Y=x-1, x^2-2y=26.

To solve the system of equations given by: 1. \( y = x - 1 \) 2. \( x^2 - 2y = 26 \) we can substitute the expression for \( y \) from the first equation into the second equation. Substituting \( y \) into the second equation: \[ x^2 - 2(x - 1) = 26 \] Now, simplify the equation: \[ x^2 - 2x + 2 = 26 \] Subtract 26 from both sides: \[ x^2 - 2x + 2 - 26 = 0 \] This simplifies to: \[ x^2 - 2x - 24 = 0 \] Next, we will factor the quadratic equation: \[ (x - 6)(x + 4) = 0 \] Setting each factor to zero gives us: 1. \( x - 6 = 0 \) ⟹ \( x = 6 \) 2. \( x + 4 = 0 \) ⟹ \( x = -4 \) Now, we can find the corresponding \( y \) values for each \( x \): 1. For \( x = 6 \): \[ y = 6 - 1 = 5 \] 2. For \( x = -4 \): \[ y = -4 - 1 = -5 \] Thus, the solutions to the system of equations are: 1. \( (6, 5) \) 2. \( (-4, -5) \) So, the pairs \((x, y)\) are: \[ (6, 5) \quad \text{and} \quad (-4, -5) \]