Question

Solve the following quadratic equations by factorization: \[\frac{3}{x + 1} - \frac{1}{2} = \frac{2}{3x - 1}, x \neq - 1, \frac{1}{3}\]

Answer 1

\[\frac{3}{x + 1} - \frac{1}{2} = \frac{2}{3x - 1}\]

\[ \Rightarrow \frac{6 - \left( x + 1 \right)}{2(x + 1)} = \frac{2}{3x - 1}\]

\[ \Rightarrow \frac{6 - x - 1}{2x + 2} = \frac{2}{3x - 1}\]

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

\[ \Rightarrow 15x - 5 - 3 x^2 + x = 4x + 4\]

\[ \Rightarrow - 3 x^2 + 16x - 5 - 4x - 4 = 0\]

\[ \Rightarrow - 3 x^2 + 12x - 9 = 0\]

\[ \Rightarrow 3 x^2 - 12x + 9 = 0\]

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

\[ \Rightarrow x^2 - 3x - x + 3 = 0\]

\[ \Rightarrow x(x - 3) - 1(x - 3) = 0\]

\[ \Rightarrow (x - 1)(x - 3) = 0\]

\[ \Rightarrow x - 1 = 0 \text { or } x - 3 = 0\]

\[ \Rightarrow x = 1 \text { or } x = 3\]

Hence, the factors are 3 and 1.