The product of a positive number and the number that is less than is . What is the value of ?

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SAT Nonlinear Equations & Systems Question 95 | Free SAT Prep | Aniko

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Question 95·Medium·Nonlinear Equations in One Variable; Systems in Two Variables

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The product of a positive number $x$ and the number that is $6$ less than $x$ is $135$ .

What is the value of $x$ ?

For word problems about the product of a number and a quantity related to it, first convert the description directly into an algebraic equation (for example, "6 less than $x$ " becomes $x-6$ , and "their product is 135" becomes $x(x-6)=135$ ). Then rewrite the equation in standard quadratic form, solve it quickly by factoring if possible, and finally apply any conditions in the problem (such as "positive number") to select the valid solution and match it to the answer choice.

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Turn the words into algebra

Write an expression for "the number that is 6 less than $x$ " and then write an equation for the product of $x$ and that expression being $135$ .

Set up a quadratic equation

After you write the product as an equation, expand the left side and move everything to one side so the equation equals $0$ .

Solve and use the positivity condition

Solve the quadratic equation you get (by factoring or using the quadratic formula). If you find more than one solution, remember that the number $x$ must be positive.

Desmos Guide

Graph the expressions

In Desmos, enter y = x(x - 6) on one line and y = 135 on another line to represent the product and the constant value.

Find the positive intersection

Look for the intersection points of the two graphs and note the positive $x$ -coordinate where they meet; that $x$ -value is the solution that answers the question.

Step-by-step Explanation

Translate the words into an equation

"The product of a positive number $x$ and the number that is $6$ less than $x$ " means we multiply $x$ by $x-6$ .

So the situation is modeled by the equation

x(x-6) = 135.

Rewrite as a standard quadratic equation

Expand the left side:

x(x-6) = x^2 - 6x.

Set the equation equal to $0$ by subtracting $135$ from both sides:

x^2 - 6x - 135 = 0.

This is a quadratic equation in standard form $ax^2+bx+c=0$ .

Factor and use the "positive" condition

Now factor the quadratic. We need two numbers that multiply to $-135$ and add to $-6$ . Those numbers are $-15$ and $9$ , so

x^2 - 6x - 135 = (x - 15)(x + 9) = 0.

Set each factor equal to $0$ :

$x - 15 = 0$ gives $x = 15$ .
$x + 9 = 0$ gives $x = -9$ .

The problem says $x$ is a positive number, so we discard $x=-9$ and keep $x=15$ . The correct answer is $15$ , which corresponds to choice C.

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Step-by-step Explanation

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