1.8 Simplification of fractions (EMAQ)
We have studied procedures for working with fractions in earlier grades.
-
\(\dfrac{a}{b}\times \dfrac{c}{d}=\dfrac{ac}{bd} \qquad \left(b\ne 0; d\ne 0\right)\)
-
\(\dfrac{a}{b}+\dfrac{c}{b}=\dfrac{a+c}{b} \qquad \left(b\ne 0\right)\)
-
\(\dfrac{a}{b}\div\dfrac{c}{d}=\dfrac{a}{b}\times \dfrac{d}{c}=\dfrac{ad}{bc} \qquad \left(b\ne
0; c\ne 0; d\ne 0\right)\)
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
\[\frac{2x + 2 + {x}^{2} + x - {x}^{2}}{x\left(x - 1\right)\left(x + 1\right)} = \frac{3x + 2}{x\left(x
- 1\right)\left(x + 1\right)}\]
Textbook Exercise 1.10
\[\frac{3a}{15} = \frac{a}{5}\]
\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: \(a \ne -2\).
\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*}
\(\dfrac{b^2+10b+21}{3(b^2-9)}
\div
\dfrac{2b^2+14b}{30b^2-90b}\)
\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: \(p \ne 0\).
\(\dfrac{x - 3}{3} - \dfrac{x + 5}{4}\)
\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: \(a \ne \pm 2b\).
\(\dfrac{1}{x-2}\)
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: \(x \ne 0 \text{ and } x \ne \pm 1\)