We are given a system of equations involving absolute values: $|x| + |y-1| = 1$ $x + 2y = 3$

1. Problem Description

We are given a system of equations involving absolute values:

$|x| + |y-1| = 1$

$x + 2y = 3$

2. Solution Steps

First, we can express $x$ in terms of $y$ from the second equation:

$x = 3 - 2y$

Now we consider different cases for the absolute values:

Case 1: $x \ge 0$ and $y-1 \ge 0$ (i.e., $y \ge 1$)

In this case, $|x| = x$ and $|y-1| = y-1$. Thus, the first equation becomes:

$x + y - 1 = 1$

$x + y = 2$

Substituting $x = 3 - 2y$, we have:

$(3 - 2y) + y = 2$

$3 - y = 2$

$y = 1$

Then, $x = 3 - 2(1) = 3 - 2 = 1$.

Since $x = 1 \ge 0$ and $y = 1 \ge 1$, this is a valid solution. So $(x, y) = (1, 1)$ is a solution.

Case 2: $x < 0$ and $y-1 \ge 0$ (i.e., $y \ge 1$)

In this case, $|x| = -x$ and $|y-1| = y-1$. Thus, the first equation becomes:

$-x + y - 1 = 1$

$-x + y = 2$

Substituting $x = 3 - 2y$, we have:

$-(3 - 2y) + y = 2$

$-3 + 2y + y = 2$

$3y = 5$

$y = \frac{5}{3}$

Then, $x = 3 - 2(\frac{5}{3}) = 3 - \frac{10}{3} = \frac{9 - 10}{3} = -\frac{1}{3}$.

Since $x = -\frac{1}{3} < 0$ and $y = \frac{5}{3} \ge 1$, this is a valid solution. So $(x, y) = (-\frac{1}{3}, \frac{5}{3})$ is a solution.

Case 3: $x \ge 0$ and $y-1 < 0$ (i.e., $y < 1$)

In this case, $|x| = x$ and $|y-1| = -(y-1) = 1-y$. Thus, the first equation becomes:

$x + 1 - y = 1$

$x - y = 0$

$x = y$

Substituting $x = 3 - 2y$, we have:

$3 - 2y = y$

$3 = 3y$

$y = 1$

Then, $x = 1$.

However, we assumed $y < 1$, so $y = 1$ contradicts this assumption. Thus, there is no solution in this case.

Case 4: $x < 0$ and $y-1 < 0$ (i.e., $y < 1$)

In this case, $|x| = -x$ and $|y-1| = -(y-1) = 1-y$. Thus, the first equation becomes:

$-x + 1 - y = 1$

$-x - y = 0$

$x = -y$

Substituting $x = 3 - 2y$, we have:

$3 - 2y = -y$

$3 = y$

Then, $x = -3$.

However, we assumed $y < 1$, so $y = 3$ contradicts this assumption. Thus, there is no solution in this case.

Therefore, the solutions are $(1, 1)$ and $(-\frac{1}{3}, \frac{5}{3})$.

3. Final Answer

The solutions are $(1, 1)$ and $(-\frac{1}{3}, \frac{5}{3})$.