After you substitute and simplify, you will get an equation involving $x^4$ and $x^2$. Think of $x^2$ as a single variable (for example, $u = x^2$) to turn it into a quadratic equation.

Once you find the possible values of $x^2$, remember that each positive value of $x^2$ can correspond to two different real $x$-values. Then use $y = x^2 - 4$ to find the matching $y$ for each.

For each value of $x^2$ that you find, determine how many distinct $x$-values and corresponding $y$-values it gives. Add these up to find the total number of solutions.

In Desmos, enter the first equation as y = x^2 - 4 to graph the parabola.

On a new line, enter the second equation as x^2 + y^2 = 8 to graph the circle.

Click on each point where the parabola and the circle cross; Desmos will highlight the intersection points and show their coordinates. Count how many distinct intersection points there are—this number matches the number of real solutions $(x, y)$ to the system.

Use the first equation, $y = x^2 - 4$, and substitute it into the second equation $x^2 + y^2 = 8$.

This gives

Now expand and simplify:

So

Now you have a single equation involving only $x$.

Let $u = x^2$. Then the equation

becomes

Solve this quadratic using the quadratic formula:

So there are two real values for $u = x^2$, and both are positive numbers.

For each positive value of $x^2 = u$, there are two real $x$-values:

• $x = +\sqrt{u}$
• $x = -\sqrt{u}$

Once you know $x^2 = u$, the corresponding $y$ is determined by the first equation:

That means each value of $u$ leads to two distinct $x$-values (positive and negative) and one shared $y$-value.

From the quadratic in $u$, there are two positive values of $x^2$.

Each positive $x^2$ value gives two distinct $x$-values ($+\sqrt{u}$ and $-\sqrt{u}$) and a corresponding $y$-value, so each $u$ produces two distinct solution points $(x, y)$.

Therefore, the system has $2 \times 2 = 4$ distinct real solutions $(x, y)$, so the correct choice is Exactly four.