Simplifying Basic Rational Expressions Part 1

Simplifying Basic Rational Expressions Part 1By L.D.

Table of ContentsSlide 3: What is a rational expression?

Slide 4: Instructions

Slide 5: x + 6/4x

Slide 8: 7/3x – 9

Slide 12: 2x/x2 - 16

Slide 16: Instructions

Slide 17: 2x/3x

Slide 19: 3x/3(x – 1)

Slide 21: 8x2 + 16x/ 32

Slide 24: 25x2 – 5x/50x

Slide 26: x + 2/ x2 – 4

Slide 28: Problem 9: x + 3/x2 + 10x + 21

Slide 31: 4x – 12/3 – x

What is a rational expression?

An expression that involves a fraction

A rule for this type of problem is that you CANNOT have 0 in the denominator.

A fraction with a 0 denominator is called “undefined”

Instructions

Below Are Different Ways of Saying the Instructions

Find the excluded value.

Find out what makes the fraction undefined.

Find out what x can’t become

Problem 1

x + 6

4x

Problem 1

x + 6

4x

The first thing to do is to find that will make the denominator become 0. In this situation the answer would simply be 0 since

(0) + 6 6

4(0) 0

Problem 1

6

0

It ends up that since x IS NOT equal to 0, the final thing to do looks like

x ≠ 0

Problem 2

7__

3x - 9

Problem 2

7__

3x – 9

In a case like this, x is not 0 as that would make the problem become 7/-9. To make this problem undefined, we can use the short cut of taking the denominator as a fraction and equaling it to 0 to find x.

3x – 9 = 0

+9 +9

3x = 9

÷3 ÷3

x = 3

Problem 2

7__

3x – 9

x = 3

In our case, as mentioned before, we are trying to find what makes the problem UNDEFINED. So we can put the 3 in and see if that is achieved.

7__ 7__ 7__

3(3)– 9 9 – 9 0

Problem 2

Now that it is know that using the 3 makes a denominator of 0, the answer can be stated as x ≠ 3.

Problem 3

2x__

x2 - 16

Problem 3

2x__

x2 – 16

The first step for this is to do what was mentioned before and have the denominator equal 0.

x2 – 16 = 0

Both of these can have their square roots taken, so that is what I will do.

x2 – 16 = 0

x – 4 = 0

x = 4

Problem 3

2x__

x2 – 16

x = 4

Now to try it out.

2(4)_ 8__ 8__

(4)2 – 16 16 – 16 0

Problem 3

2x__

x2 – 16

It works out, but here is the twist to the problem. Since there is a squared variable, the answer of 4, could also be -4. So the final answer is….

x ≠ 4, -4

Instructions

Simplify

Problem 4

2x

3x

Problem 4

2x

3x

In this case, to make it undefined, 0 would be our girl, but we aren’t looking for that. We are looking to simplify.

To simplify, we need to find the GCF between these two, its x so x will cancel out.

2x 2

3x 3

2/3 is our final answer

Problem 5

3x___

3(x – 1)

Problem 5

3x___

3(x – 1)

In this problem, you may be tempted to use distributive, DON’T. In this variety, the only things that can be divided or canceled out are things that are multiplied. The numerator is 3 multiplied with x and the denominator is 3 multiplied with (x -1). They both have 3 in common, so it can be canceled out. The final problem will be

x

x -1

Problem 6

8x2 + 16x

32

Problem 6

8x2 + 16x

32

The first thing to do to solve this problem is to factor (go to the post titled Factoring Pt. 1/2 (x^2 + bx + c) on my blog to relearn how to factor) the top. Doing this will make it look like

8x(x + 2)

32

Problem 6

8x(x + 2)

32

Lastly to simplify the above. To do this divide the top and bottom by 8 to get

x(x + 2)

4

Problem 7

25x2 – 5x

50x

Problem 7

25x2 – 5x

50x

Now the technique used on the last slide should be repeated for this problem. In this one, 5x is the thing to cancel out.

5x(5x – 1) 5x – 1

50x 10

Problem 8

x + 2

x2 – 4

Problem 8

x + 2

x2 – 4

To solve this one, factor the bottom and then cancel out a piece.

x + 2 _1_

(x + 2)(x – 2) x - 2

Problem 9

x + 3

x2 + 10x + 21

Problem 9

x + 3

x2 + 10x + 21

The first thing to do here is to factor the bottom.

x2 + 10x + 21 = (x + 3)(x + 7)

Problem 9

x + 3

(x + 3)(x + 7)

Next to cancel out. All that is left is

1____

(x + 7)

Problem 10

4x – 12

3 – x

Problem 10

4x – 12

3 – x

The first thing to do is to factor out the top.

4(x - 3)

3 – x

Problem 10

4(x - 3)

3 – x

Since 3 – x and x – 3 are to different to be canceled out by each other, we will change that by factoring the bottom by -1.

4(x - 3)

-1(x – 3)

Problem 10

4(x - 3)

-1(x – 3)

Now we can finish up by canceling out.

4(x - 3) 4

-1(x – 3) -1= -4

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