Question

Solve the following quadratic equations by factorization:

$\frac{3}{x+1}+\frac{4}{x-1}=\frac{29}{4x-1};x\ne 1,-1,\frac{1}{4}$

Answer 1

$$\begin{gathered} \frac{3}{x+1}+\frac{4}{x-1}=\frac{29}{4x-1} \\ \Rightarrow \frac{3\left(x-1\right)+4\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}=\frac{29}{4x-1} \\ \Rightarrow \frac{3x-3+4x+4}{x^2-1}=\frac{29}{4x-1} \\ \Rightarrow \frac{7x+1}{x^2-1}=\frac{29}{4x-1} \\ \Rightarrow \left(7x+1\right)\left(4x-1\right)=29\left(x^2-1\right) \\ \Rightarrow 28x^2-7x+4x-1=29x^2-29 \\ \Rightarrow 29x^2-28x^2+3x-28=0 \\ \Rightarrow x^2+3x-28=0 \\ \Rightarrow x^2+7x-4x-28=0 \\ \Rightarrow x(x+7)-4(x+7)=0 \\ \Rightarrow (x-4)(x+7)=0 \\ \Rightarrow x-4=0\mathrm{or}x+7=0 \\ \Rightarrow x=4\mathrm{or}x=-7 \end{gathered}$$

Hence, the factors are 4 and −7.