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 when we add or subtract the two equations, one of the variables will be eliminated.
To eliminate the x-term, we can multiply the second equation by 2:
(2)(-2x + 4y) = (2)(30)
This gives us:
-4x + 8y = 60
Now we can add this equation to the first equation in order to eliminate the x-term:
4x - 6y + (-4x + 8y) = -48 + 60
This simplifies to:
2y = 12
Now we can solve for y by dividing both sides of the equation by 2:
2y/2 = 12/2
y = 6
Substituting this value of y back into one of the original equations, we can solve for x.
Using the first equation:
4x - 6(6) = -48
4x - 36 = -48
4x = -12
x = -3
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 want to eliminate one variable by multiplying one or both equations by appropriate constants so that the coefficients of either x or y in both equations become additive inverses (i.e., add up to zero).
Let's start by multiplying the second equation by 2 to eliminate the x term:
2(-2x + 4y) = 2(30)
-4x + 8y = 60
Now, we have:
4x - 6y = -48
-4x + 8y = 60
Notice that the x terms have opposite coefficients: 4x and -4x. If we add the two equations, the x term will be eliminated:
(4x - 6y) + (-4x + 8y) = -48 + 60
4x - 6y - 4x + 8y = 12
Simplifying, we get:
2y = 12
Now, we need to solve for y by dividing both sides of the equation by 2:
2y/2 = 12/2
y = 6
Substituting this value of y into one of the original equations, let's use the first equation:
4x - 6(6) = -48
4x - 36 = -48
Next, let's isolate the x term by adding 36 to both sides:
4x - 36 + 36 = -48 + 36
4x = -12
Now, we divide both sides by 4 to solve for x:
4x/4 = -12/4
x = -3
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 want to eliminate one of the variables by multiplying one or both of the equations so that the coefficients of one of the variables will cancel out when adding or subtracting the equations.
Let's start by multiplying the second equation by 2 to eliminate the coefficient of "x":
Original equations:
1) 4x - 6y = -48
2) -2x + 4y = 30
Multiply equation 2 by 2:
3) -4x + 8y = 60
Now, we will add equation 1 and equation 3 together:
1) 4x - 6y = -48
3) -4x + 8y = 60
When we add these equations, the x terms will cancel out:
4) (4x + (-4x)) + (-6y + 8y) = (-48 + 60)
Simplifying equation 4, we have:
0 + 2y = 12
Now, we solve for y:
2y = 12
y = 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 - 6y = -48
Substituting y = 6:
4x - 6(6) = -48
4x - 36 = -48
Next, we isolate x by adding 36 to both sides:
4x = -48 + 36
4x = -12
Dividing by 4:
x = -12/4
x = -3
Therefore, the solution to the system of equations is x = -3 and y = 6.
To check our solution, simply substitute the values of x and y into both original equations and see if they hold true.