In the system of equations below, which negative value of satisfies both equations?

Solution available with step-by-step explanation

SAT Nonlinear Equations & Systems Question 52 | Free SAT Prep | Aniko

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Question 52·Easy·Nonlinear Equations in One Variable; Systems in Two Variables

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In the system of equations below, which negative value of $x$ satisfies both equations?

\begin{cases} x^2 + y = 13 \\ y = 4 \end{cases}

For systems where one equation gives a variable directly (like $y = 4$ ), immediately substitute that value into the other equation to reduce the system to a single-variable equation. Solve carefully, remembering that equations like $x^2 = k$ usually have two solutions, and then apply any extra conditions in the question (such as "negative value" or "greater than 0") to pick the correct one from the possible solutions.

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Use the second equation

You are given the exact value of $y$ in the second equation. How can you use this in the first equation?

Get an equation in one variable

After you replace $y$ in $x^2 + y = 13$ , you should have an equation with only $x$ . Try to isolate $x^2$ .

Remember both square roots

When you solve $x^2 = 9$ , think about all real numbers whose square is $9$ , then pay attention to which of those is negative.

Desmos Guide

Enter the two equations as graphs

In Desmos, type y = 13 - x^2 for the first equation (this is $x^2 + y = 13$ rewritten) and y = 4 for the second equation.

Find the intersection points

Look at where the graphs of y = 13 - x^2 and y = 4 intersect. Desmos will show two intersection points; note both x-coordinates and then choose the negative x-value as the solution.

Step-by-step Explanation

Substitute the given value of y

You are told that $y=4$ . Substitute this into the first equation $x^2 + y = 13$ .

That gives:

x^2 + 4 = 13

Now you have an equation with only $x$ .

Isolate $x^2$

Solve for $x^2$ by subtracting $4$ from both sides of the equation:

\begin{aligned} x^2 + 4 &= 13 \\ x^2 &= 13 - 4 \\ x^2 &= 9 \end{aligned}

Now find the values of $x$ whose square is $9$ .

Solve for x and apply the "negative" condition

If $x^2 = 9$ , then $x$ can be $3$ or $-3$ , because both $3^2$ and $(-3)^2$ equal $9$ .

The question specifically asks for the negative value of $x$ that satisfies both equations, so the correct solution is $x = -3$ .

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