Question

Solve: $\frac{x-4}{7x+1}+\frac{1}{2}=\frac{11}{2(3+2x)}$

Answer 1

Given, $\frac{x-4}{7x+1}+\frac{1}{2}=\frac{11}{2(3+2x)}$

$\Rightarrow \frac{2(x-4)+7x+1}{2(7x+1)}=\frac{11}{6+4x}$

$\Rightarrow \frac{2x-8+7x+1}{14x+2}=\frac{11}{6+4x}$

$\Rightarrow \frac{9x-7}{14x+2}=\frac{11}{6+4x}$

$\Rightarrow \frac{9x-7}{2(7x+1)}=\frac{11}{2(3+2x)}$

$\Rightarrow \frac{9x-7}{(7x+1)}=\frac{11}{(3+2x)}$

$\Rightarrow (9x-7)(3+2x)=11(7x+1)$

$\Rightarrow 27x+18x^2-21-14x=77x+11$

$\Rightarrow 18x^2-21+13x-77x-11=0$

$\Rightarrow 18x^2-64x-32=0$

$\Rightarrow 9x^2-32x-16$

$\Rightarrow 9x^2-36x+4x-16=0$

$\Rightarrow 9x(x-4)+4(x-4)=0$

$\Rightarrow (x-4)(9x+4)=0$

$\therefore x=4,\frac{-4}{9}$