8.\(\frac { 2x }{ 3 } + 1 = \frac { 7x }{ 15 } + 3\)

Given that:\(\frac { 2x }{ 3 } + 1 = \frac { 7x }{ 15 } + 3\)
15\((\frac { 2x }{ 3 } + 1)\) = 15\((\frac { 7x }{ 15 } + 3)\quad \)[multiplying both the sides by 15]
\(\Rightarrow \frac { 2x }{ 3 } × 15 + 1 ×15 = \frac { 7x }{ 15 } ×15 + 3 ×15 \)

\(\Rightarrow \) 2x × 5 + 15 = 7x + 45

\(\Rightarrow \) 10x + 15 = 7x + 45

\(\Rightarrow \) 10x – 7x = 45 – 15 [Transposing 7x to LHS and 15 to RHS]

\(\Rightarrow \)3x = 30

\(\Rightarrow \) x = 30 ÷ 3 = 10 [Transposing 3 to RHS]

Thus, the required solution is x = 10.

Checking

Substitute x=10 in the given equation, we have

\(\frac { 2×10 }{ 3 } + 1 = \frac { 7×10 }{ 15 } + 3\)

\(\Rightarrow \frac { 20 }{ 3 } + 1 = \frac { 70 }{ 15 } + 3\)

\(\Rightarrow \frac { 20 + 3}{ 3 } = \frac { 70 + 45}{ 15 } \)

\(\Rightarrow \frac { 23}{ 3 } = \frac { 115}{ 15 } \)

\(\Rightarrow \) LHS= RHS

Hence,the required solution is x = 10.