x^2+y=5 x-y=7

To solve the system of equations:

1. \( x^2 + y = 5 \) (Equation 1)
2. \( x - y = 7 \) (Equation 2)

We can start by solving Equation 2 for \( y \):

\[ y = x - 7 \]

Now, we can substitute this expression for \( y \) back into Equation 1:

\[ x^2 + (x - 7) = 5 \]

This simplifies to:

\[ x^2 + x - 7 = 5 \]

Next, we can move the 5 to the left side of the equation:

\[ x^2 + x - 12 = 0 \]

Now, we can factor the quadratic equation:

\[ (x - 3)(x + 4) = 0 \]

This gives us two potential solutions for \( x \):

1. \( x - 3 = 0 \) → \( x = 3 \)
2. \( x + 4 = 0 \) → \( x = -4 \)

Now we can find corresponding values of \( y \) for each \( x \).

For \( x = 3 \):

Substitute \( x = 3 \) into Equation 2:

\[ y = 3 - 7 = -4 \]

So, one solution is \( (3, -4) \).

For \( x = -4 \):

Substitute \( x = -4 \) into Equation 2:

\[ y = -4 - 7 = -11 \]

So, the second solution is \( (-4, -11) \).

Thus, the solutions to the system of equations are:

1. \( (3, -4) \)
2. \( (-4, -11) \)