Solve the following quadratic equation by factorization, the roots are $0, 1$
$3 (\frac{7 x + 1}{5 x - 3}) - 4 (\frac{5 x - 3}{7 x + 1}) = 11; x \neq \frac{3}{5}, - \frac{1}{7}$

True

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False

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Solution

The correct option is A True
$3 (\frac{7 x + 1}{5 x - 3}) - 4 (\frac{5 x - 3}{7 x + 1}) = 11$

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

$\Rightarrow$ $3 p - \frac{4}{p} = 11$
$\Rightarrow$ $3 p^{2} - 4 = 11 p$
$\Rightarrow$ $3 p^{2} - 11 p - 4 = 0$
$\Rightarrow$ $3 p^{2} - 12 p + p - 4 = 0$
$\Rightarrow$ $3 p (p - 4) + 1 (p - 4) = 0$
$\Rightarrow$ $(p - 4) (3 p + 1) = 0$
$\Rightarrow$ $p - 4 = 0$ and $3 p + 1 = 0$
$\Rightarrow$ $p = 4$ and $p = \frac{- 1}{3}$
$\Rightarrow$ $\frac{7 x + 1}{5 x - 3} = 4$ and $\frac{7 x + 1}{5 x - 3} = \frac{- 1}{3}$

$\Rightarrow$ $7 x + 1 = 20 x - 12$ and $21 x + 3 = - 5 x + 3$
$\Rightarrow$ $13 x = 13$ and $26 x = 0$
$∴$ $x = 1$ and $x = 0$

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

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

- 1

- 5

\frac{x + 3}{x + 2} = \frac{3 x - 7}{2 x - 3}

3 (\frac{7 x + 1}{5 x - 3}) - 4 (\frac{5 x - 3}{7 x + 1}) = 11, x \neq \frac{3}{5}, - \frac{1}{7}

4, - 7

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