A system of equations is given by If is a solution of the system above and , what is the value of ?

Solution available with step-by-step explanation

SAT Nonlinear Equations & Systems Question 83 | Free SAT Prep | Aniko

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Question 83·Medium·Nonlinear Equations in One Variable; Systems in Two Variables

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A system of equations is given by

y = x^2 + 1

y + 4x = 13

If $(x,y)$ is a solution of the system above and $x>0$ , what is the value of $xy$ ?

For systems where one equation is already solved for a variable (like $y = x^2 + 1$ ) and the other is linear, use substitution: plug the expression for $y$ into the other equation so you have one equation in a single variable. Rearrange into standard quadratic form, factor quickly if possible (or use the quadratic formula), and then apply any given condition (such as $x>0$ ) to select the correct solution. Finally, substitute back once to find the requested quantity.

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Connect the two equations

One equation already gives $y$ in terms of $x$ . How can you use that expression in the other equation so that everything is written in terms of a single variable?

Form and solve a quadratic

After you substitute for $y$ in the second equation, rearrange the result into the standard quadratic form $ax^2 + bx + c = 0$ so you can factor it.

Use the condition on $x$ and then find $y$

You will get two possible $x$ -values. Use $x>0$ to choose the right one, then plug that value into $y = x^2 + 1$ to get the specific $y$ -value.

Desmos Guide

Enter the two equations

In Desmos, type y = x^2 + 1 for the first equation. For the second equation, solve for $y$ to get $y = 13 - 4x$ , then type y = 13 - 4x as the second graph.

Find the intersection points

Look for the intersection points of the two graphs. There should be two points where the parabola $y = x^2 + 1$ and the line $y = 13 - 4x$ cross.

Use the condition $x>0$

Click or tap on each intersection and note their coordinates. Identify the intersection where the $x$ -coordinate is positive; this corresponds to the solution required by the problem.

Compute $xy$

From the intersection with positive $x$ , note $(x, y) = (2, 5)$ . Multiply the coordinates: $xy = 2 \times 5 = 10$ .

Step-by-step Explanation

Substitute to get one equation in $x$

From the first equation, we know

y = x^2 + 1.

Substitute this expression for $y$ into the second equation $y + 4x = 13$ :

x^2 + 1 + 4x = 13.

Now move 13 to the left side so the equation is set equal to 0:

x^2 + 4x + 1 - 13 = 0,

which simplifies to

x^2 + 4x - 12 = 0.

Solve the quadratic equation

We now solve

x^2 + 4x - 12 = 0.

Factor the quadratic:

x^2 + 4x - 12 = (x + 6)(x - 2) = 0.

So the possible values of $x$ are

x = -6 \text{ or } x = 2.

Use the condition $x>0$ to pick the correct $x$

The problem states that $x>0$ , so we cannot use $x=-6$ .

Therefore, the only valid solution for $x$ is

x = 2.

Find the corresponding $y$ and compute $xy$

Use $x=2$ in the first equation $y = x^2 + 1$ :

y = 2^2 + 1 = 4 + 1 = 5.

Now compute the product:

xy = 2 \times 5 = 10.

Thus, the correct choice is $10$ .

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