Use the first equation to replace $y$ in the second equation. This will give you a single equation in terms of $x$ only.

After substitution, you will get a quadratic equation in $x$. Use the discriminant $b^2 - 4ac$ to decide how many real solutions for $x$ there are.

Each real solution for $x$ gives a corresponding $y$-value from $y = 2x - 1$. How does the number of $x$-solutions relate to the number of distinct ordered pairs $(x, y)$?

In Desmos, type y = 2x - 1 on one line and x^2 + y^2 = 13 on another line. Desmos will display a line and a circle.

Look for the points where the line and the circle cross. Tap or hover over each intersection point that appears on the graph to see its coordinates.

Count how many distinct intersection points appear. That number is the number of distinct real ordered pairs $(x, y)$ that satisfy the system.

The system is:

• $y = 2x - 1$ is a straight line.
• $x^2 + y^2 = 13$ is a circle centered at the origin with radius $\sqrt{13}$.

We are looking for how many intersection points this line and this circle have. Each intersection point corresponds to one real ordered pair $(x, y)$.

Since $y = 2x - 1$, substitute this into the circle equation in place of $y$:

Now expand and simplify:

So

We now have a quadratic equation in $x$.

For the quadratic

the discriminant is

Because $\Delta > 0$, this quadratic has two distinct real solutions for $x$. That means there will be two different $x$-values that work in the system.

Now solve for the actual $x$-values:

So

• $x = \dfrac{4 + 16}{10} = 2$,
• $x = \dfrac{4 - 16}{10} = -\dfrac{6}{5}$.

Find the matching $y$-values using $y = 2x - 1$:

• If $x = 2$, then $y = 2(2) - 1 = 3$, giving $(2, 3)$.
• If $x = -\dfrac{6}{5}$, then

giving $\left(-\dfrac{6}{5}, -\dfrac{17}{5}\right)$.

There are two distinct real ordered pairs that satisfy the system, so the correct choice is Exactly two.