Topic Content:
- Equations with Fractions Involving Binomial Denominator
Worked Example 9.3.1:
Solve the following:
i. \( \frac{1}{x\: +\: 3} \:+\: \frac{5}{1}\scriptsize = 0 \)
ii. \( \frac{10}{x \: – \: 2} \scriptsize = 4 \)
iii. \( \frac{5}{2x \: – \: 7} = \frac{3}{x \: + \: 2} \)
iv. \( \frac{1}{3y \: – \: 7} = \frac{2}{4y \: – \: 8} \)
v. \( \frac{2}{1\: + \: 2x} = \frac{4}{1 \: + \: x} \)
Solution
i. \( \frac{1}{x\: +\: 3} \:+\: \frac{5}{1}\scriptsize = 0 \)
Find the L.C.M
\( \frac{1(1) \: + \:5(x \: + \:3)}{x \:+ \: 3} \scriptsize = 0 \) \( \frac{1 \:+ \:5x \: + \:15}{x \: + \: 3} \scriptsize = 0 \)Cross multiply
\( \scriptsize 1 \; + \; 5x \; + \; 15 = 0 \)Take like terms
\( \scriptsize 5x \: + \: 16 \: = 0 \) \( \scriptsize 5x = -16 \)Divide both sides by 5
\( \frac{5x}{5} = \frac{-16}{5} \) \( \frac{\not{5}x}{\not{5}} = \frac{-16}{5} \) \( \scriptsize x = \normalsize \frac{-16}{5} \)⇒ \( \scriptsize x = \scriptsize -3\frac{1}{5} \)
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