Solve the system of two equations using elimination with multiplication. 4x-6y=-48. -2x+4y=30
To solve the system of equations using elimination with multiplication, we want to eliminate one of the variables by multiplying one or both of the equations by a constant that will result in the coefficients of one of the variables being additive inverses.
To eliminate the variable "x," let's multiply the first equation by 2 and the second equation by 4:
(2)(4x - 6y) = (2)(-48)
(4)(-2x + 4y) = (4)(30)
This simplifies to:
8x - 12y = -96
-8x + 16y = 120
Now, we can add the two equations together to eliminate the variable "x":
(8x - 12y) + (-8x + 16y) = -96 + 120
This simplifies to:
4y = 24
Dividing both sides of the equation by 4 gives us:
y = 6
Now that we have the value of "y," we can substitute it back into one of the original equations to solve for "x." Let's use the first equation:
4x - 6(6) = -48
Simplifying further, we have:
4x - 36 = -48
Adding 36 to both sides of the equation gives us:
4x = -12
Dividing both sides of the equation by 4 gives us:
x = -3
So, the solution to the system of equations is x = -3 and y = 6.