How many ordered pairs of real numbers $(x,y)$ satisfy the following system of equations? \[\left\{ \begin{aligned} x+3y&=3 \\ \left| |x| - |y| \right| &= 1 \end{aligned}\right.\]
We try graphing both equations in the $xy-$plane. The graph of $x+3y=3$ is a line passing through $(3,0)$ and $(0,1).$ To graph $\left| |x|- |y| \right| = 1,$ notice that the equation does not change if we replace $x$ by $-x$ or if we replace $y$ by $-y.$ Thus, the graph of $\left| |x|- |y| \right| = 1$ is symmetric about the $y-$axis and the $x-$axis, so if we graph the equation in just the first quadrant, we can produce the remainder of the graph by reflecting it over the axes.

If $(x, y)$ lies in the first quadrant, then $x \ge 0$ and $y \ge 0,$ so the equation $\left| |x|- |y| \right| = 1$ becomes just $|x-y| = 1.$ Therefore, either $x-y = 1$ or $y-x = 1,$ whose graphs in the first quadrant are rays. This gives us the whole graph of $\left| |x|- |y| \right| = 1:$
[asy]
size(8cm);
draw((0,1)--(3,4),blue,EndArrow);
draw((1,0)--(4,3),blue,EndArrow);
draw((0,-1)--(3,-4),blue,EndArrow);
draw((1,0)--(4,-3),blue,EndArrow);
draw((0,1)--(-3,4),blue,EndArrow);
draw((-1,0)--(-4,3),blue,EndArrow);
draw((0,-1)--(-3,-4),blue,EndArrow);
draw((-1,0)--(-4,-3),blue,EndArrow);
draw((-5,0)--(5,0),EndArrow);
draw((0,-5)--(0,5),EndArrow);
draw((-4,7/3)--(4,-1/3),red,Arrows);
dot((0,1)^^(-3,2)^^(1.5,0.5));
for (int i=-4; i<=4; ++i) draw((i,-0.15)--(i,0.15)^^(-0.15,i)--(0.15,i));
label("$x$",(5,0),E);
label("$y$",(0,5),N);
[/asy]
(The graph of $\left||x|-|y|\right|=1$ is drawn in blue, and the line $x+3y=3$ is drawn in red.) We see that the two graphs intersect at $\boxed{3}$ points.
3
Level 4
Intermediate Algebra
Problem: How many ordered pairs of real numbers $(x,y)$ satisfy the following system of equations? \[\left\{ \begin{aligned} x+3y&=3 \\ \left| |x| - |y| \right| &= 1 \end{aligned}\right.\]
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To determine the number of ordered pairs of real numbers \((x,y)\) that satisfy the given system of equations, we start by analyzing each equation separately.

The first equation is \(x + 3y = 3\). This can be rearranged to express \(x\) in terms of \(y\):
\[ x = 3 - 3y. \]

Next, we substitute this expression for \(x\) into the second equation, which is \(\left| |x| - |y| \right| = 1\). This leads to:
\[
\left| |3 - 3y| - |y| \right| = 1.
\]

Now, we need to consider the absolute values involved. The expression \(|3 - 3y|\) can take different forms depending on the value of \(y\). We analyze two cases based on the critical point where \(3 - 3y = 0\), which occurs at \(y = 1\).

**Case 1: \(y < 1\)**  
In this case, \(3 - 3y > 0\), so \(|3 - 3y| = 3 - 3y\). The equation simplifies to:
\[
\left| (3 - 3y) - |y| \right| = 1.
\]
We need to consider two subcases based on the sign of \(y\).

**Subcase 1.1: \(y \geq 0\)**  
Here, \(|y| = y\), and the equation becomes:
\[
\left| (3 - 3y) - y \right| = 1 \implies \left| 3 - 4y \right| = 1.
\]
This leads to two equations:
1. \(3 - 4y = 1 \implies 4y = 2 \implies y = \frac{1}{2}\).
2. \(3 - 4y = -1 \implies 4y = 4 \implies y = 1\).

For \(y = \frac{1}{2}\), substituting back gives:
\[
x = 3 - 3\left(\frac{1}{2}\right) = \frac{3}{2}.
\]
Thus, one solution is \(\left(\frac{3}{2}, \frac{1}{2}\right)\).

**Subcase 1.2: \(y < 0\)**  
Here, \(|y| = -y\), and the equation becomes:
\[
\left| (3 - 3y) + y \right| = 1 \implies \left| 3 - 2y \right| = 1.
\]
This leads to two equations:
1. \(3 - 2y = 1 \implies 2y = 2 \implies y = 1\) (not valid since \(y < 1\)).
2. \(3 - 2y = -1 \implies 2y = 4 \implies y = 2\) (not valid since \(y < 0\)).

Thus, there are no valid solutions in this subcase.

**Case 2: \(y \geq 1\)**  
In this case, \(3 - 3y \leq 0\), so \(|3 - 3y| = -(3 - 3y) = 3y - 3\). The equation simplifies to:
\[
\left| (3y - 3) - |y| \right| = 1.
\]
Again, we consider two subcases based on the sign of \(y\).

**Subcase 2.1: \(y \geq 0\)**  
Here, \(|y| = y\), and the equation becomes:
\[
\left| (3y - 3) - y \right| = 1 \implies \left| 2y - 3 \right| = 1.
\]
This leads to two equations:
1. \(2y - 3 = 1 \implies 2y = 4 \implies y = 2\).
2. \(2y - 3 = -1 \implies 2y = 2 \implies y = 1\).

For \(y = 2\), substituting back gives:
\[
x = 3 - 3(2) = -3.
\]
Thus, another solution is \((-3, 2)\).

For \(y = 1\), substituting back gives:
\[
x = 3 - 3(1) = 0.
\]
Thus, another solution is \((0, 1)\).

**Summary of Solutions:**  
The valid ordered pairs \((x,y)\) that satisfy both equations are:
1. \(\left(\frac{3}{2}, \frac{1}{2}\right)\)
2. \((0, 1)\)
3. \((-3, 2)\)

Therefore, the total number of ordered pairs of real numbers \((x,y)\) that satisfy the system of equations is \(3\).<|END_OF_LATENT|> The final answer is $3$.