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 (−3, 6) left parenthesis negative 3 comma 6 right parenthesis (−33,−9) left parenthesis negative 33 comma negative 9 right parenthesis
To solve the system of equations using elimination with multiplication, you want to eliminate one variable by multiplying one or both equations by a constant so that when you add or subtract the equations, one of the variables cancels out.
Let's start by multiplying the second equation by 2:
-2x + 4y = 30
(2)(-2x + 4y) = (2)(30)
-4x + 8y = 60
Now, you can add the two equations together:
(4x - 6y) + (-4x + 8y) = -48 + 60
-6y + 8y = 12
2y = 12
y = 6
Substitute this value of y back into one of the original equations, let's use the first equation:
4x - 6(6) = -48
4x - 36 = -48
4x = -12
x = -3
Therefore, the solution to the system of equations is (x, y) = (-3, 6). So the correct answer is (−3, 6).
To solve the system of equations using elimination with multiplication, we want to eliminate one variable by multiplying one or both equations so that the coefficients of either x or y become additive inverses.
Let's start with the given equations:
4x − 6y = −48 (Equation 1)
−2x + 4y = 30 (Equation 2)
We can eliminate x by multiplying Equation 2 by 2:
2(-2x + 4y) = 2(30)
-4x + 8y = 60 (Multiplying Equation 2 by 2)
Now we have:
4x − 6y = −48 (Equation 1)
-4x + 8y = 60 (Equation 3, obtained by multiplying Equation 2 by 2)
Next, we can add Equation 1 and Equation 3 to eliminate x:
(4x - 6y) + (-4x + 8y) = -48 + 60
-6y + 8y = 12
2y = 12
y = 6
Substituting y = 6 into Equation 1:
4x - 6(6) = -48
4x - 36 = -48
4x = -12
x = -12/4
x = -3
Therefore, the solution to the system of equations is:
(x, y) = (-3, 6)
So, the correct response is (−3, 6)
To solve the system of equations using elimination with multiplication, we need to eliminate one variable by manipulating the equations and then solve for the remaining variable.
Given the system of equations:
1) 4x - 6y = -48
2) -2x + 4y = 30
Let's eliminate the variable 'x' by multiplying equation 2 by 2 and equation 1 by -1. This will allow us to add the equations together and eliminate 'x'.
Multiply equation 2 by 2:
2 * (-2x + 4y) = 2 * 30
-4x + 8y = 60 (equation 3)
Multiply equation 1 by -1:
-1 * (4x - 6y) = -1 * (-48)
-4x + 6y = 48 (equation 4)
Now, let's add equations 3 and 4 together:
(-4x + 8y) + (-4x + 6y) = 60 + 48
-8x + 14y = 108 (equation 5)
Next, we'll solve equation 5 for 'y':
-8x + 14y = 108
14y = 8x + 108
y = (8/14)x + 108/14
y = (4/7)x + 54/7 (equation 6)
Now substitute equation 6 into either equation 1 or equation 2 to solve for 'x'.
Let's use equation 1:
4x - 6y = -48
Substitute the value of 'y' from equation 6:
4x - 6((4/7)x + 54/7) = -48
4x - (24/7)x - 324/7 = -48
(28/7)x - (24/7)x = -48 + 324/7
(4/7)x = (-48*7 + 324)/7
(4/7)x = (336 - 48 + 324)/7
(4/7)x = 612/7
(4/7)x = 612/7
Now, solve the equation for 'x':
x = (612/7) * (7/4)
x = 612/4
x = 153
Finally, substitute the value of 'x' back into equation 1 or equation 2 to find 'y'. Let's use equation 1:
4x - 6y = -48
4(153) - 6y = -48
612 - 6y = -48
-6y = -48 - 612
-6y = -660
y = (-660)/(-6)
y = 110
Therefore, the solution to the system of equations is (x, y) = (153, 110).