Question

Find the quadrant, where point of intersection of the lines $2x-3y=-3$, $-12=-4x+y$ lies

• A. Quadrant I
• B. Quadrant II
• C. Quadrant III
• D. Quadrant IV

Answer 1

The correct option is A Quadrant I

Let $2x-3y=-3$.......(1)

$-12=-4x+y$

and $\Rightarrow -4x+y=-12$........(2)

Multiply equation (1) by $2$, we get

$4x-6y=-6$........(3)

Add equation (2) and (3), we get

$-5y=-18$

$\Rightarrow y=\frac{18}{5}=3.6$

Put the value of $y=3.6$ in equation (1), then

$2x-3\left(3.6\right)=-3$

$\Rightarrow 2x-7.2=-3$

]$\Rightarrow 2x=-3+7.2$

$\Rightarrow 2x=4.2$

$\Rightarrow x=2.1$

So point of intersection of the line is $(2.1,3.6)$

Both point in Quadrant I