Question

Solve each of the following equation and also check your result in each case:
$\frac{4x}{9}+\frac{1}{3}+\frac{13}{108}x=\frac{8x+19}{18}$

Answer 1

$$\begin{gathered} \frac{4\mathrm{x}}{9}+\frac{1}{3}+\frac{13}{108}\mathrm{x}=\frac{8\mathrm{x}+19}{18} \\ \mathrm{or}\frac{48\mathrm{x}+36+13\mathrm{x}}{108}=\frac{8\mathrm{x}+19}{18} \\ \mathrm{or}\frac{61\mathrm{x}+36}{108}=\frac{8\mathrm{x}+19}{18} \\ \mathrm{or}61\mathrm{x}+36=6(8\mathrm{x}+19)[\mathrm{Multiply}\text{ing}\mathrm{both}s\mathrm{ides}\mathrm{by}108] \\ \mathrm{or}61\mathrm{x}+36=48\mathrm{x}+114 \\ \mathrm{or}61\mathrm{x}-48\mathrm{x}=114-36 \\ \mathrm{or}13\mathrm{x}=78 \\ \mathrm{or}\mathrm{x}=\frac{78}{13} \\ \mathrm{or}\mathrm{x}=6 \\ \mathrm{Thus},\mathrm{x}=6\mathrm{is}\mathrm{the}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{equation}. \\ \mathrm{Check}: \\ \mathrm{Substituting}\mathrm{x}=6\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{equation},\mathrm{we}\mathrm{get}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{4\times 6}{9}+\frac{1}{3}+\frac{13}{108}\times 6=\frac{24}{9}+\frac{1}{3}+\frac{13}{18}=\frac{48+6+13}{18}=\frac{67}{18} \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{8\times 6+19}{18}=\frac{67}{18} \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\mathrm{for}\mathrm{x}=6. \end{gathered}$$