7. x =\( \frac { 4 }{ 5 } (x + 10)\)

Given x = \(\frac { 4 }{ 5 } (x + 10)\)
\(\Rightarrow\) 5 × x = 4(x + 10) (Transposing 5 to LHS)

\(\Rightarrow\) 5x = 4x + 40 (Solving the bracket)

\(\Rightarrow\) 5x – 4x = 40 (Transposing 4x to LHS)

\(\Rightarrow\) x = 40

Thus x = 40 is the required solution.

Checking:

x = \(\frac { 4 }{ 5 }\) (x + 10)

Substituting x = 40 in the given equation, we have

40 = \(\frac { 4 }{ 5 } \)(40 + 10)

\(\Rightarrow\) 40 =\( \frac { 4 }{ 5 }\) × 50

\(\Rightarrow\) 40 = 4× 10

\(\Rightarrow\) 40 = 40

LHS = RHS

Hence x = 40 is the required solution.