Question

Solve each of the following quadratic equations:
$\frac{3}{x+1}-\frac{1}{2}=\frac{2}{3x-1},x\ne -1,\frac{1}{3}$

Answer 1

$\frac{3}{x+1}-\frac{1}{2}=\frac{2}{3x-1}$

$\frac{3}{x+1}-\frac{2}{3x-1}=\frac{1}{2}$

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

$\frac{(9x-3-2x-2)}{(3x^2-x+3x-1)}=\frac{1}{2}$

$2(7x-5)=(3x^2+2x-1)$

$3x^2+2x-1-14x+10=0$

$3x^2-12x+9=0$

$x^2-4x+3=0$

$x^2-3x-x+3=0$

$x(x-3)-1(x-3)=0$

$(x-3)(x-1)=0$

Either

$x-3=0$

$x=3$

or

$x-1=0$

$x=1$

So x = 1, 3