After you substitute your expression for $y$ into $x^2 + y^2 = 13$, you should get an equation with only $x$ in it. What kind of equation is it, and how do you normally solve that type?

A quadratic will give you two possible $x$-values. Use $y = 5 - x$ for each one and check which pair $(x, y)$ satisfies the condition $x \ge y$.

In Desmos, enter the equation y = 5 - x to graph the line representing $x + y = 5$.

Enter the equation x^2 + y^2 = 13 to graph the circle representing $x^2 + y^2 = 13$.

Look for the intersection points of the line and the circle. Click on each intersection to see its coordinates $(x, y)$.

Compare the coordinates of the intersection points and identify which one has $x \ge y$. The x-coordinate of that point is the value of $x$ that solves the problem.

From the first equation, $x+y=5$, solve for $y$ in terms of $x$:

We will substitute this expression into the second equation.

Substitute $y = 5 - x$ into $x^2 + y^2 = 13$:

Now expand $(5 - x)^2$:

So the equation becomes

Combine like terms:

Subtract 13 from both sides:

Divide the entire equation by 2 to simplify:

Factor the quadratic $x^2 - 5x + 6$:

Set each factor equal to 0:

So the possible $x$-values from the system are $x=2$ and $x=3$.

Recall that $y = 5 - x$.

• If $x=2$, then $y = 5 - 2 = 3$, so $x = 2$ and $y = 3$, which does not satisfy $x \ge y$.
• If $x=3$, then $y = 5 - 3 = 2$, so $x = 3$ and $y = 2$, which does satisfy $x \ge y$.

Therefore, the value of $x$ that satisfies both equations and the condition $x \ge y$ is $3$.