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. \( (-4, -11) \)