To solve this system of equations using elimination with multiplication, we need to make either the x-coefficients or y-coefficients the same in both equations. Let's make the x-coefficients the same by multiplying the first equation by 2 and the second equation by 2:
8x - 12y = -96
-4x + 8y = 60
Now, we can add these two equations together to eliminate the x-variable:
(8x - 12y) + (-4x + 8y) = -96 + 60
4x - 4y = -36
x - y = -9
Now, we have a new equation x - y = -9. Let's solve for x:
x - y = -9
x = y - 9
Now, let's substitute x = y - 9 into one of the original equations. Let's use the first equation:
4(y - 9) - 6y = -48
4y - 36 - 6y = -48
-2y = 12
y = -6
Now, we can substitute y = -6 back into x = y - 9 to find the value of x:
x = -6 - 9
x = -15
Therefore, the solution to the system of equations is x = -15 and y = -6.
Solve the system of two equations using elimination with multiplication.
4x−6y=−48
−2x+4y=30(1 point)
1 answer