Solve the system of two equations using elimination with multiplication.%0D%0A4x−6y=−48%0D%0A−2x+4y=30(1 point)

Answer 1

To solve the system of equations using elimination with multiplication, we need to choose a common multiple for the coefficients of x or y in both equations. In this case, we can choose 4 as the common multiple.

First, we can multiply the first equation by 2 and the second equation by -1 to get:
8x - 12y = -96
2x - 4y = -30

Now, we can add the two equations together to eliminate x:
(8x - 12y) + (2x - 4y) = -96 + (-30)
10x - 16y = -126

Next, we can divide the equation by 2 to simplify it further:
5x - 8y = -63

Now, we have a new equation:
5x - 8y = -63

We can now solve this equation for one variable in terms of the other:
5x = 8y - 63
x = (8y - 63) / 5

We can substitute this value of x back into one of the original equations to solve for y. Let's use the first equation:
4x - 6y = -48
4((8y - 63) / 5) - 6y = -48
(32y - 252) / 5 - 6y = -48

Now, we can solve this equation for y:
(32y - 252) / 5 - 6y = -48
(32y - 252) - 30y = -240
32y - 252 - 30y = -240
2y = 12
y = 12 / 2
y = 6

Finally, we can substitute the value of y back into the equation x = (8y - 63) / 5 to solve for x:
x = (8(6) - 63) / 5
x = (48 - 63) / 5
x = -15 / 5
x = -3

Therefore, the solution to the system of equations is x = -3 and y = 6.