Once you have $y$ written in terms of $x$, substitute that expression into the other equation $x^2 - y = 7$ so that the new equation has only $x$ in it.

After substituting, you should get a quadratic equation in $x$. Rearrange it so it equals $0$, then solve it by factoring or by using the quadratic formula.

A quadratic equation can have two real solutions. Both $x$-values that satisfy the quadratic will work in the original system; the question only asks you to provide one of them.

In Desmos, enter the first equation in function form: y = 5 - x. Then enter the second equation in function form: y = x^2 - 7.

Look at where the graphs of y = 5 - x and y = x^2 - 7 intersect. Tap or click on each intersection point; Desmos will show the coordinates $(x, y)$.

The $x$-coordinates of these intersection points are the possible values of $x$ that solve the system. Choose either $x$-value shown by Desmos as your answer (the question only requires one).

Start with the first equation:

Solve for $y$ in terms of $x$ by subtracting $x$ from both sides:

Now you have $y$ written in terms of $x$.

Use the expression $y = 5 - x$ in the second equation $x^2 - y = 7$.

Substitute $5 - x$ for $y$:

Simplify the left side:

Now move all terms to one side to set the equation equal to $0$:

This is a quadratic equation in $x$.

Factor the quadratic $x^2 + x - 12$.

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

Set each factor equal to $0$:

So the solutions are:

The problem asks for one possible value of $x$, so a correct response is $3$.