Question

Solve each of the following equations and check your answers:
$\frac{1}{3}-2x=0$

Answer 1

$\frac{1}{3}-2x=0$
Subtracting $\frac{1}{3}$ from both sides, we get
⇒ $\frac{1}{3}-2x-\frac{1}{3}$ = 0 $-$$\frac{1}{3}$
⇒ $-2x=-\frac{1}{3}$
Multiplying both sides by $-$1, we get
⇒ $-2x\times (-1)=-\frac{1}{3}\times (-1)$
⇒ 2x = $\frac{1}{3}$
Dividing both sides by 2, we get
⇒ $\frac{2x}{2}=\frac{1}{3\times 2}$
⇒ x = $\frac{1}{6}$
Verification:
Substituting x = $\frac{1}{6}$ in LHS, we get
LHS = $\frac{1}{3}-\left(2\times \frac{1}{6}\right)=\frac{1}{3}-\frac{1}{3}=0$, and RHS = 0
LHS = RHS
Hence, verified.