Question

Solve the following quadratic equations by factorization:

\[3\left( \frac{3x - 1}{2x + 3} \right) - 2\left( \frac{2x + 3}{3x - 1} \right) = 5; x \neq \frac{1}{3}, - \frac{3}{2}\]

Answer 1

\[3\left( \frac{3x - 1}{2x + 3} \right) - 2\left( \frac{2x + 3}{3x - 1} \right) = 5\]

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

\[ \Rightarrow \frac{3\left( 9 x^2 + 1 - 6x \right) - 2\left( 4 x^2 + 9 + 12x \right)}{6 x^2 - 2x + 9x - 3} = 5\]

\[ \Rightarrow \frac{27 x^2 + 3 - 18x - 8 x^2 - 18 - 24x}{6 x^2 + 7x - 3} = 5\]

\[ \Rightarrow 19 x^2 - 42x - 15 = 5\left( 6 x^2 + 7x - 3 \right)\]

\[ \Rightarrow 19 x^2 - 42x - 15 = 30 x^2 + 35x - 15\]

\[ \Rightarrow 30 x^2 - 19 x^2 + 35x + 42x - 15 + 15 = 0\]

\[ \Rightarrow 11 x^2 + 77x = 0\]

\[ \Rightarrow x^2 + 7x = 0\]

\[ \Rightarrow x\left( x + 7 \right) = 0\]

\[ \Rightarrow x = 0 \text { or } x + 7 = 0\]

\[ \Rightarrow x = 0 \text { or } x = - 7\]

Hence, the factors are 0 and −7.