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McDougal Littell, Chapter 5 HS Pre-Algebra, Unit 05: Rational Numbers and Equations Page 1 of 28 Revised 2013 - CCSS

Pre-Algebra Notes – Unit Five: Rational Numbers and Equations

Rational Numbers

Rational numbers are numbers that can be written as a quotient of two integers. Since decimals are special fractions, all the rules we learn for fractions will work for decimals. The only difference is the denominators for decimals are powers of 10; i.e., 101, 102, 103, 104, etc.... Students normally think of powers of 10 in standard form: 10, 100, 1000, 10,000, etc. In a decimal, the numerator is the number to the right of the decimal point. The denominator is not written, but is implied by the number of digits to the right of the decimal point. The number of digits to the right of the decimal point is the same as the number of zeros in the power of 10: 10, 100, 1000, 10,000… Therefore, one place is tenths, two places are hundredths, and three places are thousandths.

Examples: 1) 0.56 2 places → 56100

2) 0.532 3 places → 5321000

3) 3.2 1 place → 2310

The correct way to say a decimal numeral is to:

1) Forget the decimal point (if it is less than one).

2) Say the number.

3) Then say its denominator and add the suffix “ths”.

Examples: 1) 0.53 Fifty-three hundredths 2) 0.702 Seven hundred two thousandths 3) 0.2 Two tenths 4) 5.63 Five and sixty-three hundredths When there are numbers on both sides of the decimal point, the decimal point is read as “and”. You say the number on the left side of the decimal point, and then the decimal point is read as “and”. You then say the number on the right side with its denominator.


Examples: 1) Write 15.203 in word form. Fifteen and two hundred three thousandths

2) Write 7.0483 in word form.

Seven and four hundred eighty-three ten-thousandths

3) Write 247.45 in word form. Two hundred forty-seven and forty-five hundredths

Converting Fractions to Decimals:

Terminating and Repeating Decimals

Syllabus Objective: (3.7) The student will write equivalent representations of fractions, decimals, and percents. CCSS 8.NS.1-2: Understand informally that every number has a decimal expansion; show that the decimal expansion of a rational number repeats eventually or terminates.

A rational number, a number that can be written in the form of ab

(quotient of two integers), will

either be a terminating or repeating decimal. A terminating decimal has a finite number of decimal places; you will obtain a remainder of zero. A repeating decimal has a digit or a block of digits that repeat without end. One way to convert fractions to decimals is by making equivalent fractions.

Example: Convert 12

to a decimal.

Since a decimal is a fraction whose denominator is a power of 10, look for a power of 10 that 2 will divide into evenly.

1 52 10=

Since the denominator is 10, we need only one digit to the right of the decimal point, and the answer is 0.5.


0.3758 3.000

Example: Convert 34

to a decimal.

Again, since a decimal is a fraction whose denominator is a power of 10, we look for powers of 10 that the denominator will divide into evenly. 4 won’t go into 10, but 4 will go into 100 evenly.

3 754 100=

Since the denominator is 100, we need two digits to the right of the decimal point, and the answer is 0.75.

There are denominators that will never divide into any power of 10 evenly. Since that happens, we look for an alternative way of converting fractions to decimals. Could you recognize numbers that are not factors of powers of ten? Using your Rules of Divisibility, factors of powers of ten can only have prime factors of 2 or 5. That would mean 12, whose prime factors are 2 and 3, would not be a factor of a power of ten. That means that 12 will never divide into a power of 10

evenly. For example, a fraction such as 512

will not terminate – it will be a repeating decimal.

Not all fractions can be written with a power of 10 as the denominator. We need to look at another way to convert a fraction to a decimal: divide the numerator by the denominator.

Example: Convert 38

to a decimal.

This could be done by equivalent fractions since the only prime factor of 8

is 2. 3 3 125 3758 8 125 1000→ ⋅ =

However, it could also be done by division. Doing this division problem, we get 0.375 as the equivalent decimal.

Example: Convert 512

to a decimal.

This could not be done by equivalent fractions since one of the factors of 12 is 3. We can still convert it to a decimal by division.

0.41666...

12 5.00000

Six is repeating, so we can write it as 0 416. .

The vinculum is written over the digit or digits that repeat.


Example: Convert 411

to a decimal.

This would be done by division.

0 3636...4 11 4 0000 or 0 36

11

.. .→

Converting Decimals to Fractions

Syllabus Objective: (3.7) The student will write equivalent representations of fractions, decimals, and percents. CCSS 8.NS.1-3: Convert a decimal expansion which repeats eventually into a rational number. To convert a decimal to a fraction:

1) Determine the denominator by counting the number of digits to the right of the decimal point.

