y²-x=14 x-y=-2

To solve the system of equations:

1. \( y^2 - x = 14 \) (Equation 1)
2. \( x - y = -2 \) (Equation 2)

First, we can rearrange Equation 2 for \( x \):

\[ x = y - 2 \]

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

\[ y^2 - (y - 2) = 14 \]

This simplifies to:

\[ y^2 - y + 2 = 14 \]

Now, move 14 to the left side:

\[ y^2 - y + 2 - 14 = 0 \]

\[ y^2 - y - 12 = 0 \]

Next, we will factor the quadratic equation:

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

This gives us two solutions for \( y \):

\[ y - 4 = 0 \quad \Rightarrow \quad y = 4 \]

\[ y + 3 = 0 \quad \Rightarrow \quad y = -3 \]

Now we can find the corresponding values for \( x \) using \( x = y - 2 \).

1. For \( y = 4 \):

\[ x = 4 - 2 = 2 \]

2. For \( y = -3 \):

\[ x = -3 - 2 = -5 \]

Thus, the two solutions to the system of equations are:

1. \( (x, y) = (2, 4) \)
2. \( (x, y) = (-5, -3) \)

These are the pairs that satisfy both equations.