Question

Solve the following system of equations by substitution method :
$\frac{4}{x}+3y=8$

$\frac{6}{x}-4y=-5where[x\ne 0]$

Answer 1

$\frac{4}{x}+3y=8$ ............. (1)

$\frac{6}{x}-4y=-5$ ............. (2)

Solve equation (1) to get the value of $y$ in terms of $x$ ,

$3y=8-\frac{4}{x}$

$y=\frac{8}{3}-\frac{4}{3x}$ .................... (3)

Substitute the above value in equation (2)

$\frac{6}{x}-4(\frac{8}{3}-\frac{4}{3x})=-5$

$\frac{6}{x}-\frac{32}{3}+\frac{16}{3x}=-5$

$\frac{6}{x}+\frac{16}{3x}=-5+\frac{32}{3}$

$\frac{18+16}{3x}=\frac{-15+32}{3}$

$\frac{34}{3x}=\frac{17}{3}$

$\frac{34}{x}=17$

$x=\frac{34}{17}=2$

From equation (3),

$y=\frac{8}{3}-\frac{4}{3\times 2}$

$y=\frac{8}{3}-\frac{2}{3}$

$y=\frac{6}{3}=2$