2 + 6a + 8} \times \frac{a^2 + a - 12}{3} - \frac{3}{2} &= \frac{(a-4)(a+2)}{(a+2)(a+4)} \times \frac{(a+4)(a-3)}{3} - \frac{3}{2} \\ &= \frac{(a-4)(a-3)}{3} - \frac{3}{2} \\ &= \frac{2(a-4)(a-3) - 9}{6} \\ &= \frac{2(a^2 - 7a + 12) - 9}{6} \\ &= \frac{2a^2 - 14a + 15}{6} \end{align*}
$$\dfrac{4x^2 -1}{3x^2 + 10x + 3} \div \dfrac{6x^2 + 5x + 1}{4x^2 + 7x - 3} \times \dfrac{9x^2 + 6x + 1}{8x^2 - 6x + 1}$$
\begin{align*} & \frac{4x^2 -1}{3x^2 + 10x + 3} \div \frac{6x^2 + 5x + 1}{4x^2 + 7x - 3} \times \frac{9x^2 + 6x + 1}{8x^2 - 6x + 1} \\ &= \frac{(2x-1)(2x+1)}{(x+3)(3x+1)} \times \frac{(x+3)(4x-1)}{(2x+1)(3x+1)} \times \frac{(3x+1)^2}{(2x-1)(4x-1)} \\ &= 1 \end{align*}
$$\dfrac{x+4}{3} - \dfrac{x-2}{2}$$
\begin{align*} \frac{x+4}{3} - \frac{x-2}{2} &= \frac{2(x+4) - 3(x-2)}{6} \\ &= \frac{2x+8 - 3x +6}{6} \\ &= \frac{14 - x}{6} \end{align*}

$$\dfrac{p^{3} + q^{3}}{p^{2}} \times \dfrac{3p - 3q}{p^{2} - q^{2}}$$

\begin{align*} \frac{p^{3} + q^{3}}{p^{2}} \times \frac{3p - 3q}{p^{2} - q^{2}} & = \frac{(p + q)(p^2 - pq + q^2)}{p^2} \times \frac{3(p - q)}{(p - q)(p + q)} \\ & = \frac{(p + q)(p^2 - pq + q^2)}{p^2} \times \frac{3}{p + q} \\ & = \frac{3(p^2 - pq + q^2)}{p^2} \end{align*}

Note restriction: $$p \ne 0$$.

Simplify (assume all denominators are non-zero):

$$\dfrac{x - 3}{3} - \dfrac{x + 5}{4}$$
\begin{align*} \frac{x-3}{3} - \frac{x+5}{4} &= \frac{4(x-3) - 3(x+5)}{12} \\ &= \frac{4x-12 - 3x-15}{12} \\ &= \frac{x - 27}{12} \end{align*}
$$\dfrac{2x - 4}{9} - \dfrac{x - 3}{4} + 1$$
\begin{align*} \frac{2x - 4}{9} - \frac{x-3}{4} + 1 &= \frac{4(2x - 4) - 9(x-3) + 36}{36} \\ &= \frac{8x - 16 - 9x + 27 + 36}{36}\\ &= \frac{47-x}{36} \end{align*}
$$1 + \dfrac{3x - 4}{4} - \dfrac{x + 2}{3}$$
\begin{align*} 1 + \frac{3x - 4}{4} - \frac{x + 2}{3} &= \frac{12 + 3(3x - 4) - 4(x+2)}{12} \\ &= \frac{12 + 9x - 12 - 4x - 8}{12} \\ &= \frac{5x -8}{12} \end{align*}

$$\dfrac{11}{a + 11} + \dfrac{8}{a - 8}$$

\begin{align*} \frac{11}{a + 11} + \frac{8}{a-8} & = \frac{11(a-8) + 8(a+11)}{(a+11)(a-8)}\\ &= \frac{11a-88 + 8a+88}{(a+11)(a-8)}\\ &= \frac{19a}{(a+11)(a-8)} \end{align*}

Note restrictions: $$a \ne -11$$ and $$a \ne 8$$.

$$\dfrac{12}{x-12} - \dfrac{6}{x-6}$$

\begin{align*} \frac{12}{x-12} - \frac{6}{x-6} &= \frac{12(x-6) - 6(x-12)}{(x-12)(x-6)}\\ &= \frac{12x-72 - 6x+72}{(x-12)(x-6)}\\ &= \frac{6x}{(x-12)(x-6)} \end{align*}

