Question

Solve the following quadratic equations by factorization:

$\frac{16}{x}-1=\frac{15}{x+1};x\ne 0,-1$

Answer 1

$$\begin{gathered} \frac{16}{x}-1=\frac{15}{x+1} \\ \Rightarrow \frac{16-x}{x}=\frac{15}{x+1} \\ \Rightarrow \left(16-x\right)\left(x+1\right)=15x \\ \Rightarrow 16x+16-x^2-x=15x \\ \Rightarrow -x^2+16+15x=15x \\ \Rightarrow -x^2+16=0 \\ \Rightarrow x^2-16=0 \\ \Rightarrow \left(x-4\right)\left(x+4\right)=0 \\ \Rightarrow x-4=0\mathrm{or}x+4=0 \\ \Rightarrow x=4\mathrm{or}x=-4 \end{gathered}$$

Hence, the factors are 4 and −4.