Solve the system of two equations using elimination with multiplication 4x-6y=-48 and -2x+4y=30
To solve the system of equations using elimination with multiplication, we need to eliminate one variable by multiplying one (or both) of the equations by a constant so that the coefficients of that variable in both equations become additive inverses of each other.
Let's start by multiplying the first equation by 2 and the second equation by 3 to get:
(2)(4x - 6y) = (2)(-48)
(3)(-2x + 4y) = (3)(30)
Simplifying these equations, we get:
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
Combining like terms, we have:
2x = -6
Dividing both sides by 2, we find:
x = -3
Now, we substitute this value of x into one of the original equations to find the corresponding value of y:
4x - 6y = -48
4(-3) - 6y = -48
-12 - 6y = -48
Adding 12 to both sides:
-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 multiply one or both equations by a factor(s) to make the coefficients of one of the variables equal in magnitude but opposite in signs. In this case, we'll eliminate the variable 'x'.
Let's start by multiplying the second equation by 2, which will make the coefficient of 'x' in both equations equal to -4:
2 * (-2x + 4y) = 2 * 30
-4x + 8y = 60
Now, we have the following system of equations:
4x - 6y = -48
-4x + 8y = 60
Adding the two equations together will eliminate the 'x' term:
(4x - 6y) + (-4x + 8y) = -48 + 60
4x - 4x - 6y + 8y = 12
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) into one of the original equations. Let's use the first equation:
4x - 6y = -48
4x - 6(6) = -48
4x - 36 = -48
Now, isolate 'x' by adding 36 to both sides of the equation:
4x - 36 + 36 = -48 + 36
4x = -12
Finally, divide both sides of the equation 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 need to manipulate the equations so that the coefficients of either x or y are the same (but with opposite signs when applying elimination).
The given system of equations is:
1) 4x - 6y = -48
2) -2x + 4y = 30
If we multiply both sides of equation 2 by 2, we can make the coefficient of x in the second equation a -4, which is the opposite of 4 from the first equation. So, let's perform this multiplication:
2) 2(-2x + 4y) = 2(30)
-4x + 8y = 60
Now we have the system of equations:
1) 4x - 6y = -48
2) -4x + 8y = 60
Next, we will add these equations together to eliminate the x term:
(4x - 6y) + (-4x + 8y) = -48 + 60
4x - 6y - 4x + 8y = 12
The x terms cancel out, and we are left with:
2y = 12
To solve for y, we divide both sides of the equation by 2:
2y/2 = 12/2
y = 6
Now that we have found the value of y, we can substitute it back into either of the original equations to solve for x. Let's substitute y = 6 into equation 1:
4x - 6(6) = -48
4x - 36 = -48
Then, we isolate x by moving the constant term to the other side:
4x = -48 + 36
4x = -12
Finally, we divide both sides of the equation by 4 to solve for x:
4x/4 = -12/4
x = -3
Thus, the solution to the system of equations is x = -3 and y = 6.