After you substitute your expression for $y$ into $y^2 - x = 7$, expand carefully and bring all terms to one side so you get a polynomial equation in $x$ only. Can you factor it by testing small integer values of $x$?

Once you factor the polynomial, you will see one simple root and a remaining cubic factor. The problem says the system has exactly two real solutions—how does that limit how many real roots the cubic can have?

From the first equation, express $x + y$ in terms of $x$ only. Then plug in each possible $x$-value and compare the resulting values of $x + y$ to decide which is larger.

Enter the first equation as y = x^2 - 1. Then enter the second equation as y^2 - x = 7 (Desmos will graph this implicitly). You should see two intersection points between these curves.

Click on each intersection point to see its coordinates $(x,y)$. These are the two real solutions of the system. One will have both $x$ and $y$ positive; the other will have $x$ negative and $y$ positive.

For each intersection point, compute $x + y$ (either mentally, on paper, or by entering expressions like x1 + y1 using the coordinates you found). The larger of these two sums is the answer.

Start with the system

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

Substitute this expression for $y$ into the second equation $y^2 - x = 7$:

Now expand and simplify:

so the equation becomes

which simplifies to

This is a quartic (degree 4) equation in $x$.

To factor

try simple integer values of $x$.

Check $x = 2$:

so $x = 2$ is a root, meaning $(x - 2)$ is a factor.

Divide $x^4 - 2x^2 - x - 6$ by $(x - 2)$ (using long division or synthetic division) to get

So the possible $x$-values satisfy

Thus one possible $x$ is $2$, and any other real $x$ must be a root of the cubic $x^3 + 2x^2 + 2x + 3 = 0$.

The problem states that the system has exactly two real solutions $(x,y)$. For each real $x$ that satisfies the quartic, there is a corresponding real $y = x^2 - 1$, so the quartic can have only two real roots.

We already have one real root $x = 2$. That means the cubic

must have exactly one real root (if it had more than one, the quartic would have more than two real roots).

To see where this root lies, check the cubic at two nearby integers:

• At $x = -2$:

• At $x = -1$:

Since the cubic is continuous and changes sign between $x = -2$ and $x = -1$, it has a real root in the interval $(-2,-1)$. That root is the second possible value of $x$ for the system.

For any solution of the system, the first equation gives

so

1. For $x = 2$:

2. For the other solution: we know $x$ is between $-2$ and $-1$. Consider

This is an upward-opening parabola. On the interval $[-2,-1]$, its values at the endpoints are

Because the parabola opens upward, its maximum on $[-2,-1]$ is at one of the endpoints, so for any $x$ strictly between $-2$ and $-1$ we have

Therefore, for the second solution, $x + y < 1$.

So the two possible values of $x + y$ are $5$ (from $x = 2$) and a value less than $1$ (from the negative $x$). The greater of the two possible values of $x + y$ is