After you substitute $y = x - 5$ into $x^2 - y = 7$, simplify carefully. You should get a quadratic equation in $x$ only. What standard form do quadratics usually have?

When you solve the quadratic, you will get two possible values of $x$. List both, then see which of those values appears in the answer choices.

Rewrite the first equation $x^2 - y = 7$ as $y = x^2 - 7$. In Desmos, enter the two equations as:

• y = x^2 - 7
• y = x - 5

Look at the graph where the parabola $y = x^2 - 7$ intersects the line $y = x - 5$. Click on each intersection point; Desmos will show the coordinates $(x, y)$ of each point.

Note the $x$-coordinates of the intersection points (these are the $x$-values that satisfy both equations). Compare those $x$-values with the answer choices and select the one that matches.

You are given the system

The second equation already solves for $y$ in terms of $x$. Substitute $y = x - 5$ into the first equation so that the first equation has only $x$:

$x^2 - y = 7$ becomes $x^2 - (x - 5) = 7$.

Now simplify $x^2 - (x - 5) = 7$:

• Distribute the minus sign: $x^2 - x + 5 = 7$.
• Move $7$ to the left side to get a standard quadratic form:

which simplifies to

Factor the quadratic $x^2 - x - 2 = 0$.

You need two numbers that multiply to $-2$ and add to $-1$. Those numbers are $-2$ and $1$, so

Set each factor equal to zero:

• $x - 2 = 0$ gives $x = 2$.
• $x + 1 = 0$ gives $x = -1$.

So there are two possible $x$-values that solve the system: $2$ and $-1$.

Check the answer choices: A) $-2$, B) $-1$, C) $4$, D) $5$.

From the quadratic, the possible $x$-values are $2$ and $-1$, and only $-1$ is listed as an option. Therefore, the correct choice is B) $-1$.