Question

Solve the following quadratic equations by factorization:

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

Answer 1

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

\[ \Rightarrow \frac{3(7x + 1 )^2 - 4 \left( 5x - 3 \right)^2}{\left( 5x - 3 \right)\left( 7x + 1 \right)} = 11\]

\[ \Rightarrow \frac{3\left( 49 x^2 + 1 + 14x \right) - 4\left( 25 x^2 + 9 - 30x \right)}{35 x^2 + 5x - 21x - 3} = 11\]

\[ \Rightarrow \frac{147 x^2 + 3 + 42x - 100 x^2 - 36 + 120x}{35 x^2 - 16x - 3} = 11\]

\[ \Rightarrow 47 x^2 + 162x - 33 = 11\left( 35 x^2 - 16x - 3 \right)\]

\[ \Rightarrow 47 x^2 + 162x - 33 = 385 x^2 - 176x - 33\]

\[ \Rightarrow 385 x^2 - 47 x^2 - 176x - 162x - 33 + 33 = 0\]

\[ \Rightarrow 338 x^2 - 338x = 0\]

\[ \Rightarrow x^2 - x = 0\]

\[ \Rightarrow x\left( x - 1 \right) = 0\]

\[ \Rightarrow x = 0 \text { or } x - 1 = 0\]

\[ \Rightarrow x = 0 \text { or } x = 1\]

Hence, the factors are 0 and 1.