12.Solve the equation \(\frac { x }{ 3 } + 1 = \frac { 7 }{ 15 }\)

Here we have \(\frac{x}3+1=\frac{7}{15}\)
\(\Rightarrow \frac{x}3+1-1=\frac{7}{15}-1 \quad \)(Subtracting 1 from both the sides)

\(\Rightarrow \frac{x}3=\frac{7}{15}-1 \)

\(\Rightarrow \frac{x}3=\frac{7-15}{15}\)

\(\Rightarrow \frac{x}3=\frac{-8}{15}\)

\(\Rightarrow \frac{x}3×3=\frac{-8}{15}×3\)(multiplying both the sides by 3)

\(\Rightarrow x=\frac{-8}{5}\)

So, we have \(x=\frac{-8}{5}\) as the required solution.
Check:

Keep \(x=\frac{-8}{5}\) in the given solution

L.H.S= \(\frac { x }{ 3 } + 1 =\frac{-8}{5}×\frac{1}{3}+1=\frac{-8+15}{15}=\frac{7}{15}=R.H.S\)

Hence \(x=\frac{-8}{5}\) is the required solution.