Question

Solve:
$\frac{1}{3x}+\frac{1}{5y}=1$;
$\frac{1}{5x}+\frac{1}{3y}=1\frac{2}{15}$

• A. $x=\frac{21}{3},y=\frac{6}{5}$
• B. $x=-\frac{8}{5},y=\frac{9}{4}$
• C. $x=\frac{7}{2},y=-\frac{8}{3}$
• D. $x=\frac{2}{3},y=\frac{2}{5}$

Answer 1

Answer: (D)

$x=\frac{2}{3},y=\frac{2}{5}$

The given equations are

$\frac{1}{3x}+\frac{1}{5y}=1$ .....(1)

$\frac{1}{5x}+\frac{1}{3y}=1\frac{2}{15}$

$\Rightarrow \frac{1}{5x}+\frac{1}{3y}=\frac{17}{15}$ .....(2)

Multiplying equation (1) by $\frac{1}{5}$ and equation (2) by $\frac{1}{3}$, we get

$\frac{1}{15x}+\frac{1}{25y}=\frac{1}{5}$ ....(3)

$\frac{1}{15x}+\frac{1}{9y}=\frac{17}{45}$$ ....(4)

Subtracting equations (4) from (3),

$\frac{1}{25y}\frac{1}{9y}=\frac{1}{5}\frac{17}{45}$

$\Rightarrow \frac{9-25}{225y}=\frac{9-17}{45}$

$\Rightarrow \frac{-16}{225y}=\frac{-8}{45y}$

$\Rightarrow y=\frac{16\times 45}{225\times 8}=\frac{2}{5}$

Substitute $y=\frac{2}{5}$ in equation (3), we get

$\frac{1}{3x}+\frac{1}{5\times \frac{2}{5}}=1$

$\Rightarrow \frac{1}{3x}=1\frac{1}{2}$

$\Rightarrow \frac{1}{3x}=\frac{1}{2}$

$\Rightarrow x=\frac{2}{3}$

Hence, by solving equations, we get

$x=\frac{2}{3},y=\frac{2}{5}$