Question

Solve the following simultaneous equations :

$\frac{1}{3x}+\frac{1}{5y}=\frac{1}{15};\frac{1}{2x}+\frac{1}{3y}=\frac{1}{12}$

• A. $x=5,y=3$
• B. $x=6,y=6$
• C. $x=2,y=-2$
• D. $x=1,y=-5$

Answer 1

The correct option is C $x=2,y=-2$

On putting $\frac{1}{x}=X$ and $\frac{1}{y}=Y$ in given equations, we get

$\frac{X}{3}+\frac{Y}{5}=\frac{1}{15}\Rightarrow 5X+3Y=1$ ...(i)

$\frac{X}{2}+\frac{Y}{3}=\frac{1}{12}\Rightarrow 6X+4Y=1$ ...(ii)
On multiplying (i) by 6 and (ii) by 5, we get
$30X+18Y=6$
$\underset{–}{\underset{-}{3}0X\underset{-}{+}20Y=\underset{-}{5}}$
$-2Y=1$
$\therefore Y=-\frac{1}{2}$
On putting $Y=-\frac{1}{2}$ in (i), we get

$5X+3\times -\frac{1}{2}=1\Rightarrow 5X=\frac{5}{2}\Rightarrow X=\frac{1}{2}$

$\therefore x=\frac{1}{X}=2,y=\frac{1}{Y}=-2$