Question

Solve for x:

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

Answer 1

We have:

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

Let

$$\frac{7x+1}{5x-3}=y$$

$$3y-4\left(\frac{1}{y}\right)=11$$

$$\Rightarrow 3y^2-4=11y$$

$$\Rightarrow 3y^2-11y-4=0$$

$$\Rightarrow 3y^2-12y+y-4=0$$

$$\Rightarrow 3y(y-4)+1(y-4)=0$$

$$\Rightarrow (y-4)(3y+1)=0$$

$$\Rightarrow y=4ory=-\frac{1}{3}$$

On putting the values of y in equation (1), we get:
For y = 4, we have:

$$\frac{7x+1}{5x-3}=4$$

$$\Rightarrow 7x+1=4(5x-3)$$

$$\Rightarrow 7x+1=20x-12$$

$$\Rightarrow 13x=13$$
$$\Rightarrow x=1$$

For $$y=-\frac{1}{3}$$

$$\frac{7x+1}{5x-3}=-\frac{1}{3}$$

$$\Rightarrow 3(7x+1)=-(5x-3)$$

$$\Rightarrow 21x+3=-5x+3$$

$$\Rightarrow 26x=0$$

$$\Rightarrow x=0$$

Hence, the required solutions for x are 0 and 1.