Hi Everyone!

On this page you will find some material about Lesson 15. Read through the material below, watch the videos, and follow up with your instructor if you have questions.

**Lesson 15: Solving Rational Equations**

Table of Contents

### Resources

*In this section you will find some important information about the specific resources related to this lesson: *

*the learning outcomes,**the section in the textbook,**the WeBWorK homework sets,**a link to the pdf of the lesson notes,**a link to a video lesson.*

**Learning Outcomes.**

- Solve a rational equation.
- Check the potential solutions.

**Topic**. This lesson covers Section 5.5: Solving Rational Equations.

**WeBWorK**. There is one WeBWorK assignment on today’s material:

FractionalEquations

**Lesson Notes.**

**Video Lesson. **

Video Lesson 15 (based on Lesson 15 Notes)

### Warmup Questions

*These are questions on fundamental concepts that you need to know before you can embark on this lesson. Don’t skip them! Take your time to do them, and check your answer by clicking on the “Show Answer” tab.*

#### Warmup Question 1

Solve $$\dfrac{3x+5}{x}=2.$$

#### Show Answer 1

$$\dfrac{3x+5}{x}=2$$

Multiplying both sides by $x$ gives

$$3x+5 = 2x$$

So $x=-5$.

**Check**: $$\dfrac{3(-5)+5}{-5}=\dfrac{-10}{-5} =2$$

The solution set is $\{-5\}$.

#### Warmup Question 2

What is the LCD (least common denominator) of

$$\dfrac{1}{x^4}, \dfrac{x^3}{x^2-1}, \quad\text{and}\quad\dfrac{x-1}{2x(x+1)}?$$

#### Show Answer 2

The LCD is $2x^4(x^2-1)$.

### Review

*If you are not comfortable with the Warmup Questions, don’t give up! Click on the indicated lesson for a quick catchup. A brief review will help you boost your confidence to start the new lesson, and that’s perfectly fine.*

Need a review? Check Lesson 1.

### Quick Intro

*This is like a mini-lesson with an overview of the main objects of study. It will often contain a list of key words, definitions and properties – all that is new in this lesson. We will use this opportunity to make connections with other concepts. It can be also used as a review of the lesson.*

**A Quick Intro to Solving Rational Equations**

**Key Words.** Solution to an equation, rational equation, LCD (least common denominator), checking potential solutions.

The first step in solving a rational equation is to clear the denominators. For example, in the equation

$$\dfrac{2}{x}=\dfrac{x}{2}$$

the LCD is $2x$. Multiplying both sides by $2x$ clears the denominators.

$$\dfrac{2}{x}\cdot 2x=\dfrac{x}{2}\cdot 2x$$

$$2\cdot 2 = x\cdot x$$

$$4 = x^2$$

$$x^2=4$$

$$x=\pm 2$$

The potential solutions are $x=\pm 2$. We now need to check them.

**$\bullet$ Check: $x=-2$**

$$\dfrac{2}{-2}\stackrel{?}{=}\dfrac{-2}{2} $$

$$-1 = -1 \quad\checkmark$$

**$\bullet$ Check: $x=2$**

$$\dfrac{2}{2}\stackrel{?}{=}\dfrac{2}{2}$$

$$1=1 \quad\checkmark$$

The solution set is $\{-2,2\}$.

Now watch the video lesson to see that, depending on the equation, another approach can be taken. You will see that it is important to always check the solutions.

### Video Lesson

*Many times the mini-lesson will not be enough for you to start working on the problems. You need to see someone explaining the material to you. In the video you will find a variety of examples, solved step-by-step – starting from a simple one to a more complex one. Feel free to play them as many times as you need. Pause, rewind, replay, stop… follow your pace!*

**Video Lesson**

**A description of the video**

In the video you will see how to solve

$$\dfrac{x}{x-1}-\dfrac{1}{x} = \dfrac{3}{2}$$

in two ways.

