Try squaring $x + y$ to get $(x + y)^2$. Then use the identity $(x + y)^2 = x^2 + 2xy + y^2$ and substitute the given values.

Once you know $x + y$ and $xy$, think of $x$ and $y$ as roots of a quadratic equation. What quadratic has roots with that sum and product?

In one expression line, type y = 10 - x. In another line, type x^2 + y^2 = 58. Desmos will graph the line and the circle.

Click on the points where the line and circle intersect. Desmos will display the coordinates of these intersection points; note the x-coordinates, which are the possible values of $x$ that satisfy the system.

We are given

Square the sum $x + y$:

Since $x + y = 10$, we have $(x + y)^2 = 10^2 = 100$.

Substitute the given value $x^2 + y^2 = 58$ into the identity:

Now solve for $xy$:

So we know both $x + y = 10$ and $xy = 21$.

If $x$ and $y$ are numbers with sum $10$ and product $21$, then $x$ and $y$ are the roots of the quadratic

Substitute $x + y = 10$ and $xy = 21$:

Now solve this quadratic for $t$ (which will represent the possible values of $x$ and $y$).

Factor the quadratic:

So $t = 7$ or $t = 3$. These are the possible values for $x$ (and also for $y$ in the opposite pairing). The question asks for one possible value of $x$, so a correct response is