Solve using elimination.
8x − 3y = –10
–8x + 6y = –20
To solve using elimination, let's multiply the first equation by 2 and the second equation by 3 to make the x coefficients the same but opposite in sign:
Equation 1: 16x - 6y = -20
Equation 2: -24x + 18y = -60
Now, add the two equations together:
16x - 6y + (-24x + 18y) = -20 - 60
-8x + 12y = -80
Now, let's divide the new equation by 4 to simplify:
-2x + 3y = -20
Now we have simplified the system of equations to just one equation:
-2x + 3y = -20
This equation does not represent the same system as the original equations, so there seems to be a mistake in the calculation. Let's start over and try again:
Equation 1: 8x - 3y = -10
Equation 2: -8x + 6y = -20
Let's eliminate x by adding the two equations:
(8x - 3y) + (-8x + 6y) = -10 - 20
3y = -30
y = -10
Now substitute y = -10 into one of the original equations to solve for x:
8x - 3(-10) = -10
8x + 30 = -10
8x = -40
x = -5
Therefore, the solution to the system of equations is x = -5 and y = -10.
what is the answer? Sorry got a little confused.
The solution to the system of equations is x = -5 and y = -10.
You have to use this (x,y) metthod thing it's only one answer to.
You have to use this (x,y) metthod thing it's only one answer to
I apologize for the confusion earlier. Let's solve the system of equations using the elimination method again:
Equation 1: 8x - 3y = -10
Equation 2: -8x + 6y = -20
Let's add the two equations together to eliminate x:
(8x - 3y) + (-8x + 6y) = -10 - 20
3y = -30
y = -10
Now that we have the value of y, we substitute it back into Equation 1 to find the value of x:
8x - 3(-10) = -10
8x + 30 = -10
8x = -40
x = -5
Therefore, the solution to the system of equations is x = -5 and y = -10. So, the answer is (-5, -10). Thank you for pointing that out.