Algebraic Fractions (Addition and Subtraction)

Back to: MATHEMATICS JSS 2

 

Welcome to Class !!

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In today’s Mathematics class, We will be discussing  Algebraic Fractions. We hope you enjoy the class!

 

Content

• Equivalent Fractions
• Addition and Subtraction of algebraic fraction
• Fractions with brackets

 

 

Equivalent Fraction

Equivalent fractions can be made by multiplying or dividing the numerator and denominator of a fraction by the same quantity.

For Example:

Multiplication

$\frac{3}{d} = \frac{3 \times 2b}{d \times 2b} = \frac{6b}{2bd}$

Division.

$\frac{4x}{6y} = \frac{4x ÷ 2}{6y ÷ 2} = \frac{2x}{3y}$

 

Complete the following :

(a) $\frac{3a}{2} = \frac{\left[\right]}{10}$   (b) $\frac{5ab}{12a} = \frac{\left[\right]}{12}$

Solution:

$$\begin{gathered} \mathbf{\left(}\mathit{a}\mathbf{\right)} \frac{3a}{2} = \frac{\left[\right]}{10} \\ Compare the two denomination \\ 2 \times 5 = 10 \\ The denominator of the first has been multiplied by 5 \\ The numerator can also be multiplied by 5 \\ \frac{3a}{2} = \frac{3a \times 5}{2 \times 5} = \frac{15a}{10} \end{gathered}$$

 

$$\begin{gathered} \mathbf{\left(}\mathit{b}\mathbf{\right)}\mathbf{ }\frac{5\mathrm{ab}}{12\mathrm{a}} = \frac{\left[\right]}{12} \\ The denominator of the first has been divided by a \\ Hence, divide the numerator by a. \\ \frac{5\mathrm{ab}}{12\mathrm{a}} = \frac{5\mathrm{ab} ÷ \mathrm{a}}{12\mathrm{a} ÷ \mathrm{a}} = \frac{5\mathrm{b}}{12} \end{gathered}$$

 

Evaluation: 

1. $\frac{8b}{5} = \frac{\left|\right|}{15}$
2. $\frac{9ah}{6ak} = \frac{3h}{\left[\right]}$
3. $\frac{\left[\right]}{8yz} = \frac{3x}{8y}$

 

 

Addition and Subtraction of algebraic fractions

Algebraic fractions must have common denominators before they can be added or subtracted.

Example: 

$$\begin{gathered} \mathbf{\left(}\mathit{a}\mathbf{\right)}\mathbf{ }\frac{5}{2a}+\frac{7}{2a} \mathbf{\left(}\mathit{b}\mathbf{\right)}\mathbf{ }\mathbf{ }\frac{4}{a} + b \\ \mathbf{\left(}\mathit{c}\mathbf{\right)}\mathbf{ }\frac{1}{u}\mathbf{ }\mathbf{+}\mathbf{ }\frac{\mathbf{1}}{\mathbf{v}}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{\left(}\mathit{d}\mathbf{\right)}\mathbf{ }\frac{5}{4c}\mathbf{ }\mathbf{–}\mathbf{ }\frac{4}{3d} \end{gathered}$$

 

Solution:

$\mathbf{\left(}\mathit{a}\mathbf{\right)}\mathbf{ }\frac{5}{2a}\mathbf{ }\mathbf{+}\mathbf{ }\frac{7}{2a}\mathbf{ }= \frac{5+7}{2a} = \frac{12 ÷ 2}{2a ÷ 2} = \frac{6}{a}$

 

$$\begin{gathered} \mathbf{\left(}\mathit{b}\mathbf{\right)}\mathbf{ }\frac{4}{a} + b = \frac{4}{a} + \frac{b}{1} \\ The L.C.M of a and 1 is a \\ \frac{4}{a} + \frac{b}{1} = \frac{4}{a} + \frac{ab}{a} \\ = \frac{4 + ab}{a} \end{gathered}$$

