Question

Solve the following system of equation for x and y $\frac{2}{x}+\frac{3}{y}=2;\frac{8}{x}-\frac{6}{y}=2$

• A. $\frac{1}{2},\frac{1}{3}$
• B. $\frac{1}{3},\frac{1}{2}$
• C. 2, 3
• D. 3, 2

Answer 1

The correct option is C 2, 3

Given equations are:
$\frac{2}{x}+\frac{3}{y}=2$................eq(1)
$\frac{8}{x}-\frac{6}{y}=2$.................eq(2)
on multiplying eq(1) by 6 and eq(2) by 3 we get
$\frac{12}{x}+\frac{18}{y}=12$..........eq(1) multiplied by 6
$\frac{24}{x}-\frac{18}{y}=6$..............eq(2) multiplied by 3
On adding both euqations we get
$\frac{12}{x}+\frac{18}{y}=12$
$\frac{24}{x}-\frac{18}{y}=6$
__________________________________
$\frac{36}{x}=18$
Therefore x=2
Putting x=2 in equation(1) we get
$\frac{2}{2}+\frac{3}{y}=2$
$1+\frac{3}{y}=2$
$y+3=2y$
$-y=-3$
$y=3$
Therefore x=2 and y=3