Question

Solve each of the following equation and also check your result in each case:
$\frac{(45-2x)}{15}-\frac{(4x+10)}{5}=\frac{(15-14x)}{9}$

Answer 1

$$\begin{gathered} \frac{45-2\mathrm{x}}{15}-\frac{4\mathrm{x}+10}{5}=\frac{15-14\mathrm{x}}{9} \\ \mathrm{or}\frac{45-2\mathrm{x}-12\mathrm{x}-30}{15}=\frac{15-14\mathrm{x}}{9} \\ \mathrm{or}\frac{15-14\mathrm{x}}{5}=\frac{15-14\mathrm{x}}{3}[\mathrm{Multiply}\text{ing}\mathrm{both}\mathrm{sides}\mathrm{by}3] \\ \mathrm{or}45-42\mathrm{x}=75-70\mathrm{x}[\text{After c}\mathrm{ross}\mathrm{multipl}\text{ication}] \\ \mathrm{or}70\mathrm{x}-42\mathrm{x}=75-45 \\ \mathrm{or}28\mathrm{x}=30 \\ \mathrm{or}\mathrm{x}=\frac{30}{28} \\ \mathrm{or}\mathrm{x}=\frac{15}{14} \\ \mathrm{Thus},\mathrm{x}=\frac{15}{14}\mathrm{is}\mathrm{the}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{equation}. \\ \mathrm{Check}: \\ \mathrm{Substituting}\mathrm{x}=\frac{15}{14}\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{equation},\mathrm{we}\mathrm{get}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{45-2\times {\displaystyle \frac{15}{14}}}{15}-\frac{4\times {\displaystyle \frac{15}{14}}+10}{5}=\frac{45\times 7-15}{105}-\frac{30+70}{35}=\frac{300}{105}-\frac{100}{35}=0 \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{15-14\times {\displaystyle \frac{15}{14}}}{9}=0 \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\mathrm{for}\mathrm{x}=\frac{15}{14} \end{gathered}$$