Question

If $\frac{2}{x}+\frac{3}{y}=\frac{9}{xy}$ and $\frac{4}{x}+\frac{9}{y}=\frac{21}{xy}$, where $x\ne 0,y\ne 0$, then the values of $x$ and $y$ are

• A. $0$ and $1$
• B. $1$ and $2$
• C. $2$ and $3$
• D. $1$ and $3$

Answer 1

The correct option is D $1$ and $3$

The given equations are

$\frac{2}{x}+\frac{3}{y}=\frac{9}{xy}$

$⟹\frac{2y+3x}{xy}=\frac{9}{xy}$

$⟹3x+2y=9$ ........(i)

$\frac{4}{x}+\frac{9}{y}=\frac{21}{xy}$

$⟹\frac{4y+9x}{xy}=\frac{21}{xy}$

$⟹9x+4y=21$ ........(ii)

Now $\left(i\right)\times 3⟶9x+6y=27$ .....(iii)

Subtracting (iii) from (ii),

$-2y=-6\Rightarrow y=3.$

Putting $y=3$ in (i),

$3x+2\times 3=9⟹x=1.\therefore (x,y)=(1,3)$