Question

Solve the following quadratic equations by factorization: \[\frac{16}{x} - 1 = \frac{15}{x + 1}; x \neq 0, - 1\]

Answer 1

\[\frac{16}{x} - 1 = \frac{15}{x + 1}\]

\[ \Rightarrow \frac{16 - x}{x} = \frac{15}{x + 1}\]

\[ \Rightarrow \left( 16 - x \right)\left( x + 1 \right) = 15x\]

\[ \Rightarrow 16x + 16 - x^2 - x = 15x\]

\[ \Rightarrow - x^2 + 16 + 15x = 15x\]

\[ \Rightarrow - x^2 + 16 = 0\]

\[ \Rightarrow x^2 - 16 = 0\]

\[ \Rightarrow \left( x - 4 \right)\left( x + 4 \right) = 0\]

\[ \Rightarrow x - 4 = 0 \text { or } x + 4 = 0\]

\[ \Rightarrow x = 4\text {  or } x = - 4\]

Hence, the factors are 4 and −4.