Question

The solutions set of system of equation $\frac{4}{x}+5y=7,\frac{3}{x}+4y=5$ is

• A. $(\frac{1}{3},-1)$
• B. $(\frac{-1}{3},1)$
• C. $(\frac{-1}{3},-1)$
• D. $(\frac{1}{3},1)$

Answer 1

The correct option is A $(\frac{1}{3},-1)$

Given equations are
$\frac{4}{x}+5y=7$ --- (1)
And $\frac{3}{x}+\frac{4}{y}=5$ --- (2)

Multiplying equation $\left(1\right)$ with $4$ we get, $\frac{16}{x}+20y=28$ ----- equation $\left(3\right)$

Multiplying equation $\left(2\right)$ with $5$ we get, $\frac{15}{x}+20y=25$ ----- equation $\left(4\right)$

Subtracting equation $\left(4\right)$
from $\left(3\right)$, we get $\frac{1}{x}=3=>x=\frac{1}{3}$

Substituting $x=\frac{1}{3}$
in the equation $\left(1\right)$, we get $\frac{4}{\frac{1}{3}}+5y=7=>12+5y=7=>5y=-5=>y=-1$