2) The numerator is the number to the right of the decimal point. 3) Simplify, if possible.

Examples: 1) Convert 0.52 to a fraction.

520.52100

13 =25

=

2) Convert 0.613 to a fraction.

6130.6131000

=

3) Convert 8.32 to a mixed number and improper fraction.

328.32 8100

8 208 8 or 25 25

=

=


10 3 33330 3333

9 3 0000

x .x .

x .

=− =

=

9 3 00009 9

13

x .

x

=

=

But what if we have a repeating decimal? While the decimals 0.3 and 0.3 look alike at first glance, they are different. They do not have the

same value. We know 0.3 is three tenths, 310

. How can we say or write 0.3 as a fraction?

As we often do in math, we take something we don’t recognize and make it look like a problem we have done before. To do this, we eliminate the repeating part – the vinculum (line over the 3).

Example: Convert 0.3 to a fraction.

0.3 0.333333...=

Let’s let x = 0.333333... Notice, and this is important, that only one number is repeating. If I multiply both sides of the equation above by 10 (one zero), then subtract the two equations, the repeating part disappears.

13

is the equivalent fraction for 0.3

Example: Convert 0.345 to a fraction.

The difficulty with this problem is the decimal is repeating. So we eliminate the repeating part by letting 0.345x = .

0.345 0.345345345...= Note, three digits are repeating. By multiplying both sides of the equation

by 1000 (three zeros), the repeating parts line up. When we subtract, the repeating part disappears.


1000 345.345345345... 0.345345345...

999 345999 999

345 115 999 333

xx

x

x or

=− =

=

= Example: Convert 0.13 to a fraction.

Note, one digit is repeating, but one is not. By multiplying both sides of the equation by 10, the repeating parts line up. When we subtract, the repeating part disappears.

10 1 33333

13333

9 1 2

9 1 29 9

1 2 12 2 or which simplifies to9 90 15

x .

x .

x .

x .

.x

=

− =

=

=

=

Ready for a “short cut”? Let’s look at some patterns for repeating decimals.

1 0 111 or 019

2 0 222 or 0 29

3 0 333 or ?9

4 ?9

. .

. .

.

=

=

=

=

1 0 0909 or 0 0911

2 0 1818 or 01811

3 0 2727 or ?11

4 ?11

. .

. .

.

=

=

=

=

It is easy to generate the missing decimals when you see the pattern! Let’s continue to look at a few more repeating decimals, converting back into fractional form.

Because we are concentrating on the pattern, we will choose NOT to simplify fractions where applicable. This would be a step to add later.


37220 37599900

42340 42769900

18140 20159000

59640 60249900

806230 8143799000

498070 5534190000

.

.

.

.

.

.

=

=

=

=

=

=

210 2390

320 3590

4230 427990

3220 325990

42720 42769990

2120 235900

.

.

.

.

.

.

=

=

=

=

=

=

50 59

60 69

?0 79

?0 8?

.

.

.

.

=

=

=

=

1301399

250 2599

?0 3799

?0 56?

.

.

.

.

=

=

=

=

1230123999

1540154999

?0 421999

?0 563?

.

.

.

.

=

=

=

=

The numerator of the fraction is the same numeral as the numeral under the vinculum. We can also quickly determine the denominator: it is 9ths for one place under the vinculum, 99ths for two places under the vinculum, 999ths for three places under the vinculum, and so on. But what if the decimal is of a form where not all the numerals are under the vinculum? Let’s look at a few.

Note that again we chose not to simplify fractions where applicable as we want to concentrate on the pattern.

Does ????

Do you believe it? Let's look at some reasons why it's true. Using the method we just looked at:

The numerator is generated by subtracting the number not under the vinculum from the entire number (including the digits under the vinculum). We still determine the number of nines in the denominator by looking at the number of digits under the vinculum. The number of digits not under the vinculum gives us the number of zeroes.


Surely if 9x = 9, then x = 1. But since x also equals .9999999... we get that .9999999... = 1.

But this is unconvincing to many people. So here's another argument. Most people who have trouble with this fact oddly don't have trouble with the fact that 1/3 = .3333333... . Well, consider the following addition of equations then:

This seems simplistic, but it's very, very convincing, isn't it? Or try it with some other denominator:

Which works out very nicely. Or even:

It will work for any two fractions that have a repeating decimal representation and that add up to 1. The problem, though, is BELIEVING it is true.

