Lesson 7 We All Scream for Ice Cream Practice Understanding

Learning Focus

Solve equations that contain rational expressions.

What real-life situations can be modeled with rational functions?

Open Up the Math: Launch, Explore, Discuss

The Glacier Bowl is an enormous ice cream treat sold at the neighborhood ice cream parlor. It is so large that people who can eat it within 30 minutes get a T-shirt and their picture posted on a wall. Because the Glacier Bowl is so big, it costs $60 and most people split the treat with a group.

Amera and some of her friends are planning to get together to share the bowl of ice cream. They plan to split the cost between them equally.

1.

What is an algebraic expression for the amount that each person in the group will pay?

2.

At the last minute, one of the friends couldn’t go with the group. Write an expression that represents the amount that each person in the group now pays.

3.

It turns out that each person in the group had to pay $2 more than they would have if everyone in the original group had shared the ice cream. How many people were in the original group?

4.

Explain why your answer(s) makes sense in this situation.

This story and the problem it represents provide an opportunity to model a situation that requires a rational equation. Rational equations can take many forms, but they are solved using principles we have worked with before. Try applying some of the strategies for working with rational expressions that we have used in this unit to solve these equations.

5.

$\frac{2}{x + 4} - \frac{1}{x} = \frac{2}{3 x}$

6.

$\frac{2 x - 3}{x + 1} = \frac{x + 6}{x - 2}$

7.

$x + \frac{20}{x - 4} = \frac{5 x}{x - 4} - 2$

8.

$\frac{x}{x + 3} - \frac{4}{x - 2} = \frac{- 5 x^{2}}{x^{2} + x - 6}$

Ready for More?

Try to solve this challenging equation: $\frac{3 x}{x^{2} + 5 x + 6} + \frac{2}{x^{2} + x - 2} = \frac{5 x}{x^{2} + 2 x - 3}$

Takeaways

Useful strategies for solving rational equations:

Adding Notation, Vocabulary, and Conventions

Extraneous solution:

Vocabulary

extraneous solution
Bold terms are new in this lesson.

Lesson Summary

In this lesson, we learned several strategies for solving rational equations. We found that it is often useful to combine two fractions into one expression or to multiply both sides of the equation by the common denominator of the fractions. Solving rational equations sometimes produces an extraneous solution that makes the denominator of one of the rational expressions in the original equation equal to zero and is therefore not an actual solution to the equation.

Retrieval

1.

The side lengths of right triangle $A B C$ are $8$ , $15$ , and $17$ . If $15$ is the numerator of $\sin A$ in the triangle, what is the trigonometric ratio for $\cos A$ ?

2.

a.

Add: $\frac{x^{2} - 4}{x (x + 2)} + \frac{3}{(x + 2)}$

b.

Multiply: $\frac{x^{2} - 4}{x (x + 2)} \cdot \frac{3}{(x + 2)}$