Note restriction: $$x \ne 12$$ and $$x \ne 6$$.

$$\dfrac{12}{r+12} + \dfrac{8}{r-8}$$

\begin{align*} \frac{12}{r+12} + \frac{8}{r-8} &= \frac{12(r-8) + 8(r+12)}{(r+12)(r-8)}\\ &= \frac{12r-96 + 8r+96}{(r+12)(r-8)}\\ &= \frac{20r}{(r+12)(r-8)} \end{align*}

Note restriction: $$r \ne -12$$ and $$r \ne 8$$.

$$\dfrac{2}{xy} + \dfrac{4}{xz} + \dfrac{3}{yz}$$

\begin{align*} \frac{2}{xy} + \frac{4}{xz} + \frac{3}{yz} & = \frac{2z}{xyz} + \frac{4y}{xyz} + \frac{3x}{xyz}\\ & = \frac{2z + 4y + 3x}{xyz} \end{align*}

Note restrictions: $$x \ne 0$$; $$y \ne 0$$ and $$z \ne 0$$.

$$\dfrac{5}{t - 2} - \dfrac{1}{t - 3}$$

\begin{align*} \frac{5}{t - 2} - \frac{1}{t - 3} & = \frac{(5)(t - 3)}{(t - 3)(t - 2)} - \frac{1(t - 2)}{(t - 2)(t - 3)}\\ & = \frac{5(t - 3) - (t - 3)}{(t - 2)(t - 3)} \\ & = \frac{5t - 15 - t + 3}{(t - 2)(t - 3)} \\ & = \frac{4t - 12}{(t - 2)(t - 3)} \end{align*}

Note restrictions: $$t \ne 2$$ and $$t \ne 3$$.

$$\dfrac{k + 2}{k^{2} + 2} - \dfrac{1}{k + 2}$$

\begin{align*} \frac{k + 2}{k^{2} + 2} - \frac{1}{k + 2} & = \frac{(k + 2)(k + 2)}{(k^{2} + 2)(k + 2)} - \frac{1(k^{2} + 2)}{(k^{2} + 2)(k + 2)}\\ & = \frac{(k + 2)^{2} - (k^{2} + 2)}{(k^{2} + 2)(k + 2)} \\ & = \frac{k^{2} + 4k + 4 - k^{2} - 2}{(k^{2} + 2)(k + 2)} \\ & = \frac{4k + 2}{(k^{2} + 2)(k + 2)} \\ & = \frac{2(k + 2)}{(k^{2} + 2)(k + 2)} \end{align*}

Note restrictions: $$k \ne -2$$ and $$k^2 \ne \pm \sqrt{2}$$.

$$\dfrac{t + 2}{3q} + \dfrac{t + 1}{2q}$$

\begin{align*} \frac{t + 2}{3q} + \frac{t + 1}{2q} & = \frac{(t + 2)(2q)}{(3q)(2q)} + \frac{(t + 1)(3q)}{(3q)(2q)}\\ & = \frac{(2tq + 4q) + (3tq + 3q)}{6q^{2}} \\ & = \frac{q(5t + 7)}{6q^{2}}\\ & = \frac{5t + 7}{6q} \end{align*}

Note restriction: $$q \ne 0$$.

$$\dfrac{3}{p^{2} - 4} + \dfrac{2}{(p - 2)^{2}}$$

\begin{align*} \frac{3}{p^{2} - 4} + \frac{2}{(p - 2)^{2}} & = \frac{3(p - 2)^{2}}{(p^{2} - 4)(p - 2)^{2}} + \frac{2(p^{2} - 4)}{(p^{2} - 4)(p - 2)^{2}}\\ & = \frac{3(p - 2)(p - 2) + 2(p - 2)(p + 2)}{(p + 2)(p - 2)^{3}} \\ & = \frac{[p - 2][3(p - 2) + 2(p + 2)]}{(p + 2)(p - 2)^{3}}\\ & = \frac{3p - 6 + 2p + 4}{(p + 2)(p - 2)^{2}}\\ & = \frac{5p - 2}{(p + 2)(p - 2)^{2}} \end{align*}

