Consider the system of equations: What are the possible values of ?

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SAT Nonlinear Equations & Systems Question 217 | Free SAT Prep | Aniko

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Question 217·Medium·Nonlinear Equations in One Variable; Systems in Two Variables

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Consider the system of equations:

\begin{aligned} x + y &= 5 \\ x^2 - y &= 11 \end{aligned}

What are the possible values of $x$ ?

When one equation in a system is linear and the other is quadratic, quickly solve the linear equation for one variable (whichever is easier) and substitute into the quadratic to get an equation in a single variable. Simplify to standard quadratic form $ax^2 + bx + c = 0$ , then use the quadratic formula, being very careful with signs in $-b$ and in $b^2 - 4ac$ . Finally, if time allows, you can verify by plugging your solutions back into one of the original equations or by checking the intersection points on Desmos.

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Start with the simpler equation

Look at $x + y = 5$ . Can you solve this equation for $y$ (or for $x$ ) and then use that expression in the other equation?

Substitute to get one variable

Once you express $y$ in terms of $x$ , plug that expression into $x^2 - y = 11$ . What kind of equation in $x$ do you get?

Solve the quadratic carefully

After simplifying, you should have a quadratic equation in $x$ . Identify $a$ , $b$ , and $c$ and apply the quadratic formula. Pay close attention to the signs in $-b$ and in $b^2 - 4ac$ .

Check the discriminant

When you compute $b^2 - 4ac$ , are you adding or subtracting the $4ac$ term? Make sure you use the correct sign for $c$ when calculating this.

Desmos Guide

Graph both equations in terms of y

Rewrite the system as $y = 5 - x$ and $y = x^2 - 11$ . In Desmos, enter these as two separate lines: y = 5 - x and y = x^2 - 11.

Locate the intersection points

On the graph, find where the line and the parabola intersect. Tap or hover over each intersection point; note the $x$ -coordinates of these points, since those are the possible values of $x$ that satisfy both equations.

Alternative: Use a single quadratic

After you algebraically form the quadratic $x^2 + x - 16 = 0$ , type f(x) = x^2 + x - 16 into Desmos. Then tap where the graph crosses the x-axis; the x-intercepts shown by Desmos are the same $x$ -values you would get from solving the quadratic with the quadratic formula.

Step-by-step Explanation

Use the linear equation to express one variable

From the first equation

x + y = 5

you can solve for $y$ in terms of $x$ :

y = 5 - x.

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

Substitute into the second equation and form a quadratic

The second equation is

x^2 - y = 11.

Substitute $y = 5 - x$ into this equation:

x^2 - (5 - x) = 11.

Now simplify the left side:

x^2 - 5 + x = 11.

Move all terms to one side to get a standard quadratic form:

\begin{aligned} x^2 + x - 5 - 11 &= 0 \\ x^2 + x - 16 &= 0. \end{aligned}

So you need to solve the quadratic equation $x^2 + x - 16 = 0$ .

Set up the quadratic formula and find the discriminant

For a quadratic equation $ax^2 + bx + c = 0$ , the quadratic formula is

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.

In $x^2 + x - 16 = 0$ :

$a = 1$
$b = 1$
$c = -16$

Compute the discriminant $b^2 - 4ac$ :

b^2 - 4ac = 1^2 - 4(1)(-16) = 1 + 64 = 65.

Also, $2a = 2$ . These values will go into the quadratic formula for $x$ .

Apply the quadratic formula and match the answer choice

Plug $a = 1$ , $b = 1$ , $c = -16$ into the quadratic formula:

x = \frac{-1 \pm \sqrt{65}}{2}.

So the possible values of $x$ are

\boxed{\dfrac{-1 \pm \sqrt{65}}{2}},

which corresponds to choice A.

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Step-by-step Explanation

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