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# 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.