Question

Solve the following equations by substitution method.

$\frac{3}{x}+\frac{1}{y}=7;\frac{5}{x}-\frac{4}{y}=6.(x\ne 0,y\ne 0)$

• A. $x=\frac{1}{2},y=1$
• B. $x=\frac{-1}{2},y=1$
• C. $x=\frac{1}{2},y=-1$
• D. None of these

Answer 1

The correct option is A $x=\frac{1}{2},y=1$

Given equations are,

$\frac{3}{x}+\frac{1}{y}=7$

$\frac{5}{x}-\frac{4}{y}=6$

$\frac{1}{x}=\frac{1}{3}(7-\frac{1}{y})$.................from (1)

Putting value of $\frac{1}{x}$ in (1), we get

$5\left[\frac{1}{3}\right(7-\frac{1}{y}\left)\right]-\frac{4}{y}=6$

$\frac{35}{3}-\frac{5}{3y}-\frac{4}{y}=6$

$\frac{17}{3}=\frac{5+12}{3y}$

$y=1$

Putting value of $y$ in (1)

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