Question

Solve the following system of equation
$2x+3y+8=0$
$4x+5y+14=0$

Answer 1

Given, $2x+3y+8=0$ ...(1)

$4x+5y+14=0$ ...(2)

multiply equation. (1) by 5 and equation (2) by 3 we get

$10x+15y+40=0$

$12x+15y+42=0$

$---$

___________________

$-2x-2=0$

$[x=-1]$ from equation.(1) $3y=8-2x$

$y=\frac{10}{3}$

Hence using cross multiplication method intersection point of line 1 & 2 are$(-1,\frac{10}{3})$