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Question
Solve the following quadratic equations by factorization: \[\frac{16}{x} - 1 = \frac{15}{x + 1}; x \neq 0, - 1\]
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Solution
\[\frac{16}{x} - 1 = \frac{15}{x + 1}\]
\[ \Rightarrow \frac{16 - x}{x} = \frac{15}{x + 1}\]
\[ \Rightarrow \left( 16 - x \right)\left( x + 1 \right) = 15x\]
\[ \Rightarrow 16x + 16 - x^2 - x = 15x\]
\[ \Rightarrow - x^2 + 16 + 15x = 15x\]
\[ \Rightarrow - x^2 + 16 = 0\]
\[ \Rightarrow x^2 - 16 = 0\]
\[ \Rightarrow \left( x - 4 \right)\left( x + 4 \right) = 0\]
\[ \Rightarrow x - 4 = 0 \text { or } x + 4 = 0\]
\[ \Rightarrow x = 4\text { or } x = - 4\]
Hence, the factors are 4 and −4.
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