Solve the system of equations: $|2x-1|-y=2$ $x-|4-y|=-1$

1. Problem Description

Solve the system of equations:

$|2x-1|-y=2$

$x-|4-y|=-1$

2. Solution Steps

First, we rewrite the equations:

$|2x-1| - y = 2$ (1)

$x - |4-y| = -1$ (2)

From (2), we have $|4-y| = x+1$. Since the absolute value is non-negative, we must have $x+1 \ge 0$, so $x \ge -1$.

We consider two cases for $|4-y|$:

Case 1: $4-y \ge 0$, i.e., $y \le 4$. Then $|4-y| = 4-y$, and $x - (4-y) = -1$, so $x - 4 + y = -1$, which implies $y = 3-x$.

Substituting $y = 3-x$ into (1), we have $|2x-1| - (3-x) = 2$, so $|2x-1| = 5-x$.

We now consider two subcases:

Subcase 1a: $2x-1 \ge 0$, i.e., $x \ge \frac{1}{2}$. Then $|2x-1| = 2x-1$, so $2x-1 = 5-x$, which gives $3x = 6$, so $x=2$. Then $y = 3-x = 3-2 = 1$.

Since $x=2 \ge \frac{1}{2}$ and $y=1 \le 4$, this solution is valid.

Subcase 1b: $2x-1 < 0$, i.e., $x < \frac{1}{2}$. Then $|2x-1| = -(2x-1) = 1-2x$, so $1-2x = 5-x$, which gives $-x = 4$, so $x = -4$. Then $y = 3-x = 3-(-4) = 7$.

Since $x=-4 < \frac{1}{2}$ but $y=7 > 4$, this solution is not valid because it violates $y \le 4$.

Case 2: $4-y < 0$, i.e., $y > 4$. Then $|4-y| = -(4-y) = y-4$, and $x - (y-4) = -1$, so $x - y + 4 = -1$, which implies $y = x+5$.

Substituting $y = x+5$ into (1), we have $|2x-1| - (x+5) = 2$, so $|2x-1| = x+7$.

We now consider two subcases:

Subcase 2a: $2x-1 \ge 0$, i.e., $x \ge \frac{1}{2}$. Then $|2x-1| = 2x-1$, so $2x-1 = x+7$, which gives $x = 8$. Then $y = x+5 = 8+5 = 13$.

Since $x=8 \ge \frac{1}{2}$ and $y=13 > 4$, this solution is valid.

Subcase 2b: $2x-1 < 0$, i.e., $x < \frac{1}{2}$. Then $|2x-1| = -(2x-1) = 1-2x$, so $1-2x = x+7$, which gives $-3x = 6$, so $x = -2$. Then $y = x+5 = -2+5 = 3$.

Since $x=-2 < \frac{1}{2}$ but $y=3 < 4$, this solution is not valid because it violates $y > 4$.

Therefore, the solutions are $(2, 1)$ and $(8, 13)$.

3. Final Answer

(2, 1), (8, 13)