Cogito
Integrated Math 3 · Chapter 2 · Lesson 3
Solving Rational Equations
Clearing the denominators, then checking.
12 problems · about 21 minutes · A-REI.A.2
What this lesson teaches
The student solves rational equations and rejects extraneous solutions.
- Clear denominators by multiplying every term by the LCD.
- That step can introduce extraneous solutions.
- Check every answer in the original equation and reject any that break it.
Warm Up
Straightforward practice. Get the method working first.
5 problems15 ÷ x = 5. What is x?
Answer 3
Why 3.
Why can extraneous solutions appear?
Answer Multiplying by an expression that could be zero is not reversible.
Why The clearing step can multiply by zero.
12 ÷ x = 4. What is x?
Answer 3
Why 12 ÷ 4.
10 ÷ x = 5. What is x?
Answer 2
Why 10 ÷ 5.
1 ÷ (x − 3) = 1. What is x?
Answer 4
Why x − 3 = 1.
Build It Up
The same ideas with more to keep track of.
3 problemsA candidate makes a denominator zero. Is it valid? 1 yes, 0 no.
Answer 0
Why It must be rejected.
8 ÷ (x + 1) = 2. What is x?
Answer 3
Why x + 1 = 4.
Is checking optional for rational equations? 1 yes, 0 no.
Answer 0
Why Extraneous solutions can appear.
Stretch Yourself
Mixed problems. Work out what kind of question it is before you start.
4 problemsPut the solving process in order.
Answer 1. Find the least common denominator. 2. Multiply every term by it. 3. Solve the resulting equation. 4. Check each answer in the original and reject any that break it.
Why The check is never optional here.
20 ÷ x = 4. What is x?
Answer 5
Why 20 ÷ 4.
The Rejection: x ÷ (x − 2) = 2 ÷ (x − 2). How many valid solutions?
Answer 0
Why 0.
The Simple Case: 6 ÷ x = 3. What is x?
Answer 2
Why 2.