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 (−33,−9) left parenthesis negative 33 comma negative 9 right parenthesis (3, 9) left parenthesis 3 comma 9 right parenthesis (−3, 6)
To solve the system of equations using elimination with multiplication, we must first choose a variable to eliminate. In this case, we can eliminate the variable "x" by multiplying the first equation by 2 and the second equation by 4.
By multiplying the first equation by 2, we get 8x - 12y = -96.
By multiplying the second equation by 4, we get -8x + 16y = 120.
Now we can add these two equations together to eliminate the variable "x":
(8x - 12y) + (-8x + 16y) = (-96) + 120
-12y + 16y = 24
4y = 24
y = 6
Substitute the value of "y" = 6 back into either equation to find the value of "x". Let's use the first equation:
4x - 6(6) = -48
4x - 36 = -48
4x = -48 + 36
4x = -12
x = -12/4
x = -3
Therefore, the solution to the system of equations is (-3, 6).
To solve the system of equations using elimination with multiplication, we need to eliminate one variable by multiplying one or both equations by suitable numbers.
Let's start by trying to eliminate the variable "x".
The coefficients of "x" in the given equations are 4 and -2. To eliminate the variable "x", we need to find a common multiple for these coefficients.
The common multiple of 4 and -2 is 4.
Now, multiply the first equation by 1 and the second equation by -2 so that the coefficients of "x" become opposites:
(1) 4x - 6y = -48
(2) -2x + 4y = 30
Multiplying equation (1) by 1 gives us:
(3) 4x - 6y = -48
Multiplying equation (2) by -2 gives us:
(4) 4x - 8y = -60
Now, subtract equation (4) from equation (3) to eliminate the variable "x":
(3) - (4) -> (5) -6y + 8y = -48 - (-60)
Simplifying equation (5), we get:
2y = 12
Dividing both sides of equation (5) by 2, we get:
y = 6
Now, substitute the value of y = 6 into one of the original equations to find the value of x. Let's choose equation (1):
4x - 6(6) = -48
4x - 36 = -48
Adding 36 to both sides of the equation, we get:
4x = -12
Dividing both sides of the equation by 4, we get:
x = -3
Therefore, the solution to the system of equations is x = -3 and y = 6.
Answer: (-3, 6)
To solve the given system of equations using elimination with multiplication, we need to eliminate one of the variables by manipulating the equations. Here's the step-by-step process:
Step 1: Choose one variable to eliminate. In this case, let's eliminate the variable "x".
Step 2: Multiply both sides of the first equation by a number to make the coefficients of "x" in both equations equal. We want the coefficients of "x" to be opposites of each other. To do this, we can multiply the first equation by "-2". This results in:
-2(4x - 6y) = -2(-48)
-8x + 12y = 96
Step 3: Write down the second equation as it is:
-2x + 4y = 30
Step 4: Add the two equations together. This will eliminate the "x" variable:
(-8x + 12y) + (-2x + 4y) = 96 + 30
-8x - 2x + 12y + 4y = 126
-10x + 16y = 126
Step 5: Solve the resulting equation for the remaining variable, in this case, "y". Rearrange the equation:
-10x + 16y = 126
16y = 10x + 126
y = (10/16)x + 126/16
y = (5/8)x + 63/8
Step 6: Substitute the value of "y" back into one of the original equations. Let's use the second equation:
-2x + 4((5/8)x + 63/8) = 30
-2x + (20/8)x + 63/2 = 30
-2x + (5/2)x + 63/2 = 30
(-4/2)x + (5/2)x + 63/2 = 30
(1/2)x + 63/2 = 30
(1/2)x = 30 - 63/2
(1/2)x = 60/2 - 63/2
(1/2)x = -3/2
x = (-3/2)(2/1)
x = -3
Step 7: Substitute the value of "x" into either equation to solve for "y". Let's use the first equation:
4(-3) - 6y = -48
-12 - 6y = -48
-6y = -48 + 12
-6y = -36
y = (-36)(-1/6)
y = 6
Therefore, the solution to the system of equations is (x, y) = (-3, 6).