Question

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

Answer 1

$$\begin{gathered} {\left(\frac{\mathrm{x}+1}{\mathrm{x}+2}\right)}^2=\frac{\mathrm{x}+2}{\mathrm{x}+4} \\ \mathrm{or}\frac{{\mathrm{x}}^2+2\mathrm{x}+1}{{\mathrm{x}}^2+4\mathrm{x}+4}=\frac{\mathrm{x}+2}{\mathrm{x}+4} \\ \mathrm{or}{\mathrm{x}}^3+2{\mathrm{x}}^2+\mathrm{x}+4{\mathrm{x}}^2+8\mathrm{x}+4={\mathrm{x}}^3+4{\mathrm{x}}^2+4\mathrm{x}+2{\mathrm{x}}^2+8\mathrm{x}+8[\text{After c}\mathrm{ross}\mathrm{multipl}\text{ication}] \\ \mathrm{or}{\mathrm{x}}^3-{\mathrm{x}}^3+6{\mathrm{x}}^2-6{\mathrm{x}}^2+9\mathrm{x}-12\mathrm{x}=8-4 \\ \mathrm{or}-3\mathrm{x}=4 \\ \mathrm{or}\mathrm{x}=\frac{4}{-3}=\frac{-4}{3} \\ \mathrm{Thus},\mathrm{x}=\frac{-4}{3}\mathrm{is}\mathrm{the}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{equation}. \\ \mathrm{Check}: \\ \mathrm{Substituting}\mathrm{x}=\frac{-4}{3}\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{equation},\mathrm{we}\mathrm{get}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.={\left(\frac{\frac{-4}{3}+1}{\frac{-4}{3}+2}\right)}^2={\left(\frac{-4+3}{-4+6}\right)}^2=\frac{1}{4} \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{\frac{-4}{3}+2}{\frac{-4}{3}+4}=\frac{-4+6}{-4+12}=\frac{2}{8}=\frac{1}{4} \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\mathrm{for}\mathrm{x}=\frac{-4}{3} \end{gathered}$$