After you substitute for $y$, you will get a quadratic equation in $x$. Factor or use the quadratic formula to find the possible $x$-values.

For each $x$-value you find, compute the corresponding $y$ using $y = x + 1$. Then ask: which point has both $x$ and $y$ positive?

First quadrant means both coordinates are positive. Which answer choices have both $x$ and $y$ positive? Test just those in both equations.

In Desmos, type y = x + 1 to graph the straight line representing the first equation.

On a new line in Desmos, type x^2 + y^2 = 25 to graph the circle centered at the origin with radius 5.

Use Desmos to click on the intersection points of the line and the circle; you should see two intersection coordinates. Identify which intersection has both $x$ and $y$ positive (first quadrant), then match that ordered pair to one of the answer choices.

The equation $y = x + 1$ is a straight line with slope $1$ and $y$-intercept $1$.

The equation $x^2 + y^2 = 25$ is a circle centered at the origin $(0, 0)$ with radius $5$.

The solutions to the system are the points where the line and the circle intersect. The question also says the point must be in the first quadrant, meaning both $x > 0$ and $y > 0$.

Since $y = x + 1$, substitute $x + 1$ for $y$ in the circle equation:

Now expand $(x + 1)^2$:

So the equation becomes:

Combine like terms:

Move $25$ to the left side:

Divide everything by $2$ to simplify:

Factor the quadratic:

Look for two numbers that multiply to $-12$ and add to $1$. Those numbers are $4$ and $-3$, so:

Set each factor equal to $0$:

Now find the corresponding $y$-values using $y = x + 1$:

• If $x = -4$, then $y = -4 + 1 = -3$.
• If $x = 3$, then $y = 3 + 1 = 4$.

So the system has two intersection points: $(-4, -3)$ and $(3, 4)$.

The first quadrant consists of points where $x > 0$ and $y > 0$.

• $(-4, -3)$ has both coordinates negative, so it is in the third quadrant, not the first.
• $(3, 4)$ has both coordinates positive, so it is in the first quadrant.

Therefore, the ordered pair that satisfies the system and lies in the first quadrant is $(3, 4)$, which corresponds to choice D.