Find the solution of the system of equations.
15, x, minus, 4, y, equals, minus, 50
15x−4y=−50
3, x, minus, 2, y, equals, minus, 16
3x−2y=−16
Answer
Attempt 1 out of 2
To solve the system of equations:
1) Start by multiplying the first equation by 2 and the second equation by 3 to create coefficients that can be canceled out when the equations are added together.
2 * (15x - 4y) = 2 * (-50)
30x - 8y = -100
3 * (3x - 2y) = 3 * (-16)
9x - 6y = -48
2) Now, add the two equations together to cancel out the y term:
(30x - 8y) + (9x - 6y) = -100 + (-48)
39x - 14y = -148
3) Solve for x by isolating it:
39x = -148 + 14y
39x = 14y - 148
x = (14y - 148)/39
4) Substitute this value of x into one of the original equations to solve for y.
15x - 4y = -50
15((14y - 148)/39) - 4y = -50
(210y - 2220)/39 - (4y * 39)/39 = -50
(210y - 4y * 39 - 2220)/39 = -50
(210y - 156y - 2220)/39 = -50
(54y - 2220)/39 = -50
54y - 2220 = -50 * 39
54y - 2220 = -1950
54y = 270
y = 270/54
y = 5
5) Substitute the value of y back into one of the original equations to solve for x:
15x - 4(5) = -50
15x - 20 = -50
15x = -30
x = -30/15
x = -2
Therefore, the solution to the system of equations is x = -2 and y = 5.