Solve the systems of equations by elimination
3x-5y = 29
2x+3y = -6
To solve the system of equations by elimination, we can multiply each equation by a constant in order to create terms that cancel each other out when we add them together.
First, let's multiply the second equation by 5:
5 * (2x + 3y) = 5 * (-6)
10x + 15y = -30
Now we have the following system of equations:
3x - 5y = 29
10x + 15y = -30
Next, we can add the two equations together to eliminate the y terms:
(3x - 5y) + (10x + 15y) = 29 + (-30)
3x + 10x - 5y + 15y = -1
13x + 10y = -1
Now we have a new equation:
13x + 10y = -1
To eliminate the x terms, we can multiply the first equation by 10:
10 * (3x - 5y) = 10 * 29
30x - 50y = 290
Now we have the following system of equations:
30x - 50y = 290
13x + 10y = -1
Next, we can add the two equations together to eliminate the x terms:
(30x - 50y) + (13x + 10y) = 290 + (-1)
30x + 13x - 50y + 10y = 289
43x - 40y = 289
Now we have a new equation:
43x - 40y = 289
So, the system of equations has been transformed into the single equation:
43x - 40y = 289
Now, we can solve this equation to find the values of x and y. However, since it is not possible to solve for x and y with this equation alone, we need more information or another equation to find the unique solution to the system.