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.