# 1.6 Fractions  (Page 3/2)

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## Simplification of fractions

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 in the numerator and a binomial in the denominator. You can apply the different factorisation methods to simplify the expression.

$\begin{array}{cccc}& \phantom{\rule{4pt}{0ex}}& \frac{{x}^{2}+3x}{x+3}\hfill & \\ & =& \frac{x\left(x+3\right)}{x+3}\hfill & \\ & =& x\hfill & \mathrm{provided}\phantom{\rule{2pt}{0ex}}x\ne -3\hfill \end{array}$

If $x$ were 3 then the denominator, $x-3$ , would be 0 and the fraction undefined.

Simplify: $\frac{2x-b+x-ab}{a{x}^{2}-abx}$

1. Use grouping for numerator and common factor for denominator in this example.

$\begin{array}{ccc}& =& \frac{\left(ax-ab\right)+\left(x-b\right)}{a{x}^{2}-abx}\hfill \\ & =& \frac{a\left(x-b\right)+\left(x-b\right)}{ax\left(x-b\right)}\hfill \\ & =& \frac{\left(x-b\right)\left(a+1\right)}{ax\left(x-b\right)}\hfill \end{array}$

$\begin{array}{ccc}& =& \frac{a+1}{ax}\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\hfill \end{array}$

Simplify: $\frac{{x}^{2}-x-2}{{x}^{2}-4}÷\frac{{x}^{2}+x}{{x}^{2}+2x}$

1. $\begin{array}{ccc}& =& \frac{\left(x+1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}÷\frac{x\left(x+1\right)}{x\left(x+2\right)}\hfill \end{array}$
2. $\begin{array}{ccc}& =& \frac{\left(x+1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}×\frac{x\left(x+2\right)}{x\left(x+1\right)}\hfill \end{array}$

$\begin{array}{ccc}& =& 1\hfill \end{array}$

## Simplification of fractions

1. Simplify:
 (a) $\frac{3a}{15}$ (b) $\frac{2a+10}{4}$ (c) $\frac{5a+20}{a+4}$ (d) $\frac{{a}^{2}-4a}{a-4}$ (e) $\frac{3{a}^{2}-9a}{2a-6}$ (f) $\frac{9a+27}{9a+18}$ (g) $\frac{6ab+2a}{2b}$ (h) $\frac{16{x}^{2}y-8xy}{12x-6}$ (i) $\frac{4xyp-8xp}{12xy}$ (j) $\frac{3a+9}{14}÷\frac{7a+21}{a+3}$ (k) $\frac{{a}^{2}-5a}{2a+10}÷\frac{3a+15}{4a}$ (l) $\frac{3xp+4p}{8p}÷\frac{12{p}^{2}}{3x+4}$ (m) $\frac{16}{2xp+4x}÷\frac{6{x}^{2}+8x}{12}$ (n) $\frac{24a-8}{12}÷\frac{9a-3}{6}$ (o) $\frac{{a}^{2}+2a}{5}÷\frac{2a+4}{20}$ (p) $\frac{{p}^{2}+pq}{7p}÷\frac{8p+8q}{21q}$ (q) $\frac{5ab-15b}{4a-12}÷\frac{6{b}^{2}}{a+b}$ (r) $\frac{{f}^{2}a-f{a}^{2}}{f-a}$
2. Simplify: $\frac{{x}^{2}-1}{3}×\frac{1}{x-1}-\frac{1}{2}$

Using the concepts learnt in simplification of fractions, we can now add and subtract simple fractions. To add or subtract fractions we note that we can only add or subtract fractions that have the same denominator. So we must first make all the denominators the same and then perform the addition or subtraction. This is called finding the lowest common denominator or multiple.

For example, if you wanted to add: $\frac{1}{2}$ and $\frac{3}{5}$ we would note that the lowest common denominator is 10. So we must multiply the first fraction by 5 and the second fraction by 2 to get both of these with the same denominator. Doing so gives: $\frac{5}{10}$ and $\frac{6}{10}$ . Now we can add the fractions. Doing so, we get $\frac{11}{10}$ .

Simplify the following expression: $\frac{x-2}{{x}^{2}-4}+\frac{{x}^{2}}{x-2}-\frac{{x}^{3}+x-4}{{x}^{2}-4}$

1. $\frac{x-2}{\left(x+2\right)\left(x-2\right)}+\frac{{x}^{2}}{x-2}-\frac{{x}^{3}+x-4}{\left(x+2\right)\left(x-2\right)}$
2. We make all the denominators the same so that we can add or subtract the fractions. The lowest common denominator is $\left(x-2\right)\left(x+2\right)$ .

$\frac{x-2}{\left(x+2\right)\left(x-2\right)}+\frac{\left({x}^{2}\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{{x}^{3}+x-4}{\left(x+2\right)\left(x-2\right)}$
3. Since the fractions all have the same denominator we can write them all as one fraction with the appropriate operator

$\frac{x-2+\left({x}^{2}\right)\left(x+2\right)-{x}^{3}+x-4}{\left(x+2\right)\left(x-2\right)}$
4. $\frac{2{x}^{2}+2x-6}{\left(x+2\right)\left(x-2\right)}$
5. $\frac{2\left({x}^{2}+x-3\right)}{\left(x+2\right)\left(x-2\right)}$

## Two interesting mathematical proofs

We can use the concepts learnt in this chapter to demonstrate two interesting mathematical proofs. The first proof states that ${n}^{2}+n$ is even for all $n\in {Z}$ . The second proof states that ${n}^{3}-n$ is divisible by 6 for all $n\in {Z}$ . Before we demonstrate that these two laws are true, we first need to note some other mathematical rules.

If we multiply an even number by an odd number, we get an even number. Similarly if we multiply an odd number by an even number we get an even number. Also, an even number multiplied by an even number is even and an odd number multiplied by an odd number is odd. This result is shown in the following table:

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