# Solve each of the following equation and also check your result in each case:

Question:

Solve each of the following equation and also check your result in each case:

$\frac{(45-2 x)}{15}-\frac{(4 x+10)}{5}=\frac{(15-14 x)}{9}$

Solution:

$\frac{45-2 \mathrm{x}}{15}-\frac{4 \mathrm{x}+10}{5}=\frac{15-14 \mathrm{x}}{9}$

or $\frac{45-2 \mathrm{x}-12 \mathrm{x}-30}{15}=\frac{15-14 \mathrm{x}}{9}$

or $\frac{15-14 \mathrm{x}}{5}=\frac{15-14 \mathrm{x}}{3}[$ Multiply ing both sides by 3$]$

or $45-42 \mathrm{x}=75-70 \mathrm{x}[$ After c ross multiplication $]$

or $70 \mathrm{x}-42 \mathrm{x}=75-45$

or $28 \mathrm{x}=30$

or $\mathrm{x}=\frac{30}{28}$

or $\mathrm{x}=\frac{15}{14}$

Thus, $\mathrm{x}=\frac{15}{14}$ is the solution of the given equation. Check:

Substituting $\mathrm{x}=\frac{15}{14}$ in the given equation, we get:

L. H. S. $=\frac{45-2 \times \frac{15}{14}}{15}-\frac{4 \times \frac{15}{14}+10}{5}=\frac{45 \times 7-15}{105}-\frac{30+70}{35}=\frac{300}{105}-\frac{100}{35}=0$

R.H. S. $=\frac{15-14 \times \frac{15}{14}}{9}=0$

$\therefore$ L.H.S. $=$ R. H. S. for $\mathrm{x}=\frac{15}{14}$