# 6.2.3 Algebraic Expressions (III), PT3 Practice

Question 11:
(a)
Expand: x (2 + y)

(b)

Solution:
(a)
x (2 + y) = 2x + xy

(b)
$\begin{array}{l}\frac{5}{6y}-\frac{3x-5}{12y}\\ =\frac{5×2}{6y×2}-\frac{3x-5}{12y}\\ =\frac{10-\left(3x-5\right)}{12y}\\ =\frac{10-3x+5}{12y}\\ =\frac{15-3x}{12y}\\ =\frac{\overline{)3}\left(5-x\right)}{{\overline{)12}}^{4}y}\\ =\frac{5-x}{4y}\end{array}$

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

# 6.2.1 Algebraic Expressions (III), PT3 Practice

6.2.1 Algebraic Expressions (III), PT3 Practice

Question 1:
(a)(i) Factorise 18a + 3
(a)(ii) Expand –3 (–y + 5)
(b) Express $\frac{5}{6y}-\frac{3x-5}{12y}$ as a single fraction in its simplest form.

Solution:
(a)(i) 18+ 3 = 3(6a + 1)

(a)(ii) –3 (–+ 5) = 3y – 15

(b)
$\begin{array}{l}\frac{5}{6y}-\frac{3x-5}{12y}=\frac{5×2}{6y×2}-\frac{\left(3x-5\right)}{12y}\\ \text{}=\frac{10-3x+5}{12y}\\ \text{}=\frac{15-3x}{12y}\\ \text{}=\frac{\overline{)3}\left(5-x\right)}{4\overline{)12}y}\\ \text{}=\frac{5-x}{4y}\end{array}$

Question 2:
(a) Expand:
(i) 3 (–a + c)
(ii) –5 (c)
(b) Factorise 4+ 2
(c) Simplify: $\frac{3x+6}{{x}^{2}-4}÷\frac{x+2}{x-2}$

Solution:
(a)(i) 3 (–+ c) = –3a + 3c

(a)(ii) –5 (c) = –5a + 5c

(c)
$\begin{array}{l}\frac{3x+6}{{x}^{2}-4}÷\frac{x+2}{x-2}=\frac{3\left(\overline{)x+2}\right)}{\left(x+2\right)\overline{)\left(x-2\right)}}×\frac{\overline{)x-2}}{\overline{)x+2}}\\ \text{}=\frac{3}{x+2}\end{array}$

Question 3:
(a) Factorise:
(i) 5m + 25
(ii) 7x + 9xy
(b) Simplify: $\frac{4x-12}{4y}÷\frac{{x}^{2}-9}{yz}$

Solution:
(a)(i)5m + 25 = 5 (m + 5)

(a)(ii)7x + 9xy = x (7 + 9y)

$\begin{array}{l}\text{(b)}\frac{4x-12}{4y}÷\frac{{x}^{2}-9}{yz}=\frac{\overline{)4}\overline{)\left(x-3\right)}}{\overline{)4}\overline{)y}}×\frac{\overline{)y}z}{\left(x+3\right)\overline{)\left(x-3\right)}}\\ \text{}=\frac{z}{x+3}\end{array}$

Question 4:
(a) Factorise completely:
4 – 100n2
(b) Express $\frac{4}{5x}-\frac{7-10y}{15x}$  as a single fraction in its simplest form.

Solution:
(a)
4 – 100n2= (2 + 10n)(2 – 10n)
(b)
$\begin{array}{l}\frac{4}{5x}-\frac{7-10y}{15x}=\frac{4×3}{5x×3}-\frac{\left(7-10y\right)}{15x}\\ \text{}=\frac{12-7+10y}{15x}\\ \text{}=\frac{5+10y}{15x}\\ \text{}=\frac{\overline{)5}\left(1+2y\right)}{3\overline{)15}x}\\ \text{}=\frac{1+2y}{3x}\end{array}$

Question 5:
(a) Simplify:
(m – 4n)(m + 4n) – m2
(b) Simplify: $\frac{3x-3y}{x+y}×\frac{2x+2y}{6x}$

Solution:
(a)
(m – 4n)(+ 4n) – m2
= m2 + 4mn – 4mn – 4n2m2
= 0

(b)
$\begin{array}{l}\frac{3x-3y}{x+y}×\frac{2x+2y}{6x}=\frac{\overline{)3}\left(x-y\right)}{\overline{)x+y}}×\frac{\overline{)2}\overline{)\left(x+y\right)}}{\overline{)6}x}\\ \text{}=\frac{x-y}{x}\end{array}$

# 6.1 Algebraic Expressions III

6.1 Algebraic Expressions III

6.1.1 Expansion
1. The product of an algebraic term and an algebraic expression:
• a(b + c) = ab + ac
•  a(bc) = ab ac
2. The product of an algebraic expression and another algebraic expression:
• (a + b) (c + d)  = ac + ad + bc + bd
• (a + b)2= a2 + 2ab + b2
• (ab)2= a2 – 2ab + b2
• (a + b) (ab) = a2b2

6.1.2 Factorization
1. Factorize algebraic expressions:
•  ab + ac = a(b + c)
• a2b2 = (a + b) (ab)
• a2+ 2ab + b2 = (a + b)2
• ac + ad + bc + bd = (a + b) (c + d)
2. Algebraic fractions are fractions where both the numerator and the denominator or either the numerator or the denominator are algebraic terms or algebraic expressions.
Example:
$\frac{3}{b},\frac{a}{7},\frac{a+b}{a},\frac{b}{a-b},\frac{a-b}{c+d}$

3(a) Simplification of algebraic fractions by using common factors:
$\begin{array}{l}•\text{}\frac{{}^{1}\overline{)4}\overline{)b}c}{{}^{{}_{{}^{3}}}\overline{)12}\overline{)b}d}=\frac{c}{3d}\\ •\text{}\frac{bm+bn}{em+en}=\frac{b\overline{)\left(m+n\right)}}{e\overline{)\left(m+n\right)}}\\ \text{}=\frac{b}{e}\end{array}$

3(b) Simplification of algebraic fractions by using difference of two squares:
$\begin{array}{l}\frac{{a}^{2}-{b}^{2}}{an+bn}=\frac{\overline{)\left(a+b\right)}\left(a-b\right)}{n\overline{)\left(a+b\right)}}\\ \text{}=\frac{a-b}{n}\end{array}$

6.1.3 Addition and Subtraction of Algebraic Fractions
1. If they have a common denominator:
$\frac{a}{m}+\frac{b}{m}=\frac{a+b}{m}$

2.
If they do not have a common denominator:
$\frac{a}{m}+\frac{b}{n}=\frac{an+bm}{nm}$

6.1.4 Multiplication and Division of Algebraic Fractions
1. Without simplification:
$\begin{array}{l}•\text{}\frac{a}{m}×\frac{b}{n}=\frac{ab}{mn}\\ •\text{}\frac{a}{m}÷\frac{b}{n}=\frac{a}{m}×\frac{n}{b}\\ \text{}=\frac{an}{bm}\end{array}$

2.
With simplification:
$\begin{array}{l}•\text{}\frac{a}{c\overline{)m}}×\frac{b\overline{)m}}{d}=\frac{ab}{cd}\\ •\text{}\frac{a}{cm}÷\frac{b}{dm}=\frac{a}{c\overline{)m}}×\frac{d\overline{)m}}{b}\\ \text{}=\frac{ad}{bc}\end{array}$