The system of equations is Which ordered pairs satisfy the system?

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SAT Nonlinear Equations & Systems Question 131 | Free SAT Prep | Aniko

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Question 131·Medium·Nonlinear Equations in One Variable; Systems in Two Variables

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The system of equations is

\begin{cases} y = 3x - 2 \\ x^2 + y^2 = 20 \end{cases}

Which ordered pairs $(x, y)$ satisfy the system?

For systems where one equation is linear and the other is a circle or another nonlinear curve, use substitution: solve the linear equation for one variable, substitute into the nonlinear equation, and solve the resulting quadratic (or higher-degree) equation. Expect up to two real solutions for a line–circle system, and always plug each solution back into the original linear equation to find the corresponding coordinate and to avoid sign mistakes or overlooking a second valid solution.

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Use substitution

You know $y$ in terms of $x$ from the first equation. How can you use that in the second equation, which has $x^2 + y^2$ ?

Recognize the resulting equation type

After you substitute for $y$ in $x^2 + y^2 = 20$ , you should get an equation with only $x$ . What kind of equation is it, and how do you usually solve that kind (for example, factoring or quadratic formula)?

Find matching y-values

Once you find possible $x$ -values, plug each one back into $y = 3x - 2$ to get the corresponding $y$ -values. Then check that each pair works in both original equations.

Desmos Guide

Graph both equations

In Desmos, enter the two equations as separate lines: y = 3x - 2 and x^2 + y^2 = 20. This will graph a line and a circle.

Locate intersection points

Adjust the zoom so you can clearly see where the line crosses the circle. There should be two intersection points where the graphs meet.

Read off the solutions

Click or tap each intersection point; Desmos will display its coordinates. Those coordinate pairs are the solutions of the system. Compare both pairs you see with the answer choices to decide which option matches.

Step-by-step Explanation

Substitute the line equation into the circle equation

The system is

\begin{cases} y = 3x - 2 \\ x^2 + y^2 = 20 \end{cases}

Use substitution: wherever you see $y$ in the second equation, replace it with $3x - 2$ .

This gives

x^2 + (3x - 2)^2 = 20.

Now expand $(3x - 2)^2$ to get $9x^2 - 12x + 4$ , so the equation becomes

x^2 + 9x^2 - 12x + 4 = 20.

Combine like terms:

10x^2 - 12x + 4 = 20.

Subtract 20 from both sides:

10x^2 - 12x - 16 = 0.

You can simplify by dividing everything by 2:

5x^2 - 6x - 8 = 0.

Solve the quadratic equation for x

Now solve

5x^2 - 6x - 8 = 0.

You can factor this quadratic. Look for two numbers that multiply to $5 \cdot (-8) = -40$ and add to $-6$ . Those numbers are $4$ and $-10$ .

Rewrite the middle term and factor by grouping:

$5x^2 - 6x - 8 = 5x^2 + 4x - 10x - 8$
Group: $(5x^2 + 4x) + (-10x - 8)$
Factor each group: $x(5x + 4) - 2(5x + 4)$
Factor out $(5x + 4)$ : $(5x + 4)(x - 2) = 0$

Set each factor equal to zero:

$5x + 4 = 0 \Rightarrow x = -4/5$
$x - 2 = 0 \Rightarrow x = 2$

So there are two possible $x$ -values: $x = 2$ and $x = -4/5$ .

Find the corresponding y-values and identify the solutions

Use $y = 3x - 2$ to find $y$ for each $x$ :

For $x = 2$ :
- $y = 3(2) - 2 = 6 - 2 = 4$ , so one solution pair is $(2, 4)$ .
For $x = -4/5$ :
- $y = 3(-4/5) - 2 = -12/5 - 2 = -12/5 - 10/5 = -22/5$ , so the other solution pair is $(-4/5, -22/5)$ .

Both of these pairs satisfy the circle equation as well:

$(2, 4)$ : $2^2 + 4^2 = 4 + 16 = 20$
$(-4/5, -22/5)$ : $(-4/5)^2 + (-22/5)^2 = 16/25 + 484/25 = 500/25 = 20$

Therefore, the ordered pairs that satisfy the system are $(2, 4)$ and $(-4/5, -22/5)$ , which corresponds to choice D.

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Step-by-step Explanation

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