Question 36·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

After writing $y$ in terms of $x$ from the linear equation, substitute that expression for $y$ into the other equation so that you have only $x$ left.

You should end up with a quadratic equation in $x$. Use the quadratic formula to find both possible $x$-values, then see which one(s) match the answer choices.

Rewrite the system as $y = 2x - 9$ (from $2y - 4x = -18$) and $y = x^2 - 29$ (from $x^2 - y = 29$). These are easier to graph.

In Desmos, enter y = 2x - 9 as one equation and y = x^2 - 29 as the second. You will see a straight line and a parabola that intersect at two points.

Click or tap each intersection point on the graph to see its coordinates. Note the two $x$-values. Then compare those $x$-values to the answer choices. If the choices are in radical form, you can type each choice (for example, 1 - sqrt(21)) into Desmos as a separate expression to see its decimal value and match it to one of the intersection $x$-values.

Start with the linear equation:

Divide every term by 2:

Now solve for $y$:

This expression for $y$ can be substituted into the other equation.

The second equation is

Substitute $y = 2x - 9$ into this equation:

Distribute the minus sign and simplify:

Now you have a quadratic equation in $x$.

Solve

using the quadratic formula $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a = 1$, $b = -2$, and $c = -20$:

So there are two possible $x$-values: $1 + \sqrt{21}$ and $1 - \sqrt{21}$.

The problem asks for one possible value of $x$ where the graphs intersect, and the choices are

• $-8$
• $1 - \sqrt{21}$
• $2$
• $6$.

From the quadratic, the possible $x$-values are $1 + \sqrt{21}$ and $1 - \sqrt{21}$. Only $1 - \sqrt{21}$ appears among the answer choices, so one possible value of $x$ is $1 - \sqrt{21}$.