Question

Given $\frac{5}{8}x+\frac{7}{2}y=\frac{3}{2}$ and $\frac{1}{6}x-\frac{2}{3}y=1$ are the system of equations.If $(x,y)$ satisfies the system of equations, the sum of values of $x$ and $y$ is:

• A. $\frac{5}{24}$
• B. $\frac{5}{2}$
• C. $\frac{29}{8}$
• D. $\frac{33}{8}$

Answer 1

The correct option is D $\frac{33}{8}$

Given, $\frac{5}{8}x+\frac{7}{2}y=\frac{3}{2}$

Multiply equation by $8$, we get

$\frac{5}{8}\times 8x+\frac{7}{2}\times 8y=\frac{3}{2}\times 8$

$\Rightarrow 5x+28y=12$.......(1)

consider second equation given in the question

$\frac{1}{6}x-\frac{2}{3}y=1$

Multiply equation by $6$, we get

$\frac{1}{6}\times 6x-\frac{2}{3}\times 6=1\times 6$

$\Rightarrow x-4y=6$.......(2)

Multiply equation (2) by $7$, we get

$7x-28y=42$..........(3)

Add equation (1) and (3)

$12x=54$

$\Rightarrow x=\frac{54}{12}=\frac{9}{2}$

$\Rightarrow \frac{9}{2}-4y=6$.........(4)

Multiply equation (4) by $2$, we get

$9-8y=12$

$\Rightarrow -8y=12-9$

$\Rightarrow -8y=3$

$\Rightarrow y=-\frac{3}{8}$

$\Rightarrow x+y=\frac{9}{2}+(-\frac{3}{8})$

$\Rightarrow x+y=\frac{9}{2}-\frac{3}{8}=\frac{36-3}{8}=\frac{33}{8}$