The ordered pair satisfies the system of equations and . What is the value of ?

Solution available with step-by-step explanation

SAT Nonlinear Equations & Systems Question 80 | Free SAT Prep | Aniko

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Question 80·Medium·Nonlinear Equations in One Variable; Systems in Two Variables

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The ordered pair $(x,y)$ satisfies the system of equations

\begin{cases} x^2 + y = 38 \\ y - 3x = -2 \end{cases}

and $x > 0$ . What is the value of $y$ ?

Your answer

(Express the answer as an integer)

For systems where one equation is linear and the other is nonlinear (like a quadratic), solve the linear equation for one variable (usually $y$ ) first, then substitute that expression into the nonlinear equation to get a single-variable equation. Solve the resulting quadratic carefully (factoring when possible), use any given conditions (such as $x>0$ ) to choose the correct root, and finally substitute back to find the other variable. This substitution-first approach is fast and reduces the chance of algebra mistakes on the SAT.

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Use the simpler equation first

Look at the linear equation $y - 3x = -2$ . Can you rewrite it so that it gives $y$ in terms of $x$ ?

Substitute into the nonlinear equation

Once you have $y$ expressed in terms of $x$ , plug that expression for $y$ into the equation $x^2 + y = 38$ to get an equation with only $x$ .

Solve the quadratic carefully

After substituting, you should get a quadratic equation. Factor it, find both solutions for $x$ , and then remember to apply the condition $x>0$ .

Back-substitute to find $y$

Use the valid value of $x$ in your expression for $y$ (from the linear equation) to compute the corresponding $y$ .

Desmos Guide

Enter both equations in $y=$ form

Rewrite the equations as $y = 38 - x^2$ and $y = 3x - 2$ . In Desmos, type these on two separate lines: y = 38 - x^2 and y = 3x - 2.

Find the intersection with $x>0$

On the graph, you will see two intersection points. Click or tap each intersection; Desmos will display their coordinates. Identify the intersection where the $x$ -coordinate is positive, and read off its $y$ -coordinate—that $y$ -value is the solution.

Step-by-step Explanation

Solve the linear equation for one variable

Use the simpler (linear) equation first:

y - 3x = -2

Solve for $y$ by adding $3x$ to both sides:

y = 3x - 2

Now you have $y$ written in terms of $x$ .

Substitute into the other equation

Substitute $y = 3x - 2$ into the first equation $x^2 + y = 38$ :

x^2 + (3x - 2) = 38

Simplify the left side:

x^2 + 3x - 2 = 38

Then move $38$ to the left to set the equation equal to $0$ :

x^2 + 3x - 40 = 0

Solve the quadratic and use the condition $x>0$

Factor the quadratic:

x^2 + 3x - 40 = (x + 8)(x - 5) = 0

So the possible values of $x$ are

x = -8 \text{ or } x = 5.

But the problem states that $x>0$ , so only $x = 5$ is allowed.

Find the corresponding value of $y$

Use $y = 3x - 2$ and substitute $x = 5$ :

y = 3(5) - 2 = 15 - 2 = 13

So the value of $y$ is 13.

Strategy

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Desmos Guide

Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation