Question

Solve the following system of equations by elimination method.

$8x-3y=5xy,6x-5y=-2xy,x\ne 0,y\ne 0$

• A. $(\frac{11}{23},\frac{22}{31})$
• B. $(\frac{1}{23},\frac{22}{31})$
• C. $(\frac{11}{23},\frac{2}{31})$
• D. None of these

Answer 1

Answer: (A)

$(\frac{11}{23},\frac{22}{31})$

Consider the given equations.

$8x-3y=5xy$ $-------\left(1\right)$

$6x-5y=-2xy$ $-------\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$

Multiply by $\left(8\right)$ in $\left(2\right)$ and by $\left(6\right)$ in $\left(1\right)$

$48x-18y=30xy$

$48x-40y=-16xy$

—————————

$22y=46xy$

$x=\frac{11}{23}$ $[$ put in $\left(2\right)]$

$6\times \frac{11}{23}-5y=-2\times \frac{11}{23}y$

$\frac{66-115y}{23}=-\frac{22}{23}y$

$66-115y=-22y$

$93y=66$

$y=\frac{66}{93}$

$y=\frac{22}{31}$

Hence, the value of $x$ is $\frac{11}{23}$ and the value of $y$ is $\frac{22}{31}$.