solve the system of two equations using elimination with multiplication
4x - 6y= -48
-2x + 4y = 30
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-terms or y-terms are the same. Then, we can add or subtract the equations to eliminate one of the variables.
In this case, let's multiply the first equation by 2 to make the coefficients of the x-terms the same:
2(4x - 6y) = 2(-48)
8x - 12y = -96
Now, let's write down the two equations after multiplying the first equation:
8x - 12y = -96
-2x + 4y = 30
To eliminate the y-term, we can multiply the second equation by 3 so that the coefficients of the y-terms are the same:
3(-2x + 4y) = 3(30)
-6x + 12y = 90
Now, let's write down the two equations after multiplying the second equation:
8x - 12y = -96
-6x + 12y = 90
Now, we can add the two equations together to eliminate the y-variable:
(8x - 12y) + (-6x + 12y) = (-96) + (90)
8x - 6x - 12y + 12y = -6
2x = -6
Dividing both sides by 2, we get:
x = -3
Substituting this value of x back into one of the original equations, let's use the first equation:
4x - 6y = -48
4(-3) - 6y = -48
-12 - 6y = -48
-6y = -48 + 12
-6y = -36
Dividing both sides by -6, we get:
y = 6
Therefore, the solution to the system of equations is x = -3 and y = 6.
To solve the system of equations using elimination with multiplication, we need to eliminate one of the variables when we add or subtract the two equations.
Let's eliminate the variable x by multiplying both sides of the second equation by 2.
Original equations:
1) 4x - 6y = -48
2) -2x + 4y = 30
Multiply equation 2 by 2:
3) -4x + 8y = 60
Now, add equation 1 and equation 3 together:
4) (4x - 6y) + (-4x + 8y) = -48 + 60
-6y + 8y = 12
2y = 12
Next, solve for y by dividing both sides of the equation by 2:
5) y = 6
Now, substitute the value of y back into one of the original equations. Let's use equation 1:
4x - 6(6) = -48
4x - 36 = -48
Solve for x:
4x = -48 + 36
4x = -12
x = -12/4
x = -3
The solution to the system of equations is x = -3 and y = 6.
To solve the system of equations using elimination with multiplication, we want to manipulate one or both of the equations by multiplying them by a constant so that when we add or subtract them, one of the variables will be eliminated.
1. Let's start by choosing a variable to eliminate. Looking at the equations, it seems like eliminating the variable "x" would be easier. To do that, we need to make the x-coefficient in one of the equations the same or opposite as the x-coefficient in the other equation.
2. Looking at the equations again, we see that the coefficient of x in the first equation is 4, while the coefficient of x in the second equation is -2. To make these two coefficients opposite, we need to multiply both sides of the second equation by 2.
Multiplying the second equation by 2, we have:
-2x + 4y = 30
2 * (-2x + 4y) = 2 * 30
-4x + 8y = 60
3. Now that we have two equations with opposite coefficients for x, we can add the equations together to eliminate x.
Adding the two equations, we have:
(4x - 6y) + (-4x + 8y) = (-48) + 60
4x - 6y - 4x + 8y = 12
(4x - 4x) + (-6y + 8y) = 12
0x + 2y = 12
2y = 12
4. Now, solve for y. Divide both sides of the equation by 2:
2y/2 = 12/2
y = 6
5. We have found the value of y, which is 6. Now, substitute this value back into either of the original equations to find the value of x. Let's substitute it into the first equation:
4x - 6y = -48
4x - 6(6) = -48
4x - 36 = -48
4x = -48 + 36
4x = -12
6. Solve for x by dividing both sides of the equation by 4:
4x/4 = -12/4
x = -3
7. So, the solution to the system of equations is x = -3 and y = 6.