9. \(2y + \frac { 5 }{ 3 } =\frac { 26 }{ 3 } – y\).

Given that, \(2y + \frac { 5 }{ 3 } = \frac { 26 }{ 3 } – y\)
\(\Rightarrow 2y + y = \frac { 26 }{ 3 } – \frac { 5 }{ 3 }\quad \)[Transposing -y to LHS and \(\frac{5}3\) to RHS]

\(\Rightarrow 3y = \frac { 21 }{ 3 }\)

\(\Rightarrow 3y =7\)

\(\Rightarrow \frac{3y}{3} = \frac { 7 }{ 3 }\quad \)[dividing both the sides by 3]

\(\Rightarrow y = \frac {7 }{ 3 }\)

Hence, \(y=\frac {7 }{ 3 }\) is the required value

Checking

Substitute \(y=\frac {7 }{ 3 }\) in the given equation, we have

LHS=\(2×\frac{7}3+\frac{5}3=\frac{14}3+\frac{5}3=\frac{19}3\)

RHS=\(\frac{26}3-\frac{7}3=\frac{19}3\)

LHS=RHS

Hence,\(y=\frac {7 }{ 3 }\) is the required value.