To solve the system of equations, we will use the method of substitution.
From the first equation, we can isolate x by adding 4y to both sides:
3x - 4y = 40
3x = 4y + 40
x = (4y + 40)/3
Now we can substitute x in the second equation with (4y + 40)/3:
5((4y + 40)/3) - 5y = 55
Multiply both sides of the equation by 3 to eliminate the fraction:
5(4y + 40) - 15y = 165
20y + 200 - 15y = 165
5y + 200 = 165
5y = 165 - 200
5y = -35
y = -35/5
y = -7
Now we can substitute y back into the first equation to solve for x:
3x - 4(-7) = 40
3x + 28 = 40
3x = 40 - 28
3x = 12
x = 12/3
x = 4
Therefore, the solution to the system of equations is (4, -7).
Solve the system of equations.
3x−4y=40
5x−5y=55 (1 point)
Responses
(4,−7)
left parenthesis 4 comma negative 7 right parenthesis
(8,−3)
left parenthesis 8 comma negative 3 right parenthesis
(18, 7)
left parenthesis 18 comma 7 right parenthesis
(14, 3)
1 answer
1 answer