Solve the system of two equations using elimination with multiplication. 4x−6y=−48 −2x+4y=30 (1 point) Responses (3, 9) left parenthesis 3 comma 9 right parenthesis (−21,−6) left parenthesis negative 21 comma negative 6 right parenthesis (−33,−9) left parenthesis negative 33 comma negative 9 right parenthesis (−3, 6)
To solve the system of equations using elimination with multiplication, we need to multiply one or both equations by a constant so that when we add or subtract the equations, one of the variables will cancel out.
Let's first multiply the second equation by 2 to eliminate the x terms:
2(-2x+4y)=2(30)
-4x+8y=60
Now we have the following system of equations:
4x-6y=-48
-4x+8y=60
When we add these two equations together, the x terms will cancel out:
(4x-4x) + (-6y+8y) = (-48+60)
0 + 2y = 12
2y = 12
Now we can solve for y by dividing both sides of the equation by 2:
2y/2 = 12/2
y = 6
We have found the value of y, which is 6. Now we can substitute this value back into one of the original equations to solve for x. Let's use the first equation:
4x - 6(6) = -48
4x - 36 = -48
To isolate x, we add 36 to both sides of the equation:
4x - 36 + 36 = -48 + 36
4x = -12
Now we can solve for x by dividing both sides of the equation by 4:
4x/4 = -12/4
x = -3
Therefore, the solution to the system of equations is (x, y) = (-3, 6).
To solve the system of equations using elimination with multiplication, we need to eliminate one of the variables by manipulating one or both equations.
First, let's multiply the second equation by 2 to eliminate the x-term:
-2x + 4y = 30
2(-2x + 4y) = 2(30)
-4x + 8y = 60
Now, we can add the two equations together to eliminate the x-term:
4x - 6y + -4x + 8y = -48 + 60
-6y + 8y = 12
2y = 12
Next, we solve for y by dividing both sides of the equation by 2:
2y/2 = 12/2
y = 6
Now that we have the value of y, we can substitute it back into either of the original equations to solve for x. Let's use the first equation:
4x - 6y = -48
Substitute y = 6 into the equation:
4x - 6(6) = -48
4x - 36 = -48
Now, solve for x:
4x - 36 + 36 = -48 + 36
4x = -12
x = -12/4
x = -3
Therefore, the solution to the system of equations is (x, y) = (-3, 6).
To solve the system of equations using elimination with multiplication, we need to eliminate one of the variables by multiplying one or both of the equations by a suitable constant.
Let's start by multiplying the second equation by 2 to eliminate the coefficients of x:
-2x + 4y = 30
2*(-2x + 4y) = 2*30
-4x + 8y = 60 (equation 2)
Now we have two equations:
4x - 6y = -48 (equation 1)
-4x + 8y = 60 (equation 2)
We can add both equations to cancel out the x terms:
(4x - 6y) + (-4x + 8y) = -48 + 60
This gives us:
2y = 12
To isolate y, we divide both sides of the equation by 2:
2y/2 = 12/2
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 equation 1:
4x - 6(6) = -48
Simplifying:
4x - 36 = -48
We then add 36 to both sides:
4x = -48 + 36
4x = -12
Lastly, we divide by 4 to isolate x:
4x/4 = -12/4
x = -3
Therefore, the solution to the system of equations is x = -3 and y = 6, or (-3, 6) in coordinate form.
The correct answer is: (−3, 6)