To solve the system of equations using elimination with multiplication, we need to multiply one or both equations by a constant so that the coefficients of either x or y are the same in both equations. Let's multiply the first equation by 2 and the second equation by -3:
(2)(4x - 6y) = (2)(-48)
(-3)(-2x + 4y) = (-3)(30)
Simplifying these equations gives us:
8x - 12y = -96
6x - 12y = -90
Now, we can subtract the second equation from the first equation:
(8x - 12y) - (6x - 12y) = -96 - (-90)
8x - 12y - 6x + 12y = -96 + 90
2x = -6
Dividing both sides of this equation by 2 gives us:
x = -3
Now, we can substitute this value of x into either of the original equations to solve for y. Let's use the first equation:
4x - 6y = -48
4(-3) - 6y = -48
-12 - 6y = -48
-6y = -48 + 12
-6y = -36
y = -36 / -6
y = 6
Therefore, the solution to the system of equations is (x, y) = (-3, 6). The correct answer choice is (−3, 6).
Solve the system of two equations using elimination with multiplication.
4x−6y=−48
−2x+4y=30(1 point)
Responses
(−21,−6)
(−33,−9)
(3, 9)
(−3, 6
1 answer