Question

Solve the following pair of simultaneous equations:

$\frac{6}{x}-\frac{2}{y}=1;\frac{9}{x}-\frac{6}{y}=0$

• A. $x=7,y=-4$
• B. $x=6,y=-4$
• C. $x==1,y=2$
• D. $x=3,y=2$

Answer 1

Answer: (D)

$x=3,y=2$

The equation $\frac{6}{x}-\frac{2}{y}=1$ can be solved as:

$\frac{6}{x}-\frac{2}{y}=1\Rightarrow \frac{6y-2x}{xy}=1\Rightarrow 6y-2x=xy.........\left(1\right)$

The equation $\frac{9}{x}-\frac{6}{y}=0$ can be solved as:

$\frac{9}{x}-\frac{6}{y}=0\Rightarrow \frac{9y-6x}{xy}=0\Rightarrow 9y-6x=0.........\left(2\right)$

Multiply the equation 1 by $3$.Then we get:

$18y-6x=3xy.........\left(3\right)$

Subtract equation 2 from equation 3 to eliminate $x$, because the coefficients of $x$ are same. So, we get

$(18y-9y)+(-6x+6x)=3xy-0$

i.e. $9y=3xy$

i.e. $9=3x$

i.e. $x=3$

Substituting this value of $x$ in the equation 1, we get

$6y-6=3y$

i.e. $3y=6$

i.e. $y=2$

Hence, the solution of the equations is $x=3,y=2$.