Question

Solve the following quadratic equations by factorization: \[\frac{3}{x + 1} + \frac{4}{x - 1} = \frac{29}{4x - 1}; x \neq 1, - 1, \frac{1}{4}\]

Answer 1

\[\frac{3}{x + 1} + \frac{4}{x - 1} = \frac{29}{4x - 1}\]

\[ \Rightarrow \frac{3\left( x - 1 \right) + 4\left( x + 1 \right)}{\left( x + 1 \right)\left( x - 1 \right)} = \frac{29}{4x - 1}\]

\[ \Rightarrow \frac{3x - 3 + 4x + 4}{x^2 - 1} = \frac{29}{4x - 1}\]

\[ \Rightarrow \frac{7x + 1}{x^2 - 1} = \frac{29}{4x - 1}\]

\[ \Rightarrow \left( 7x + 1 \right)\left( 4x - 1 \right) = 29\left( x^2 - 1 \right)\]

\[ \Rightarrow 28 x^2 - 7x + 4x - 1 = 29 x^2 - 29\]

\[ \Rightarrow 29 x^2 - 28 x^2 + 3x - 28 = 0\]

\[ \Rightarrow x^2 + 3x - 28 = 0\]

\[ \Rightarrow x^2 + 7x - 4x - 28 = 0\]

\[ \Rightarrow x(x + 7) - 4(x + 7) = 0\]

\[ \Rightarrow (x - 4)(x + 7) = 0\]

\[ \Rightarrow x - 4 = 0 \text { or } x + 7 = 0\]

\[ \Rightarrow x = 4 \text { or } x = - 7\]

Hence, the factors are 4 and −7.