Question

Solve the following equation and verify your answer:
${\left(\frac{x+1}{x-4}\right)}^2=\frac{x+8}{x-2}$

Answer 1

$$\begin{gathered} {\left(\frac{\mathrm{x}+1}{\mathrm{x}-4}\right)}^2=\frac{\mathrm{x}+8}{\mathrm{x}-2} \\ \mathrm{or}\frac{{\mathrm{x}}^2+2\mathrm{x}+1}{{\mathrm{x}}^2-8\mathrm{x}+16}=\frac{\mathrm{x}+8}{\mathrm{x}-2}\left[\right(a+b{)}^2=a^2+b^2+2aband(a-b{)}^2=a^2+b^2-2ab] \\ \mathrm{or}{\mathrm{x}}^3+2{\mathrm{x}}^2+\mathrm{x}-2{\mathrm{x}}^2-4\mathrm{x}-2={\mathrm{x}}^3-8{\mathrm{x}}^2+16\mathrm{x}+8{\mathrm{x}}^2-64\mathrm{x}+128[\text{After c}\mathrm{ross}\mathrm{multipl}\text{ication}] \\ \mathrm{or}{\mathrm{x}}^3-{\mathrm{x}}^3-3\mathrm{x}+48\mathrm{x}=128+2 \\ \mathrm{or}45\mathrm{x}=130 \\ \mathrm{or}\mathrm{x}=\frac{130}{45}=\frac{26}{9} \\ \mathrm{Thus}\mathrm{x}=\frac{26}{9}\mathrm{is}\mathrm{the}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{equation}. \\ \mathrm{Check}: \\ \mathrm{Substituting}\mathrm{x}=\frac{26}{9}\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{equation},\mathrm{we}\mathrm{get}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.={\left(\frac{\frac{26}{9}+1}{\frac{26}{9}-4}\right)}^2={\left(\frac{26+9}{26-36}\right)}^2=\frac{1225}{100}=\frac{49}{4} \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\left(\frac{\frac{26}{9}+8}{\frac{26}{9}-2}\right)=\left(\frac{26+72}{26-18}\right)=\frac{98}{8}=\frac{49}{4} \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\mathrm{for}\mathrm{x}=\frac{26}{9} \end{gathered}$$