X-y=1 xy=12

To solve the system of equations given by: 1. \( x - y = 1 \) 2. \( xy = 12 \) We can use substitution or elimination. Here, I'll use substitution: From the first equation, we can express \( x \) in terms of \( y \): \[ x = y + 1 \] Now, substitute \( x \) into the second equation: \[ (y + 1)y = 12 \] Expanding this gives: \[ y^2 + y = 12 \] Rearranging to form a standard quadratic equation: \[ y^2 + y - 12 = 0 \] Next, we can factor this quadratic equation: \[ (y - 3)(y + 4) = 0 \] Setting each factor equal to zero gives us the solutions for \( y \): 1. \( y - 3 = 0 \) → \( y = 3 \) 2. \( y + 4 = 0 \) → \( y = -4 \) Now we can find the corresponding values of \( x \) for each value of \( y \): 1. For \( y = 3 \): \[ x = y + 1 = 3 + 1 = 4 \] So one solution is \( (x, y) = (4, 3) \). 2. For \( y = -4 \): \[ x = y + 1 = -4 + 1 = -3 \] So the other solution is \( (x, y) = (-3, -4) \). Thus, the two pairs \((x, y)\) that solve the system of equations are: 1. \((4, 3)\) 2. \((-3, -4)\)