First, let's isolate one variable in one of the equations. Let's multiply the first equation by 5 and the second one by 3 to make the coefficients of y the same:
15x - 20y = 200
15x - 15y = 165
Now we can subtract the second new equation from the first new equation:
15x - 20y - 15x + 15y = 200 - 165
-5y = 35
y = -7
Now, substitute y = -7 into one of the original equations to solve for x:
3x - 4(-7) = 40
3x + 28 = 40
3x = 12
x = 4
Therefore, the solution to the system of equations is (4, -7).
Solve the system of equations.
3x−4y=40
5x−5y=55
Responses
(8,−3)
open paren 8 comma negative 3 close paren - no response given
(18, 7)
(18, 7)
- no response given
(14, 3)
(14, 3)
- no response given
(4,−7)
show your work
1 answer