A system of equations is given by One solution to the system is with . What is the value of ?

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SAT Nonlinear Equations & Systems Question 181 | Free SAT Prep | Aniko

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Question 181·Medium·Nonlinear Equations in One Variable; Systems in Two Variables

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A system of equations is given by

y = 2x + 1 \qquad\text{and}\qquad y = (x - 3)^2.

One solution to the system is $(x, y)$ with $x < 3$ . What is the value of $x$ ?

For systems involving a line and a parabola, set the two expressions for $y$ equal to each other to get a single equation in $x$ , simplify to a standard quadratic, and solve using the quadratic formula or factoring if possible. When the problem adds a condition like $x < 3$ or $y > 0$ , use it to pick the correct solution from the two quadratic roots—often you can quickly compare by estimating square roots instead of doing exact decimal calculations.

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Eliminate $y$

Both equations give $y$ in terms of $x$ . How can you combine them to get an equation with only $x$ ?

Form a quadratic equation

After you set $2x + 1$ equal to $(x - 3)^2$ , expand the square and move all terms to one side so the equation equals $0$ .

Solve the quadratic carefully

Use the quadratic formula with the correct values of $a$ , $b$ , and $c$ . Be careful with the sign of $b$ when you plug into $-b \pm \sqrt{b^2 - 4ac}$ .

Apply the condition $x < 3$

You will get two values for $x$ . Estimate each one (you can approximate $\sqrt{2}$ ) to see which one is less than 3.

Desmos Guide

Graph both equations

In Desmos, enter y = 2x + 1 on one line and y = (x - 3)^2 on another. You should see a straight line and a parabola on the graph.

Find the intersection points

Tap or click where the line and the parabola intersect. Desmos will display the coordinates of the intersection points. There should be two intersection points with different $x$ -values.

Choose the intersection with $x < 3$

Look at the $x$ -coordinates of the intersection points. Identify the one that is less than 3; this $x$ -value is the solution the question is asking for. Match that value to the correct answer choice.

Step-by-step Explanation

Set the equations equal

Both equations equal $y$ , so set them equal to each other:

2x + 1 = (x - 3)^2

This removes $y$ and gives you an equation with only $x$ .

Expand and rearrange into a quadratic

First expand $(x - 3)^2$ :

(x - 3)^2 = x^2 - 6x + 9

So the equation becomes

2x + 1 = x^2 - 6x + 9

Move all terms to one side to get $0$ on the other side:

\begin{aligned} 0 &= x^2 - 6x + 9 - 2x - 1 \\ 0 &= x^2 - 8x + 8 \end{aligned}

So you need to solve the quadratic equation $x^2 - 8x + 8 = 0$ .

Solve the quadratic equation

Use the quadratic formula for $ax^2 + bx + c = 0$ :

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

Here, $a = 1$ , $b = -8$ , and $c = 8$ .

Compute the discriminant:

b^2 - 4ac = (-8)^2 - 4(1)(8) = 64 - 32 = 32

Now plug into the formula:

x = \frac{-(-8) \pm \sqrt{32}}{2(1)} = \frac{8 \pm \sqrt{32}}{2}

Since $\sqrt{32} = \sqrt{16 \cdot 2} = 4\sqrt{2}$ , this simplifies to

x = \frac{8 \pm 4\sqrt{2}}{2} = 4 \pm 2\sqrt{2}

So there are two possible $x$ -values: $4 + 2\sqrt{2}$ and $4 - 2\sqrt{2}$ .

Use the condition $x < 3$ to pick the correct solution

Now use the condition from the problem: $x < 3$ .

Approximate $\sqrt{2} \approx 1.4$ :

$4 + 2\sqrt{2} \approx 4 + 2(1.4) = 4 + 2.8 = 6.8$ , which is greater than 3.
$4 - 2\sqrt{2} \approx 4 - 2(1.4) = 4 - 2.8 = 1.2$ , which is less than 3.

Therefore, the $x$ -value that satisfies the system and the condition $x < 3$ is $4 - 2\sqrt{2}$ .

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Step-by-step Explanation

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