6.2.2 Algebraic Expressions (III), PT3 Practice

Posted on February 22, 2018 by user

6.2.2 Algebraic Expressions (III), PT3 Practice

Question 6:
(a) Simplify each of the following:

$\begin{array}{l} (i) \frac{12 m n}{32} \\ (ii) \frac{x^{2} - x y}{x} \end{array}$
(b) Express

$\frac{1}{2 q} - \frac{2 p - 7}{6 q}$ as a single fraction in its simplest form.

Solution:

$\begin{array}{l} (a)(i) \frac{12 m n}{32} = \frac{3 m n}{8} \\ (a)(ii) \frac{x^{2} - x y}{x} = \frac{x (x - y)}{x} = x - y \end{array}$

(b)

$\begin{array}{l} \frac{1}{2 q} - \frac{2 p - 7}{6 q} = \frac{1 \times 3}{2 q \times 3} - \frac{(2 p - 7)}{6 q} \\ = \frac{3 - 2 p + 7}{6 q} \\ = \frac{10 - 2 p}{6 q} \\ = \frac{2 (5 - p)}{36 q} \\ = \frac{5 - p}{3 q} \end{array}$

Question 7:

(a) Factorise 2ae + 3af – 6de – 9df

(b) Simplify

$\frac{a^{2} - b^{2}}{{(a + b)}^{2}}$

Solution:

(a)

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

= (2e + 3f ) (a – 3d)

(b)

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

Question 8:

(a) Factorise –8c² – 12ac.

(b) Simplify

$\frac{a e + a d - 2 b e - 2 b d}{a^{2} - 4 b^{2}} .$

Solution:

(a)

–8c²– 12ac

= –4c (2c + 3a)

(b)

$\begin{array}{l} \frac{a e + a d - 2 b e - 2 b d}{a^{2} - 4 b^{2}} = \frac{a (e + d) - 2 b (e + d)}{(a + 2 b) (a - 2 b)} \\ = \frac{(e + d) (a - 2 b)}{(a + 2 b) (a - 2 b)} \\ = \frac{e + d}{a + 2 b} \end{array}$

Question 9:

(a) Factorise 12x² – 27y²

(b) Simplify

$\frac{3 m^{2} - 10 m + 3}{m^{2} - 9} \div \frac{3 m - 1}{m + 3} .$

Solution:

(a)

12x²– 27y²= 3 (4x² – 9y²)

= 3(2x – 3y) (2x + 3y)

(b)

$\begin{array}{l} \frac{3 m^{2} - 10 m + 3}{m^{2} - 9} \div \frac{3 m - 1}{m + 3} = \frac{(3 m - 1) (m - 3)}{(m + 3) (m - 3)} \times \frac{m + 3}{3 m - 1} \\ = 1 \end{array}$

Question 10:

$Simplify: \frac{8 m + m n}{3 m} \div \frac{n^{2} - 64}{24}$

Solution:

$\begin{array}{l} \frac{8 m + m n}{3 m} \div \frac{n^{2} - 64}{24} \\ = \frac{8 m + m n}{3 m} \times \frac{24}{n^{2} - 64} \\ = \frac{m (8 + n)}{3 m} \times \frac{24}{n^{2} - 8^{2}} \\ = \frac{m (8 + n)}{3 m} \times \frac{24_{8}}{(n - 8) (n + 8)} \\ = \frac{8}{n - 8} \end{array}$