2x+y=13

3x²-2y²=125

To solve this system of equations, we can use the method of substitution.

First, let's solve the first equation for y.
2x + y = 13
Subtract 2x from both sides:
y = 13 - 2x

Now we substitute this expression for y into the second equation.
3x² - 2(13 - 2x)² = 125
Simplify the expression inside the parentheses:
3x² - 2(-4x² + 52x - 169) = 125
Expand and simplify:
3x² - (-8x² + 104x - 338) = 125
3x² + 8x² - 104x + 338 = 125
Combine like terms:
11x² - 104x + 213 = 125
Subtract 125 from both sides:
11x² - 104x + 88 = 0

Now we can solve this quadratic equation by factoring or using the quadratic formula.
If we factor it, we get:
(11x - 8)(x - 11) = 0

Setting each factor equal to zero, we have:
11x - 8 = 0 or x - 11 = 0

For the first equation, adding 8 to both sides:
11x = 8
Dividing both sides by 11:
x = 8/11

For the second equation, adding 11 to both sides:
x = 11

So, we have two possible solutions:
x = 8/11 and x = 11

Now, substitute each value of x back into either of the original equations to solve for y.

Using x = 8/11:
2(8/11) + y = 13
Multiplying through by 11:
16 + 11y = 143
Subtracting 16 from both sides:
11y = 127
Dividing both sides by 11:
y = 127/11

Using x = 11:
2(11) + y = 13
22 + y = 13
Subtracting 22 from both sides:
y = -9

So, the two solutions to the system of equations are:
(8/11, 127/11) and (11, -9)