Question

Solve each of the following equations. Also, verify the result in each case.
$\frac{6x-2}{9}+\frac{3x+5}{18}=\frac{1}{3}$

Answer 1

$\frac{6x-2}{9}+\frac{3x+5}{18}=\frac{1}{3}$
=> $\frac{6x\times 2-2\times 2+3x+5}{18}=\frac{1}{3}$
=> $\frac{12x-4+3x+5}{18}=\frac{1}{3}$
=> $\frac{15x+1}{18}=\frac{1}{3}$
Multiplying both sides by 18, we get
=> $\frac{15x+1}{18}\times \left(18\right)=\frac{1}{3}\times \left(18\right)$
=> 15x + 1 = 6
Transposing 1 to RHS, we get
=> 15x = 6$-$1
=> 15x = 5
Dividing both sides by 15, we get
=> $\frac{15x}{15}=\frac{5}{15}$
=> x = $\frac{1}{3}$
Verification:
Substituting x = $\frac{1}{3}$ on both sides, we get
$\frac{6\left({\displaystyle \frac{1}{3}}\right)-2}{9}+\frac{3\left({\displaystyle \frac{1}{3}}\right)+5}{18}=\frac{1}{3}$
$\frac{2-2}{9}+\frac{1+5}{18}=\frac{1}{3}$
$0+\frac{6}{18}=\frac{1}{3}$
$\frac{1}{3}=\frac{1}{3}$
LHS = RHS
Hence, verified.