 

$$\begin{gathered} \left(\mathit{c}\right) \frac{1}{u}+\frac{1}{v} \\ The L.C.M of u and v is uv \\ Hence, \\ \frac{1}{u}+\frac{1}{v} = \frac{\left(1\times v\right) + \left(1\times u\right)}{uv} = \frac{1\times v}{uv} + \frac{1\times u}{uv} \\ = \frac{v }{uv}+ \frac{u}{uv} = \frac{v+u}{uv} \end{gathered}$$

 

$$\begin{gathered} \mathbf{\left(}\mathit{d}\mathbf{\right)}\mathbf{ }\frac{5}{4c} – \frac{4}{3d} \\ The L.C.M of 4c and 3d is 12cd \\ \frac{5}{4c} – \frac{4}{3d} = \frac{5\times 3d}{12cd} – \frac{4\times 4c}{12cd} \\ = \frac{15d}{12cd} – \frac{16c}{12cd} \\ = \frac{15d – 16c}{12cd} \end{gathered}$$

 

Evaluation

Simplify the following:

(a) $\frac{4}{3a}– \frac{1}{3\mathrm{a}}$

(b) $\frac{5}{4\mathrm{a}} – \frac{2}{3b}$

(c) $2b + \frac{3}{4}$

 

Fractions with brackets

Examples

Simplify:

(a) $\frac{x+3}{5} + \frac{4x – 2}{5}$

(b) $\frac{7a – 3}{6} – \frac{3a + 5}{4}$

 

Solution:

$$\begin{gathered} \mathbf{\left(}\mathit{a}\mathbf{\right)}\mathbf{ }\frac{x+3}{5} + \frac{4x – 2}{5} = \frac{\left(x + 3\right) + \left(4x – 2\right)}{5} \\ = \frac{x+3 + 4x –2 }{5} \\ = \frac{5x + 1}{5} \end{gathered}$$

 

$$\begin{gathered} \left(b\right) \frac{7a – 3}{6} – \frac{3a + 5}{4} \\ The L.C.M of 6 and 4 is 12 \\ \frac{7a – 3}{6} – \frac{3a + 5}{4} = \frac{2\left(7a–3\right)}{2\times 6} – \frac{3\left(3a +5\right)}{3\times 4} \\ = \frac{2(7a – 3)}{12} – \frac{3(3a+5)}{12} \\ = \frac{14a – 6 – 9a – 15}{12} \\ collecting like terms \\ = \frac{14a – 9a –6–15}{12 } = \frac{5a–21}{12} \end{gathered}$$

 

 

 

 

 

 

Evaluation

Simplify the following

(a)$\frac{2a – 3}{2} + \frac{a+4}{2}$

(b) $\frac{3x – 2d}{10} + \frac{2c – 3d}{15}$

(c) $\frac{2a + 3b}{a} + \frac{a–4b}{6}$

 

Weekend Assignment

1. $if \frac{3}{12a} = \frac{?}{4a} find ? \left(A\right) 0 \left(B\right) 1 \left(C\right) 3$
2. Simplify $\frac{1}{x}–\frac{1}{y} \left(A\right) \frac{x+y}{xy} \left(B\right) \frac{y–x}{xy} \left(C\right) \frac{x–y}{xy} \left(D\right) \frac{y+x}{xy}$
3. $if \frac{2c}{a} = \frac{6c^2}{?} find ? \left(A\right) 3c \left(B\right) 3ac \left(C\right) 3a \left(D\right) 2c$
4. $Simplify \frac{4x}{a} + \frac{8x}{a} \left(A\right) \frac{32x}{a} \left(B\right) \frac{4x}{3} \left(C\right) \frac{12x}{18} \left(D\right) \frac{4x}{a}$
5. $\frac{7}{8y} + \frac{9}{8y} \left(A\right) \frac{y}{2} \left(B\right) \frac{8y}{3} \left(C\right) \frac{y}{8} \left(D\right) \frac{2}{y}$

 

THEORY

1. $Simplify \frac{5}{4a} – \frac{2}{3b}$
2. $Simplify \frac{2a + 3b}{7} + \frac{9 – 4b}{6}$

 

 

 

We have come to the end of this class. We do hope you enjoyed the class?

Should you have any further question, feel free to ask in the comment section below and trust us to respond as soon as possible.

In our next class, we will be talking about Simple Algebraic Equations. We are very much eager to meet you there.

 

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