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 make the coefficients of either x or y in both equations the same. In this case, we'll focus on making the coefficients of x the same.
Using multiplication, we can make the coefficient of x in the first equation equal to -4 (since -4 * -2 = 8, which is the coefficient of x in the second equation).
Multiplying both sides of the first equation by -2:
-2(4x - 6y) = -2(-48)
-8x + 12y = 96
Now we have the system of equations:
-8x + 12y = 96
-2x + 4y = 30
Next, we can multiply both sides of the second equation by 2 to make the coefficients of x in both equations the same:
2(-2x + 4y) = 2(30)
-4x + 8y = 60
Now we have the system of equations:
-8x + 12y = 96
-4x + 8y = 60
The coefficients of x are now the same, so we can subtract the second equation from the first equation to eliminate x:
(-8x + 12y) - (-4x + 8y) = 96 - 60
-8x + 12y + 4x - 8y = 36
-4x + 4y = 36
Divide both sides of this equation by -4 to solve for y:
(-4x + 4y) / -4 = 36 / -4
y = -9
Now that we know y = -9, we can substitute this value back into one of the original equations to solve for x. Let's use the first equation:
4x - 6(-9) = -48
4x + 54 = -48
4x = -48 - 54
4x = -102
x = -102 / 4
x = -25.5
Therefore, the solution to the system of equations is x = -25.5 and y = -9.
To solve the system of equations using elimination with multiplication, we need to eliminate one of the variables by multiplying both equations by appropriate constants so that the coefficients of one of the variables becomes the same in both equations.
In this case, let's eliminate the variable x by multiplying the first equation by -2 and the second equation by 4:
-2(4x - 6y) = -2(-48)
4(-2x + 4y) = 4(30)
This gives us:
-8x + 12y = 96
-8x + 16y = 120
Now we have two equations with the same coefficient for x, which allows us to subtract one equation from the other to eliminate x. Subtracting the second equation from the first gives:
(-8x + 12y) - (-8x + 16y) = 96 - 120
-8x + 12y + 8x - 16y = -24
-4y = -24
Dividing both sides of the equation by -4, we get:
y = -24 / -4
y = 6
Now, substitute the value of y back into one of the original equations, let's use the first equation:
4x - 6(6) = -48
4x - 36 = -48
Add 36 to both sides:
4x = -12
Divide both sides by 4:
x = -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 will multiply one or both of the equations by suitable coefficients to create cancelling terms when adding or subtracting the equations. Let's begin:
Let's start by multiplying the second equation by 2 to eliminate the coefficient of 'x':
2 * (-2x + 4y) = 2 * 30
-4x + 8y = 60
Now, we have the system of equations:
4x - 6y = -48
-4x + 8y = 60
Adding the two equations together, we can eliminate the 'x' terms:
(4x - 6y) + (-4x + 8y) = -48 + 60
-6y + 8y = 12
Simplifying and combining like terms:
2y = 12
Now, we can solve for 'y' by dividing both sides of the equation by 2:
2y/2 = 12/2
y = 6
To find the value of 'x', substitute the value of 'y' (6) back into one of the original equations. Let's use the first equation:
4x - 6y = -48
4x - 6(6) = -48
4x - 36 = -48
Now, add 36 to both sides of the equation:
4x - 36 + 36 = -48 + 36
4x = -12
Finally, 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 = -3 and y = 6.