This gives us two possible values for x, x = -6 and x = 3. We can now substitute these values back into the expression we found for y in terms of x to find the corresponding y values:
For x = -6: y = 3(-6) - 5 y = -18 - 5 y = -23
For x = 3: y = 3(3) - 5 y = 9 - 5 y = 4
Therefore, the solutions to the system of equations are x = -6, y = -23 and x = 3, y = 4.
To solve this system of equations, we can start by solving the first equation for y in terms of x:
3x - y = 5
y = 3x - 5
Next, substitute this expression for y into the second equation:
x^2 + (3x - 5) = 13
x^2 + 3x - 5 = 13
x^2 + 3x - 18 = 0
This quadratic equation can be factored as:
(x + 6)(x - 3) = 0
This gives us two possible values for x, x = -6 and x = 3. We can now substitute these values back into the expression we found for y in terms of x to find the corresponding y values:
For x = -6:
y = 3(-6) - 5
y = -18 - 5
y = -23
For x = 3:
y = 3(3) - 5
y = 9 - 5
y = 4
Therefore, the solutions to the system of equations are x = -6, y = -23 and x = 3, y = 4.