After you substitute $y = x + 1$ into $x^2 + y^2 = 25$, you will get an equation involving only x. Expand $(x + 1)^2$ and combine like terms.

Once you simplify, you should get a quadratic equation of the form $x^2 + x - 12 = 0$. Think of two numbers that multiply to $-12$ and add to $1$ so you can factor and solve for x.

In Desmos, type x^2 + y^2 = 25 on one line (this is the circle) and y = x + 1 on another line (this is the line). Both graphs should appear on the coordinate plane.

Zoom or pan as needed until you can see where the line crosses the circle. Click on each intersection point that Desmos highlights to see its coordinates.

For each intersection point, note the x-coordinate shown by Desmos; these x-values are the solutions to the system. Any one of these x-values that satisfies both equations is an acceptable answer to the problem.

We are given the system:

Use substitution: replace y in the first equation with the expression from the second equation, $y = x + 1$.

That gives:

Now expand $(x+1)^2$ and combine like terms.

$(x + 1)^2 = x^2 + 2x + 1$, so

Combine like terms:

which simplifies to

Subtract 25 from both sides:

Divide the whole equation by 2 to make it simpler:

Now factor the quadratic $x^2 + x - 12 = 0$.

Look for two numbers that multiply to $-12$ and add to $1$. These numbers are $4$ and $-3$, because $4 \cdot (-3) = -12$ and $4 + (-3) = 1$.

So the factoring is:

Set each factor equal to zero:

So

Both of these values, when paired with $y = x + 1$, satisfy $x^2 + y^2 = 25$. Since the question asks for one possible value of $x$, $x = 3$ is a correct answer.