Question

1/(x-1)-1/(x+5)=6/7 solve by completing the square

Answer 1

Given,

$\longrightarrow\dfrac{1}{x-1}-\dfrac{1}{x+5}=\dfrac{6}{7}$

Making LHS a unique fraction,

$\longrightarrow\dfrac{(x+5)-(x-1)}{(x-1)(x+5)}=\dfrac{6}{7}$

$\longrightarrow\dfrac{x+5-x+1}{(x-1)(x+5)}=\dfrac{6}{7}$

$\longrightarrow\dfrac{6}{(x-1)(x+5)}=\dfrac{6}{7}$

Dividing by 6,

$\longrightarrow\dfrac{1}{(x-1)(x+5)}=\dfrac{1}{7}$

Taking the reciprocal,

$\longrightarrow (x-1)(x+5)=7$

$\longrightarrow x^2+4x-5=7$

$\longrightarrow x^2+4x=5+7$

$\longrightarrow x^2+4x=12$

Adding $\left(\dfrac{4}{2}\right)^2=4$ to both sides,

$\longrightarrow x^2+4x+4=12+4$

$\longrightarrow x^2+2\times x\times2+2^2=16$

$\longrightarrow (x+2)^2=4^2$

$\longrightarrow x+2=\pm4$

$\longrightarrow x=\pm4-2$

$\longrightarrow x=4-2\quad OR\quad x=-4-2$

$\longrightarrow \underline{\underline{x=2}}\quad OR\quad\underline{\underline{x=-6}}$