Question

Solve the following quadratic equation by factorization.
$\frac{3}{x+1}+\frac{4}{x-1}=\frac{29}{4x-1};x\ne 1,-1,\frac{1}{4}$

Answer 1

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

$\frac{3(x-1)+4(x+1)}{(x+1)(x-1)}=\frac{29}{4x-1}$

$\frac{3x-3+4x+4}{(x^2-1)}=\frac{29}{4x-1}$

$\frac{7x+1}{(x^2-1)}=\frac{29}{4x-1}$

$(7x+1)(4x-1)=29(x^2-1)$

$28x^2-7x+4x-1=29x^2-29$

$28x^2-3x-1=29x^2-29$
$x^2+3x-28=0$

$x^2-4x+7x-28=0$

$x(x-4)+7(x-4)=0$

$(x+7)(x-4)=0$

$x=-7,4$