Question

6x-2y=33 4x+3y=9 What would x and y be?

Answer 1

$6x-2y=33 \\\\ 6x=33+2y \\\\ \boxed{x=\frac{2y+33}{6}} \\\\ 4x+3y=9 \\\\ 4*\frac{2y+33}{6}+3y=9 \\\\ 2*\frac{2y+33}{3}+3y^{(3}=9^{(3} \\\\ 2(2y+33)+9y=27 \\\\ 4y+66+9y=27 \\\\ 13y=27-66 \\\\ 13y=- 39 \\\\ \boxed{y=-\frac{39}{13}=-3} \\\\ x=\frac{2*(-3)+33}{6} \\\\ x=\frac{-6+33}{6} \\\\ x=\frac{27}{6} \\\\ \boxed{x=\frac{9}{2}}$

Answer 2

To find the variables of the equations, use systems of equations.

First get rid of one variable. Let's get rid of the y by multiplyin both equations nad then adding them.
$(6x-2y=33)3$
$(4x+3y=9)2$

$18x-6y=99$
$8x+6y=18$
$26x=117$

Solve for x:
$\frac{26x}{26} = \frac{117}{26}$
$x=4.5$

Now you have x! Plug x into one quation and solve for y:
$6(4.5)-2y=33$
$27-2y=33$
$\frac{-2y}{-2} = \frac{6}{-2}$
$y=-3$

$x=4.5, y=-3$