6.2.2 Algebraic Expressions (III), PT3 Practice

6.2.2 Algebraic Expressions (III), PT3 Practice

Question 6:

(a) Simplify each of the following:
$\begin{array}{l}\text{(i)}\frac{12mn}{32}\\ \text{(ii)}\frac{{x}^{2}-xy}{x}\end{array}$
(b) Express $\frac{1}{2q}-\frac{2p-7}{6q}$  as a single fraction in its simplest form.

Solution:

$\begin{array}{l}\text{(a)(i)}\frac{12mn}{32}=\frac{3mn}{8}\\ \text{(a)(ii)}\frac{{x}^{2}-xy}{x}=\frac{\overline{)x}\left(x-y\right)}{\overline{)x}}=x-y\end{array}$

(b)

$\begin{array}{l}\frac{1}{2q}-\frac{2p-7}{6q}=\frac{1×3}{2q×3}-\frac{\left(2p-7\right)}{6q}\\ \text{}=\frac{3-2p+7}{6q}\\ \text{}=\frac{10-2p}{6q}\\ \text{}=\frac{\overline{)2}\left(5-p\right)}{3\overline{)6}q}\\ \text{}=\frac{5-p}{3q}\end{array}$

Question 7:
(a) Factorise 2ae + 3af – 6de – 9df
(b)   Simplify $\frac{{a}^{2}-{b}^{2}}{{\left(a+b\right)}^{2}}$

Solution:
(a)
2ae + 3af – 6de – 9df = (2+ 3f ) – 3d (2e + 3f)
= (2+ 3f ) (a – 3d)

(b)
$\begin{array}{l}\frac{{a}^{2}-{b}^{2}}{{\left(a+b\right)}^{2}}=\frac{\overline{)\left(a+b\right)}\left(a-b\right)}{\overline{)\left(a+b\right)}\left(a+b\right)}\\ \text{}=\frac{a-b}{a+b}\end{array}$

Question 8:
(a) Factorise –8c2 – 12ac.
(b)   Simplify $\frac{ae+ad-2be-2bd}{{a}^{2}-4{b}^{2}}.$

Solution:
(a)
–8c2– 12ac
= –4c (2c + 3a)

(b)
$\begin{array}{l}\frac{ae+ad-2be-2bd}{{a}^{2}-4{b}^{2}}=\frac{a\left(e+d\right)-2b\left(e+d\right)}{\left(a+2b\right)\left(a-2b\right)}\\ \text{}=\frac{\left(e+d\right)\overline{)\left(a-2b\right)}}{\left(a+2b\right)\overline{)\left(a-2b\right)}}\\ \text{}=\frac{e+d}{a+2b}\end{array}$

Question 9:
(a) Factorise 12x2 – 27y2
(b)   Simplify $\frac{3{m}^{2}-10m+3}{{m}^{2}-9}÷\frac{3m-1}{m+3}.$

Solution:
(a)
12x– 27y2 = 3 (4x2 – 9y2)
= 3(2x – 3y) (2x + 3y)

(b)
$\begin{array}{l}\frac{3{m}^{2}-10m+3}{{m}^{2}-9}÷\frac{3m-1}{m+3}=\frac{\overline{)\left(3m-1\right)}\overline{)\left(m-3\right)}}{\overline{)\left(m+3\right)}\overline{)\left(m-3\right)}}×\frac{\overline{)m+3}}{\overline{)3m-1}}\\ \text{}=1\end{array}$

Question 10:

Solution:
$\begin{array}{l}\frac{8m+mn}{3m}÷\frac{{n}^{2}-64}{24}\\ =\frac{8m+mn}{3m}×\frac{24}{{n}^{2}-64}\\ =\frac{m\left(8+n\right)}{3m}×\frac{24}{{n}^{2}-{8}^{2}}\\ =\frac{\overline{)m}\overline{)\left(8+n\right)}}{\overline{)3}\overline{)m}}×\frac{{\overline{)24}}_{8}}{\left(n-8\right)\overline{)\left(n+8\right)}}\\ =\frac{8}{n-8}\end{array}$