To solve the system of equations using elimination with multiplication, we need to make the coefficients of either x or y in both equations the same. In this case, we'll focus on making the coefficients of x the same.
Using multiplication, we can make the coefficient of x in the first equation equal to -4 (since -4 * -2 = 8, which is the coefficient of x in the second equation).
Multiplying both sides of the first equation by -2:
-2(4x - 6y) = -2(-48)
-8x + 12y = 96
Now we have the system of equations:
-8x + 12y = 96
-2x + 4y = 30
Next, we can multiply both sides of the second equation by 2 to make the coefficients of x in both equations the same:
2(-2x + 4y) = 2(30)
-4x + 8y = 60
Now we have the system of equations:
-8x + 12y = 96
-4x + 8y = 60
The coefficients of x are now the same, so we can subtract the second equation from the first equation to eliminate x:
(-8x + 12y) - (-4x + 8y) = 96 - 60
-8x + 12y + 4x - 8y = 36
-4x + 4y = 36
Divide both sides of this equation by -4 to solve for y:
(-4x + 4y) / -4 = 36 / -4
y = -9
Now that we know y = -9, we can substitute this value back into one of the original equations to solve for x. Let's use the first equation:
4x - 6(-9) = -48
4x + 54 = -48
4x = -48 - 54
4x = -102
x = -102 / 4
x = -25.5
Therefore, the solution to the system of equations is x = -25.5 and y = -9.
Solve the system of two equations using elimination with multiplication.
4x−6y=−48
−2x+4y=30
1 answer