Solve the system of two equations using elimination with multiplication
4x - 6y = -48
-2x + 4y = 30
To solve this system of equations using elimination with multiplication, we first need to make the coefficients of either x or y the same in both equations. Looking at the coefficients of y in the two equations, we can see that we can make them the same by multiplying the first equation by 2 and the second equation by -3.
Multiplying the first equation by 2:
8x - 12y = -96
Multiplying the second equation by -3:
6x - 12y = -90
Now, we have the two equations:
8x - 12y = -96
6x - 12y = -90
Now, we can subtract the second equation from the first equation to eliminate y:
(8x - 12y) - (6x - 12y) = -96 - (-90)
2x = -6
x = -3
Now that we have solved for x, we can substitute this value back into one of the original equations to solve for y. Using the first equation:
4(-3) - 6y = -48
-12 - 6y = -48
-6y = -36
y = 6
Therefore, the solution to the system of equations is x = -3 and y = 6.