To solve the system of equations using elimination with multiplication, we need to multiply one or both of the equations by a constant so that the coefficients of either the x or y terms in both equations are opposite or equal.
In this case, we can multiply the first equation by 2 and the second equation by 3 to make the coefficients of x opposite:
8x - 12y = -96
-6x + 12y = 90
Now, we can add the two equations together:
8x - 6x - 12y + 12y = -96 + 90
2x = -6
x = -3
Substituting the value of x into the second equation:
-2(-3) + 4y = 30
6 + 4y = 30
4y = 30 - 6
4y = 24
y = 6
Therefore, the solution to the system of equations is (-3, 6).
Solve the system of two equations using elimination with multiplication.
4x−6y=−48
−2x+4y=30 (1 point)
Responses
(−21,−6)
left parenthesis negative 21 comma negative 6 right parenthesis
(3, 9)
left parenthesis 3 comma 9 right parenthesis
(−33,−9)
left parenthesis negative 33 comma negative 9 right parenthesis
(−3, 6)
1 answer