Solve the systems of equations by elimination
3x-5y = 29
2x+3y = -6
To solve the system of equations by elimination, we can multiply the first equation by 2 and the second equation by 3 to get their coefficients for x to be the same:
2(3x-5y) = 2(29)
3(2x+3y) = 3(-6)
This simplifies to:
6x - 10y = 58
6x + 9y = -18
Now, we can subtract the second equation from the first equation to eliminate x:
(6x - 10y) - (6x + 9y) = 58 - (-18)
6x - 10y - 6x - 9y = 58 + 18
-19y = 76
y = 76 / (-19)
y = -4
Substitute this value of y back into one of the original equations, let's choose the second equation:
2x + 3(-4) = -6
2x - 12 = -6
2x = -6 + 12
2x = 6
x = 6 / 2
x = 3
So, the solution to the system of equations is x = 3 and y = -4.