We think you are located in South Africa. Is this correct?

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

\[= \frac{a + 1}{ax}\]

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

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

Write the final answer

\[=1\]

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

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

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

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

Write the final answer

\[\frac{2x + 2 + {x}^{2} + x - {x}^{2}}{x\left(x - 1\right)\left(x + 1\right)} = \frac{3x + 2}{x\left(x - 1\right)\left(x + 1\right)}\]
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: \(x \ne 0 \text{ and } x \ne \pm 1\)