After substituting $y$ into $x^2 + y^2 = 25$, notice that $x^2 = (|x|)^2$. It can help to temporarily replace $|x|$ with a new variable like $t$ to avoid confusion.

Once you find $|x|$, remember that $|x| = a$ usually corresponds to two $x$-values: $x = a$ and $x = -a$. Consider what this means when you add all possible $x$-coordinates.

In Desmos, enter x^2 + y^2 = 25 to graph the circle with radius 5 centered at the origin.

Enter y = 7 - abs(x) to graph the V-shaped line corresponding to $|x| + y = 7$.

Click on each point where the circle and the V-shaped graph intersect. Desmos will display the coordinates of these intersection points; note down all of their $x$-coordinates.

Add the $x$-values from all intersection points (you can type an expression like x1 + x2 + x3 + x4 using the recorded $x$-values into a new line in Desmos) and use this result as your final answer.

The first equation $x^2 + y^2 = 25$ is a circle centered at the origin with radius $5$.

The second equation $|x| + y = 7$ can be rewritten as $y = 7 - |x|$, which is a V-shaped graph opening downward and symmetric about the $y$-axis.

Both graphs are symmetric with respect to the $y$-axis. This means that if $(a,b)$ is a solution, then $(-a,b)$ will also be a solution, as long as $a \neq 0$.

From $|x| + y = 7$, solve for $y$:

Substitute this into the circle equation $x^2 + y^2 = 25$:

Notice that $x^2 = (|x|)^2$, so the equation depends only on $|x|$, not on the sign of $x$.

Let $t = |x|$ (so $t \ge 0$). Then

Expand and simplify:

Set this equal to $25$:

Subtract $25$ from both sides:

Divide by $2$:

Factor:

So $t = 3$ or $t = 4$. Since $t = |x|$, this gives $|x| = 3$ or $|x| = 4$, so the possible $x$-values are $x = -4, -3, 3, 4$ (all of which you can quickly check satisfy both original equations).

Now add all the possible $x$-coordinates:

So, the sum of all possible $x$-coordinates of the solutions to the system is 0.