Question

For the equations $\frac{2}{x}+\frac{3}{y}=13$ and
$\frac{5}{x}-\frac{4}{y}=-2$ where $x\ne 0\&y\ne 0$, the value of

• A. $x=\frac{1}{3}$
• B. $y=\frac{1}{3}$
• C. $x=\frac{1}{4}$
• D. $y=\frac{1}{3}$

Answer 1

The correct option is D $y=\frac{1}{3}$

$Let\frac{1}{x}be{}^{\prime }{a}^{\prime }and\frac{1}{y}be{}^{\prime }{b}^{\prime }$
$(Asx\ne 0andy\ne 0)$.

We get our equations as

$2a+3b=13------\left(i\right)$

$5a-4b=-2----\left(ii\right)$

Multiplying the $\left(i\right)$ by 4 and the $\left(ii\right)$ by 3, we get

$8a+12b=52----\left(iii\right)$

$15a-12b=-6---\left(iv\right)$

Now adding the two equations, we get
$23a=46$
$\Rightarrow a=2$

Substituting a = 2 in the $\left(i\right)$, we get

2 × 2 + 3b = 13
3b = 9
$\Rightarrow$b = 3

Earlier we have assumed that $a=\frac{1}{x}$ & $b=\frac{1}{y}$

Therefore, $x=\frac{1}{2}$ and $y=\frac{1}{3}$.