Problem 19 - Entrance Test

We are given the system of equations: 1) x + 1/y = 4 2) y + 1/x = 1 We want to find the value of xy. From equation (1), multiply by y (assuming y is not 0): xy + 1 = 4y xy = 4y - 1 (Equation 3) From equation (2), multiply by x (assuming x is not 0): xy + 1 = x xy = x - 1 (Equation 4) Now we have two expressions for xy. We can set them equal to each other: 4y - 1 = x - 1 4y = x. Substitute x = 4y into Equation 2: y + 1/(4y) = 1. To eliminate the fraction, multiply the entire equation by 4y (assuming y is not 0): 4y^2 + 1 = 4y. Rearrange this into a quadratic equation: 4y^2 - 4y + 1 = 0. This is a perfect square trinomial: (2y - 1)^2 = 0. Solving for y: 2y - 1 = 0 2y = 1 y = 1/2. Now find x using x = 4y: x = 4 * (1/2) = 2. Finally, find the value of xy: xy = 2 * (1/2) = 1. We can check these values in the original equations: 1) x + 1/y = 2 + 1/(1/2) = 2 + 2 = 4 (Correct) 2) y + 1/x = 1/2 + 1/2 = 1 (Correct) Both equations are satisfied.