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serg [7]
4 years ago
11

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

Mathematics
2 answers:
Romashka-Z-Leto [24]4 years ago
8 0
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}}
laila [671]4 years ago
4 0
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.
<u></u>(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
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