Note restriction: $$p \ne \pm 2$$.

$$\dfrac{x}{x + y} + \dfrac{x^{2}}{y^{2} - x^{2}}$$

\begin{align*} \frac{x}{x + y} + \frac{x^{2}}{y^{2} - x^{2}} & = \frac{x}{x + y} + \frac{x^{2}}{(x + y)(x - y)}\\ & = \frac{x(x - y) + x^{2}}{(x + y)(x - y)} \\ & = \frac{x^{2} - xy + x^{2}}{(x + y)(x - y)}\\ & = \frac{2x^{2} - xy}{(x + y)(x - y)} \end{align*}

Note restriction: $$x \ne \pm y$$.

$$\dfrac{1}{m + n} + \dfrac{3mn}{m^{3} + n^{3}}$$

\begin{align*} \frac{1}{m + n} + \frac{3mn}{m^{3} + n^{3}} & = \frac{1}{m + n} + \frac{3mn}{(m + n)(m^{2} - mn + n^{2})}\\ & = \frac{1(m^{2} - mn + n^{2}) + 3mn}{(m + n)(m^{2} - mn + n^{2})} \\ & = \frac{m^{2} + 2mn + n^{2}}{(m + n)(m^{2} - mn + n^{2})}\\ & = \frac{m + n}{m^{2} - mn + n^{2}} \end{align*}

$$\dfrac{h}{h^{3} - f^{3}} - \dfrac{1}{h^{2} + hf + f^{2}}$$

\begin{align*} \frac{h}{h^{3} - f^{3}} - \frac{1}{h^{2} + hf + f^{2}} & = \frac{h}{(h - f)(h^{2} + hf + f^{2})} - \frac{1}{h^{2} + hf + f^{2}}\\ & = \frac{h - h + f}{(h + f)(h^{2} + hf + f^{2})} \\ & = \frac{f}{(h + f)(h^{2} + hf + f^{2})} \end{align*}

$$\dfrac{x^{2} - 1}{3} \times \dfrac{1}{x - 1} - \dfrac{1}{2}$$

\begin{align*} \frac{x^{2} - 1}{3} \times \frac{1}{x - 1} - \frac{1}{2} & = \frac{(x^{2} - 1)(1)}{(3)(x - 1)} - \frac{1}{2}\\ & = \frac{x^{2} - 1}{3x - 3} - \frac{1}{2} \\ & = \frac{(x^{2} - 1)(2)}{2(3x - 3)} - \frac{3x - 3}{2(3x - 3)} \\ & = \frac{2x^{2} - 2 - 3x + 3}{6x - 6}\\ & = \frac{(x - 1)(2x - 1)}{6(x - 1)} \\ & = \frac{2x - 1}{6} \end{align*}

$$\dfrac{x^{2} - 2x + 1}{(x - 1)^{3}} - \dfrac{x^{2} + x + 1}{x^{3} - 1}$$

\begin{align*} \frac{x^2 - 2x + 1}{(x - 1)^3} - \frac{x^2 + x + 1}{x^3 - 1} & = \frac{(x - 1)^2}{(x - 1)^3} - \frac{x^2 + x + 1}{x^3 - 1} \\ & = \frac{1}{(x-1)} - \frac{x^2 + x + 1}{(x - 1)(x^2 + x + 1)}\\ & = \frac{1}{(x - 1)} - \frac{1}{(x - 1)} \\ & = 0 \end{align*}