### Try Questions

*Now that you have read the material and watched the video, it is your turn to put in practice what you have learned. We encourage you to try the Try Questions on your own. When you are done, click on the “Show answer” tab to see if you got the correct answer.*

#### Try Question 1

Solve

$$\dfrac{12}{x}-\dfrac{12}{x-5}=\dfrac{2}{x}.$$

#### Show Answer 1

$$\dfrac{12}{x}-\dfrac{12}{x-5}=\dfrac{2}{x} $$

The LCD is $x(x-5)$. We multiply both sides of the equation by $x(x-5)$.

$$\dfrac{12x(x-5)}{x}-\dfrac{12x(x-5)}{x-5}=\dfrac{2x(x-5)}{x} $$

$$12(x-5) – 12x= 2(x-5) $$

$$12x-60-12x = 2x-10 $$

$$-60 =2x-10 $$

$$-50 = 2x$$

$$2x = -50$$

$$x=-25$$

The potential solution is $x=-25$. The next step is to check it.

$\bullet$ **Check:** $x=-25$

$$\dfrac{12}{x}-\dfrac{12}{x-5}=\dfrac{2}{x} $$

$$\dfrac{12}{-25}-\dfrac{12}{-25-5}\stackrel{?}{=}\dfrac{2}{-25} $$

$$-\dfrac{12}{25} + \dfrac{12}{30} \stackrel{?}{=} – \dfrac{2}{25} $$

$$-\dfrac{12\cdot 6}{150} + \dfrac{12\cdot 5}{150} \stackrel{?}{=} – \dfrac{2}{25} $$

$$\dfrac{-72+60}{150} \stackrel{?}{=} – \dfrac{2}{25}$$

$$\dfrac{-12}{150} \stackrel{?}{=} – \dfrac{2}{25} $$

$$-\dfrac{2}{25} \stackrel{?}{=} – \dfrac{2}{25} \quad\checkmark $$

Therefore the solution set is $\{-25\}$.

### WeBWorK

*You should now be ready to start working on the WeBWorK problems. Doing the homework is an essential part of learning. It will help you practice the lesson and reinforce your knowledge.*

**WeBWorK**

It is time to do the homework on WeBWork:

FractionalEquations

When you are done, come back to this page for the Exit Questions.

### Exit Questions

*After doing the WeBWorK problems, come back to this page. The Exit Questions include vocabulary checking and conceptual questions. Knowing the vocabulary accurately is important for us to communicate. You will also find one last problem. All these questions will give you an idea as to whether or not you have mastered the material. Remember: the “Show Answer” tab is there for you to check your work!*

#### Exit Questions

- What is the difference between an expression and an equation?
- What does it mean to solve an equation?
- Is it necessary to check your answer if you know you have not made a mistake? Explain.

$\bigstar$ Solve $$\quad\dfrac{y}{y+3}+\dfrac{3}{y-3} = \dfrac{18}{y^2-9}.$$

#### Show Answer

The LCD is $y^2-9$ or $(y-3)(y+3)$.

$$\dfrac{y}{y+3}+\dfrac{3}{y-3} = \dfrac{18}{y^2-9}$$

$$(y-3)(y+3)\left(\dfrac{y}{y+3}+\dfrac{3}{y-3}\right) = (y-3)(y+3)\dfrac{18}{y^2-9}$$

$$\dfrac{y(y-3)(y+3)}{y+3}+\dfrac{3(y-3)(y+3)}{y-3} =\dfrac{18(y-3)(y+3)}{y^2-9} $$

$$y(y-3)+3(y+3) = 18 $$

$$y^2-3y+3y+9 & = & 18$$

$$y^2 = 9$$

$$y = \pm\sqrt 9$$

$$y=\pm 3$$

The potential solutions are $y=\pm 3$. We will now check them.

$\bullet$ **Check:** $y=3$

$$\dfrac{3}{3+3}+\dfrac{3}{3-3} \stackrel{?}{=} \dfrac{18}{3^2-9}$$

Two denominators are zero for the value of $y=3$.

$\bullet$ **Check:** $y=-3$

\[\dfrac{-3}{-3+3}+\dfrac{3}{-3-3} \stackrel{?}{=} \dfrac{18}{(-3)^2-9}\]

Two denominators are zero for the value of $y=-3$.

The solution set is the empty set, $\{\}$.

### Need more help?

*Don’t wait too long to do the following.*

- Watch the additional video resources.

- Talk to your instructor.
- Form a study group.
- Visit a tutor. For more information, check the tutoring page.