From $x^2 + y = 7$, you can write $y$ in terms of $x$. Substitute that expression into $y^2 + x = 11$ so that the new equation involves only $x$.

After you get an equation in $x$ only, remember that both $x$ and $y$ are positive and that $x^2 + y = 7$. That means $x^2$ must be less than $7$, so $x$ is between $0$ and a little more than $2.5$. Try small positive integers in that range to see which one works in both equations.

Once you find a positive value of $x$ that fits the system, plug it back into one of the original equations to get $y$, and report $y$, not $x$, as your final answer.

In Desmos, type the first equation as y = 7 - x^2. For the second equation, since $y$ is positive, type y = sqrt(11 - x) to represent the upper branch that satisfies $y^2 + x = 11$.

Look for the point where the graph of $y = 7 - x^2$ intersects the graph of y = sqrt(11 - x) in the region where both $x$ and $y$ are positive. Click that intersection point; Desmos will show its coordinates $(x, y)$.

From the intersection’s coordinates, note the $y$-value (the second number). That positive $y$-value is the answer to the question.

Start with the first equation:

Solve for $y$:

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

Use $y = 7 - x^2$ in the second equation $y^2 + x = 11$:

Expand and simplify:

so the equation becomes

Move everything to one side:

This is a polynomial equation in $x$.

From $y = 7 - x^2$ and $y > 0$, we know

so $0 < x < \sqrt{7} \approx 2.65$.

It is reasonable on the SAT to check small positive integers in this range. Test $x = 1$ and $x = 2$ in the polynomial $x^4 - 14x^2 + x + 38$:

• For $x = 1$:

So $x = 1$ is not a solution.

• For $x = 2$:

So $x = 2$ is a solution with $x > 0$.

Thus the positive $x$ that fits the system is $x = 2$.

Use the expression $y = 7 - x^2$ with $x = 2$:

Check quickly in the second equation:

so $(x, y) = (2, 3)$ satisfies both equations and both values are positive.

The question asks for the value of $y$, so the answer is $3$.