Question

Solve each of the following quadratic equations:
$3\left(\frac{7x+1}{5x-3}\right)-4\left(\frac{5x-3}{7x+1}\right)=11,x\ne \frac{3}{5},\frac{-1}{7}$

Answer 1

3(7x+1/5x−3)−4(5x−3/7x+1)=11

[3(7x+1)^2−4(5x−3)^2]/(5x−3)(7x+1)=11

3(49x^2+14x+1)−4(25x^2−30x+9)=11
(35x^2−16x−3)
147x^2+42x+3−100x^2+120x−36=385x2−176x−33

x2−x=0
x(x−1)=0
x=0 or x=1