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10.4. Solving Rational Equations. Example 1: a basic fraction problem from chapter 1. What is the common Denominator?. 12. Factor. 3 * 2 * 2 = 12. 3 2*3 2*2. The key to finding the Least Common Denominator is –. You need the factors of every denominator BUT you

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Presentation Transcript
slide1

10.4

Solving Rational

Equations

slide2

Example 1: a basic fraction problem from chapter 1

What is the common

Denominator?

12

Factor

3 * 2 * 2 = 12

3 2*3 2*2

The key to finding the Least Common Denominator is –

You need the factors of every denominator BUT you

don’t have to repeat factors that you’ve already used.

(Pause right here, write this down, and understand it)

So for the LCD we only need the 2 from the 6 because We have already gotten the 3 from the 1st fraction. We also need the 2 from the 4 because we have used One 2 from the 6, but the 4 had two 2’s so we needed Two 2’s.

This may make no sense, but don’t give up on me. Let’s finish this problem

And do a few examples and MAYBE it will become more clear.

slide3

Once you have the common denominator, you can rid

Yourself of the fractions by multiplying both sides of

The equation by the LCD.

(12)

(12)

So the 3, 6 and 4 will reduce out of the problem like this

2

4

3

pause to make sure you understand where this equation comes from.

10x = 3x + 3

7x = 3

x = 3/7

slide4

Example 2

factor

(x + 5)(x - 2)

Remember the LEAST common denominator is all

of the denominators BUT you don’t have to repeat

ones that you’ve already used.

So the LCD is (x – 2) and (x + 5), multiply that to both sides

(x-2)(x+5)

(x-2)(x+5)

2(x + 5)

= x(x + 5)

+ 6

Take time to realize that when the (x-2)’s cancelled out, that left (x+5) to multiply to the numerator. In the last fraction all of the LCD cancelled out so you don’t have to multiply anything to the 6.

slide5

2(x + 5)

= x(x + 5)

+ 6

2x + 10 = x2 + 5x + 6

Combine like terms

-2x - 10

0 = x2 +3x - 4

Solve this equation

0 = (x + 4)(x – 1)

x – 1 = 0

x + 4 = 0

x = -4

x = 1

You should check your answers to make sure the answers are

in the domain of the problem.

slide6

Example 3: #38 from page 546

factor

(x+2)(x-2)

LCD: (x + 2)(x – 2) Multiply that to both sides

(x+2)(x-2)

(x+2)(x-2)

8(x – 1)

= 4(x + 2)

8x – 8 = 4x + 8

4x = 16

x = 4

slide7

Example 4: (#30 page 546)

(2x+5)(3x)

(2x+5)(3x)

Multiply both sides by the LCD: (2x+5)(3x)

(2x+5)(3x)

(2x+5)(3x)

+ (2x+5)(3x)(1) =

Pause here to make sure everyone gets this written down

21x2 + 3x

+ 6x2 + 15x

= 20x2 +50x – 6x - 15

Combine like terms and set equal to 0

7x2 – 26x + 15 = 0

(7x – 5)(x – 3) = 0

x = 5/7

x = 3

7x – 5 = 0 or x – 3 = 0

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Main Idea: Multiply each side of the equation by the Least Common Denominator to cancel out the denominators.

How do you find the LCD?

The least common denominator must include all the factors from each of the denominators but you don’t have to repeat factors that you have used from other fractions

Do assignment #4

* If you didn’t understand these examples to enough to even attempt assignment #4, then do the Mid-Chapter Self-Test on page 540