Question

Solve for x:

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

Answer 1

We have:

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

Let

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

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

\[\Rightarrow 3 y^2 - 4 = 11y\]

\[\Rightarrow 3 y^2 - 11y - 4 = 0\]

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

\[\Rightarrow 3y\left( y - 4 \right) + 1\left( y - 4 \right) = 0\]

\[\Rightarrow \left( y - 4 \right) \left( 3y + 1 \right) = 0\]

\[\Rightarrow y = 4 or y = - \frac{1}{3}\]

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 \[y = - \frac{1}{3}\]

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

\[\Rightarrow 3\left( 7x + 1 \right) = - \left( 5x - 3 \right)\]

\[\Rightarrow 21x + 3 = - 5x + 3\]

\[\Rightarrow 26x = 0\]

\[\Rightarrow x = 0\]

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