Question

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

Answer 1

$\frac{\left(5x-1\right)}{3}-\frac{\left(2x-2\right)}{3}=1$
$$\begin{gathered} \Rightarrow \frac{5x-1-2x-(-2)}{3}=1 \\ \Rightarrow \frac{5x-1-2x+2}{3}=1 \\ \Rightarrow \frac{5x-2x+2-1}{3}=1 \\ \Rightarrow \frac{3x+1}{3}=1 \end{gathered}$$
Multiplying both sides by 3, we get
$\Rightarrow \left(\frac{3x+1}{3}\right)\times 3=1\times 3$
=> 3x + 1 = 3
Subtracting 1 from both sides, we get
=> 3x + 1 $-$ 1 = 3 $-$ 1
=> 3x = 2
Dividing both sides by 3, we get
=> $\frac{3x}{3}=\frac{2}{3}$
=> x = $\frac{2}{3}$
Verification:
Substituting x =$\frac{2}{3}$ in LHS, we get
$$\begin{gathered} =\frac{5\left({\displaystyle \frac{2}{3}}\right)-1}{3}-\frac{2\left({\displaystyle \frac{2}{3}}\right)-2}{3} \\ =\frac{{\displaystyle \frac{10}{3}}-1}{3}-\frac{{\displaystyle \frac{4}{3}}-2}{3}=\frac{{\displaystyle \frac{10-3}{3}}}{3}-\frac{{\displaystyle \frac{4-6}{3}}}{3}=\frac{7}{3\times 3}-\left({\displaystyle \frac{-2}{3\times 3}}\right)=\frac{7}{9}+\frac{2}{9}=\frac{9}{9}=1=RHS \end{gathered}$$
LHS = RHS
Hence, verified.