The system of equations above has a solution with . What is the value of ?

Solution available with step-by-step explanation

SAT Nonlinear Equations & Systems Question 49 | Free SAT Prep | Aniko

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Question 49·Easy·Nonlinear Equations in One Variable; Systems in Two Variables

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$y = -3$

$x^2 + y = 13$

The system of equations above has a solution $(x, y)$ with $x > 0$ . What is the value of $x$ ?

When one equation in a system directly gives a variable (like $y=-3$ ), substitute that value into the other equation to reduce the system to a single equation in one variable. Solve carefully—especially if it becomes a quadratic, which usually gives two solutions—and then apply any extra conditions given in the problem (such as $x>0$ ) to choose the correct solution. Always double-check that your chosen value satisfies both the equations and any stated inequalities.

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Substitute for $y$

You are told that $y=-3$ . Try replacing $y$ with $-3$ in the equation $x^2 + y = 13$ so you have an equation in terms of $x$ only.

Isolate $x^2$

After substituting $y=-3$ , simplify the equation and solve for $x^2$ by moving constants to the other side.

Remember both square roots

When you have an equation like $x^2 = 16$ , think about all possible values of $x$ that make it true. Then use the condition $x>0$ to decide which one fits the problem.

Desmos Guide

Graph both equations

In Desmos, enter the two equations on separate lines:

y = -3
x^2 + y = 13

Desmos will display a horizontal line for $y=-3$ and a curve for $x^2 + y = 13$ .

Find the intersection points

Look for the points where the graph of $y=-3$ intersects the graph of $x^2 + y = 13$ . There should be two intersection points, one with a negative $x$ -value and one with a positive $x$ -value.

Use the condition $x > 0$

Click on the intersection point that has a positive $x$ -coordinate. The $x$ -value shown for that point is the value of $x$ that satisfies both equations and the condition $x>0$ .

Step-by-step Explanation

Use substitution to get an equation in one variable

From the first equation, we know $y=-3$ .

Substitute $y=-3$ into the second equation $x^2 + y = 13$ :

x^2 + (-3) = 13

which simplifies to

x^2 - 3 = 13

Solve for $x^2$ and find possible $x$ values

Add 3 to both sides of $x^2 - 3 = 13$ :

x^2 = 16

To solve $x^2 = 16$ , take the square root of both sides. Remember, this gives two solutions:

x = 4 \text{ or } x = -4

Apply the condition $x > 0$

The problem states that $x > 0$ , so we must choose the positive solution.

Out of $x = 4$ and $x = -4$ , only $x = 4$ is greater than 0.

So, the value of $x$ is 4.

Strategy

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Desmos Guide

Step-by-step Explanation

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Strategy

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Step-by-step Explanation