Solving Systems of Equations: An Algebra Pre-Calc Challenge

In the realm of algebra and pre-calculus, solving systems of equations can be a complex yet fascinating challenge. This article will guide you through the process of solving the given system of equations: xy 2 and y x 1. We will explore different cases and explain the reasoning behind each step.

Graphical Insight

Let's start by visualizing the system graphically. The line y x 1 is a simple straight line with a slope of 1 that passes through the point (0, 1).

The equation xy 2 is more complex. It represents all points (x, y) where the taxicab distance (the sum of the absolute differences in the x and y directions) is 2. This is a unique way to understand the equation and can be visualized as the boundary of a diamond shape centered at the origin.

Case by Case Analysis

Case 1: x 0

Let's start by examining the case when x 0.

Substitute x 0 into the equation y x 1, giving us y 1. Now, substitute x 0 and y 1 into the equation xy 2: 0 * 1 2, which is obviously false. Thus, there is no solution in this case.

Case 2: x ≠ 0 and y 0

Now, let's consider the case where x ≠ 0 and y 0.

Substitute y 0 into the equation y x 1, which gives us 0 x 1, or x -1. Now, substitute x -1 and y 0 into the equation xy 2: -1 * 0 2, which is also false. Therefore, there are no solutions in this case either.

Case 3: x ≠ 0 and y ≠ 0

Finally, let's analyze the general case where x ≠ 0 and y ≠ 0.

We have the system of equations: xy 2 y x 1 Substitute y x 1 into the first equation: x(x 1) 2. This simplifies to: x^2 x - 2 0. Solving this quadratic equation using the quadratic formula, we get: x (-b ± √(b^2 - 4ac)) / 2a Where a 1, b 1, c -2. x (-1 ± √(1^2 - 4 * 1 * -2)) / 2 * 1 x (-1 ± √9) / 2 x (-1 ± 3) / 2 This gives us two solutions: x 1 and x -2 For x 1: y 1 1 2. For x -2: y -2 1 -1. Thus, we have two solutions: (1, 2) and (-2, -1).

Conclusion

In summary, solving the system of equations xy 2 and y x 1 through a case-by-case analysis, we find two solutions: (1, 2) and (-2, -1). The first case considered x 0, but it yielded no solutions. The second case examined x ≠ 0 and y 0, which also yielded no solutions. The general case, where x ≠ 0 and y ≠ 0, provided the complete solution set.