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# 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.