Both equations have a $+y$ term. What happens if you subtract one equation from the other? Which variable disappears?

After you eliminate $y$, you should get an equation involving only $x$ and $x^2$. Rearrange it into the form $ax^2 + bx + c = 0$ and see if it factors nicely.

Once you find a value for $x$, substitute it back into both original equations to make sure it works in both at the same time.

In Desmos, type y = 10 - x^2 for the first equation and y = 11 - 2x for the second equation. These represent the same equations written with $y$ isolated.

Look for the point where the parabola $y = 10 - x^2$ and the line $y = 11 - 2x$ intersect. Click on the intersection point Desmos shows.

From the intersection point’s coordinates, note the $x$-coordinate. That $x$-value is the solution to the system and answers the question.

We have the system:

• $x^2 + y = 10$
• $2x + y = 11$

Since both equations have $+y$, subtract the second equation from the first to eliminate $y$:

On the left, $y - y = 0$, so you get:

We want all terms on one side to get a standard quadratic equation. Add $1$ to both sides of

to get:

Now you have a quadratic equation in $x$.

Notice that $x^2 - 2x + 1$ is a perfect square trinomial:

So the equation becomes:

If $(x - 1)^2 = 0$, then $x - 1 = 0$, so $x = 1$.

Check quickly in the original equations:

• If $x = 1$, then from $2x + y = 11$ you get $2(1) + y = 11$, so $y = 9$.
• Substitute into the first equation: $x^2 + y = 1^2 + 9 = 10$, which matches.

So the value of $x$ that satisfies the system is $\boxed{1}$.