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

Sagot :

mariamikayla
[tex]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}}[/tex]
Johnweshas
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.
[tex](6x-2y=33)3[/tex]
[tex](4x+3y=9)2[/tex]

[tex]18x-6y=99[/tex]
[tex]8x+6y=18[/tex]
[tex]26x=117[/tex]

Solve for x:
[tex] \frac{26x}{26} = \frac{117}{26} [/tex]
[tex]x=4.5[/tex]

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

[tex]x=4.5, y=-3[/tex]
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