Question

Solve $(1+\frac{1}{x+1})(1-\frac{x}{x-1})=\frac{7}{8}$

Answer 1

Given that,

$\begin{array}{c}\left(1+\frac{1}{x+1}\right)\left(1-\frac{x}{x-1}\right)=\frac{7}{8}\\ \Rightarrow \left(\frac{x+1+1}{x+1}\right)\left(\frac{x-1-x}{x-1}\right)=\frac{7}{8}\\ \Rightarrow \left(\frac{x+2}{x+1}\right)\left(\frac{-1}{x-1}\right)=\frac{7}{8}\\ \Rightarrow \frac{-(x+2)}{x^2-1^2}=\frac{7}{8}\\ \Rightarrow \frac{x+2}{x^2-1}=\frac{-7}{8}\\ \Rightarrow -7\left(x^2-1\right)=8(x+2)\\ \Rightarrow -7x^2+7=8x+16\\ \Rightarrow -7x^2+7-8x-16=0\\ \Rightarrow -7x^2-8x-9=0\\ \Rightarrow 7x^2+8x+9=0\\ x=\frac{-8\pm \sqrt{8^2-4\times 7\times 4}}{2\times 7}\\ \Rightarrow x=\frac{-8\pm \sqrt{-188}}{14}\end{array}$

which is imaginary.