Question

Solve: $\frac{3}{x}-\frac{2}{y}=0$ and $\frac{2}{x}+\frac{5}{y}=19$. Hence, find $a$ if $y=ax+3$.

• A. $x=\frac{2}{3};y=\frac{5}{3}$ and $a=6\frac{6}{5}$
• B. $x=\frac{3}{4};y=\frac{1}{3}$ and $a=-4\frac{7}{3}$
• C. $x=\frac{7}{2};y=\frac{8}{3}$ and $a=-7\frac{2}{9}$
• D. $x=\frac{1}{2};y=\frac{1}{3}$ and $a=-5\frac{1}{3}$

Answer 1

The correct option is D $x=\frac{1}{2};y=\frac{1}{3}$ and $a=-5\frac{1}{3}$

Let $\frac{1}{x}=u$ and $\frac{1}{y}=v$

So, the equations will reduce to

$3u-2v=0$ ...$\left(1\right)$

$2u+5v=19$ ...$\left(2\right)$

Multiplying $\left(1\right)$ by $5$ and $\left(2\right)$ by $2$

$15u-10v=0$ ...$\left(3\right)$

$4u+10v=38$ ...$\left(4\right)$

Adding $\left(3\right)$ and $\left(4\right)$

$19u=28$

$u=2$

Substituting $u=2$ in $\left(1\right)$, we get

$6-2v=0$

$v=3$

$u=\frac{1}{x}$ So, $x=\frac{1}{u}=\frac{1}{2}$

$v=\frac{1}{y}$ So, $y=\frac{1}{v}=\frac{1}{3}$

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

$y=ax+3$

$\frac{1}{3}=\frac{a}{2}+3\frac{a}{2}=-\frac{8}{3}a=\frac{-16}{3}$

Option $D$