$$\dfrac{1}{(x - 1)^{2}} - \dfrac{2x}{x^{3} - 1}$$

\begin{align*} \frac{1}{(x - 1)^{2}} - \frac{2x}{x^{3} - 1} & = \frac{1}{(x - 1)^2} - \frac{2x}{(x - 1)(x^2 + x + 1)}\\ & = \frac{x^2 + x + 1 - 2x(x - 1)}{(x - 1)^2(x^2 + x + 1)}\\ & = \frac{x^2 + x + 1 - 2x^2 + 2x}{(x - 1)^2(x^2 + x + 1)}\\ & = \frac{-x^2 + 3x + 1}{(x - 1)^2(x^2 + x + 1)} \end{align*}
$$\dfrac{t^2 + 2t - 8}{t^2 + t - 6} + \dfrac{1}{t^2 - 9} +\dfrac{t + 1}{t - 3}$$
\begin{align*} \frac{t^2 + 2t - 8}{t^2 + t - 6} + \frac{1}{t^2- 9} +\frac{t+1}{t-3} &= \frac{(t+4)(t-2)}{(t+3)(t-2)} + \frac{1}{(t-3)(t+3)} +\frac{t+1}{t-3} \\ &= \frac{t+4}{t+3} + \frac{1}{(t-3)(t+3)} +\frac{t+1}{t-3} \\ &= \frac{(t-3)(t+4) + 1 + (t+1)(t+3)}{(t-3)(t+3)} \\ &= \frac{t^2 + t - 12 + 1 + t^2 + 4t + 3}{(t-3)(t+3)} \\ &= \frac{2t^2 + 5t - 8}{(t-3)(t+3)} \\ &= \frac{2t^2 + 5t - 8}{t^2 - 9} \end{align*}

Note restriction: $$t \ne \pm 3$$.

$$\dfrac{x^2 - 3x + 9}{x^3 + 27} + \dfrac{x-2}{x^2 + 4x + 3} - \dfrac{1}{x-2}$$
\begin{align*} \frac{x^2 - 3x + 9}{x^3 + 27} + \frac{x-2}{x^2 + 4x + 3} - \frac{1}{x-2} &= \frac{x^2 - 3x + 9}{(x+3)(x^2 - 3x + 9)} + \frac{x-2}{(x+3)(x+1)} - \frac{1}{x-2} \\ &= \frac{(x+1)(x-2) + (x-2)^2 - (x+3)(x+1)}{(x+3)(x+1)(x-2)} \\ &= \frac{x^2 - x -2 + x^2 - 4x + 4 - x^2 - 4x - 3}{(x+3)(x+1)(x-2)} \\ &= \frac{x^2 -9x - 1}{(x+3)(x+1)(x-2)} \end{align*}

Note restrictions: $$x \ne -3$$; $$x \ne -1$$ and $$x \ne 2$$.

$$\dfrac{1}{a^2 - 4ab + 4b^2} + \dfrac{a^2 + 2ab + b^2}{a^3 - 8b^3} - \dfrac{1}{a^2 - 4b^2}$$

\begin{align*} & \frac{1}{a^2 - 4ab + 4b^2} + \frac{a^2 + 2ab + b^2}{a^3 - 8b^3} - \frac{1}{a^2 - 4b^2} \\ & = \frac{1}{(a - 2b)(a - 2b)} + \frac{a^2 + 2ab + 4b^2}{(a - 2b)(a^2 + 2ab + 4b^2} - \frac{1}{(a - 2b)(a + 2b)} \\ & = \frac{(a + 2b) + (a - 2b)(a + 2b) - (a - 2b)}{(a - 2b)^{2}(a + 2b)} \\ & = \frac{a + 2b + a^2 - 4b^2 - a + 2b}{(a - 2b)^{2}(a + 2b)} \\ & = \frac{a^2 + 4b - 4b^2}{(a - 2b)^{2}(a + 2b)} \end{align*}

Note restriction: $$a \ne \pm 2b$$.

What are the restrictions in the following:

$$\dfrac{1}{x-2}$$

We need to find the value of $$x$$ that will make the denominator equal to $$\text{0}$$. Therefore:

\begin{align*} x - 2 & \ne 0 \\ x & \ne 2 \end{align*}
$$\dfrac{3x -9}{4x + 4}$$

First simplify the fraction:

$\frac{3x -9}{4x + 4} = \frac{3(x -1)}{4(x + 1)}$

Now we can determine the restriction:

\begin{align*} 4(x + 1) & \ne 0 \\ x + 1 & \ne 0 \\ x &\ne -1 \end{align*}
$$\dfrac{3}{x} - \dfrac{1}{x^2 - 1}$$

First simplify the fraction:

$\frac{3}{x} - \frac{1}{x^2 - 1} = \frac{3}{x} - \frac{1}{(x - 1)(x + 1)}$

Now we can determine the restrictions. There are three restrictions in this case:

\begin{align*} x & \ne 0 \\ x - 1 & \ne 0 \\ x + 1 & \ne 0 \end{align*}

Therefore: $$x \ne 0 \text{ and } x \ne \pm 1$$