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## SS2: MATHEMATICS - 2ND TERM

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Lesson 4, Topic 3
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# Equations with Fractions

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#### Topic Content:

• Equations with Fractions

To solve equations involving fractions, the first step is to clear the fractions by multiplying every term on both sides of the equation by the LCM of the denominators.

### Example 4.3.3:

Solve the following equations:

(i) $$\scriptsize 2y \: + \: 3 \: – \: \normalsize \frac {y}{3} \scriptsize = y \:+ \:7$$

(ii) $$\frac{x \: – \: 3}{3} \: – \: \frac{x \: – \: 5}{5} \scriptsize = 2$$

(iii) $$\frac{2}{5} \scriptsize (x\: +\: 5) = \frac{1}{4} \scriptsize (5x \: – \:3)$$

(iv) $$\frac{2y \: – \: 1}{3} \: – \: \frac{3 \: – \: y}{2} = \frac {y}{4}$$

(WAEC)

Solution:

(i) $$\scriptsize 2y \: + \: 3 \: – \: \normalsize \frac {y}{3} \scriptsize = y \:+ \:7$$

Clear the fraction: multiply both sides by 3

$$\scriptsize 2y \: \times \: 3 \: + \: 3 \: \times \: 3\; – \: \normalsize \frac {y}{3} \scriptsize \: \times \: 3 \scriptsize = y \: \times \: 3 \:+ \:7 \: \times \: 3$$

i.e. 6y + 9 – y = 3y + 21

Collect like terms

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