Question

Solve the following quadratic equations by factorization:

$3\left(\frac{7x+1}{5x-3}\right)-4\left(\frac{5x-3}{7x+1}\right)=11;x\ne \frac{3}{5},-\frac{1}{7}$

Answer 1

$$\begin{gathered} 3\left(\frac{7x+1}{5x-3}\right)-4\left(\frac{5x-3}{7x+1}\right)=11 \\ \Rightarrow \frac{3(7x+1{)}^2-4{\left(5x-3\right)}^2}{\left(5x-3\right)\left(7x+1\right)}=11 \\ \Rightarrow \frac{3\left(49x^2+1+14x\right)-4\left(25x^2+9-30x\right)}{35x^2+5x-21x-3}=11 \\ \Rightarrow \frac{147x^2+3+42x-100x^2-36+120x}{35x^2-16x-3}=11 \\ \Rightarrow 47x^2+162x-33=11\left(35x^2-16x-3\right) \\ \Rightarrow 47x^2+162x-33=385x^2-176x-33 \\ \Rightarrow 385x^2-47x^2-176x-162x-33+33=0 \\ \Rightarrow 338x^2-338x=0 \\ \Rightarrow x^2-x=0 \\ \Rightarrow x\left(x-1\right)=0 \\ \Rightarrow x=0\mathrm{or}x-1=0 \\ \Rightarrow x=0\mathrm{or}x=1 \end{gathered}$$

Hence, the factors are 0 and 1.