{3х-у=5 {х²+у=13

To solve this system of equations, we can use the method of substitution or the method of elimination. Let's use the method of substitution.

From the first equation, we can isolate x:

3x - y = 5 => 3x = y + 5 => x = (y + 5)/3

Now we substitute this value of x into the second equation:

x² + y = 13 => ((y + 5)/3)² + y = 13

Expanding the square:

((y² + 10y + 25)/9) + y = 13 => (y² + 10y + 25) + 9y = 117 => y² + 19y + 25 = 117 => y² + 19y - 92 = 0

Now we can solve this quadratic equation. Using factoring, the equation can be written as:

(y + 23)(y - 4) = 0

Setting each term to zero, we have:

y + 23 = 0 or y - 4 = 0 => y = -23 or y = 4

Substituting these values back into the expression for x, we get:

x = (y + 5)/3 For y = -23: x = (-23 + 5)/3 = -6 For y = 4: x = (4 + 5)/3 = 3

So the solutions to the system of equations are x = -6, y = -23 and x = 3, y = 4.