To solve the system of equations using elimination, we can multiply the first equation by 3 to make the coefficients of the x terms equal.
3(2−2x−2y)=3⋅2
6−6x−6y=6
-6x-6y=0.
Now, we can add the two equations together to eliminate the x terms:
(-6x-6y) + (6x+9y) = 0 + (-33)
-6x + 6x - 6y + 9y = -33
3y = -33
Next, we can solve for y:
y = -33/3
y = -11.
Finally, we substitute the value of y back into one of the original equations to solve for x:
2−2x−2(-11)=2
2−2x+22=2
20−2x=2
-2x = 2 - 20
-2x = -18
x = -18/-2
x = 9.
Therefore, the solution to the system of equations is x = 9 and y = -11.
Solve the system of equations using elimination: minus, 2, x, minus, 2, y, equals, 2−2x−2y=2 and 6, x, plus, 9, y, equals, minus, 33 6x+9y=−33.
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