So, you might think of 0.9999.... as another name for 1, just as 0.333... is another name for 1/3.


Comparing and Ordering Rational Numbers

Syllabus Objectives: (3.6) The student will compare and order rational numbers. (3.7) The student will write equivalent representations of fractions, decimals, and percents. (3.8) The student will explain the relationship among equivalent representations of rational numbers. We will now have fractions, mixed numbers and decimals in ordering problems. Sometimes you can simply think of (or draw) a number line and place the numbers on the line. Numbers increase as you go from left to right on the number line, so this is particularly helpful when you are asked to go from least to greatest. If placement is not obvious (for instance, when values are very close together), it may be advantageous to write all the number in the same form (decimal or fractional equivalents), and then compare.

Example: Order the numbers 5 5 130 2 4 3 34 2 3

, . , . , , ,− − − − from least to greatest.

Let’s first rewrite all improper fractions as mixed numbers.

5 1 5 1 13 11 ; 2 ; 44 4 2 2 3 3

− = − = − = −

Now let’s place the values on the number line. From least to greatest, the order would be 13 5 13 0 2 2 4 3

3 4 2, , , . , , .− − − − .

Sometimes writing the numbers in the same form will assist you in ordering.

122

114

−

5− 5 0

143

− 4 3. 0 2.− 3−


Example: Order 7 50 25 1 1 18 11

, . , , , . from least to greatest.

Find the decimal equivalents, then compare.

7 0 8758

0 25 0 250

1 1 000

1 0 5002

1 1 1 100

.

. .

.

.

. .

=

=

=

=

=

Adding and Subtracting Fractions with Like Denominators

Let’s add 1 1 to

4 4. Will it be 2

8? Why not? If we did, the fraction 2

8 would indicate that we

have two equal size pieces and that 8 of these equal size pieces made one whole unit. That’s not true. Let’s draw a picture to represent this:

14

14

+

Lining up the decimals, the order from least to greatest is: 0.250 0.500 0.875 1.000 1.100 Using the original forms:

1 70 25 1 1 12 8

. , , , , .

Or find the fractional equivalents, then compare.

7 358 40

25 1 100 25100 4 40

40140

1 202 40

11 441 110 40

.

.

=

= = =

=

=

= =

Having found a common denominator, the order from least to greatest is: 10 20 35 40 4440 40 40 40 40

, , , ,

Using the original forms: 1 70 25 1 1 12 8

. , , , , .


Notice the pieces are the same size. That will allow us to add the pieces together. Each rectangle has 4 equally sized pieces. Mathematically, we say that 4 is the common denominator. Now let’s count the number of shaded pieces. Adding the numerators, a total of 2 equally sized pieces are shaded and 4 pieces make one unit. We can now show:

Example: Find the sum of 4 19 9+ .

Since the fractions have the same denominator, we write the sum over 9. 4 1 59 9 9+ =

Example: Find the difference of 4 15 5

and .

Since the fractions have the same denominator, we write the difference over 5. 4 1 35 5 5− =

Writing these problems with variables does not change the strategy.

Example: Simplify the variable expression. 5 212 12

x x+

5 2 5 2 712 12 12 12

x x x x x++ = =

14

14

+

2 14 2

or


Adding and Subtracting Fractions with Unlike Denominators

Let’s first review the ways to find a common denominator. We find the least common denominator by determining the least common multiple.

Strategy 1: Multiply the numbers. This is a quick, easy method to use when the numbers are relatively prime (have no factors in common).

Example: Find the LCM of 4 and 5. Since 4 and 5 are relatively prime, LCM would be 4 5 or 20⋅ .

Strategy 2: List the multiples. Write multiples of each number until there is a common multiple.

Example: Find the LCM of 12 and 16.

12, 24, 36, 48, 60, … 16, 32, 48, 64, … 48 is the smallest multiple of both numbers; therefore, 48 is the LCM.

Strategy 3: Prime factorization. Write the prime factorization of both numbers. The

LCM must contain all the factors of both numbers. Write all prime factors, using the highest exponent.


260 2 3 5= ⋅ ⋅ and 3 272 2 3= ⋅ The LCM is 3 22 3 5 360⋅ ⋅ =

This strategy can also be shown by using a Venn diagram. Example: Find the LCM of 36 and 45.

Draw a Venn diagram, placing common factors in the intersection. The LCM is the product of all the factors in the diagram.

As the numbers in the denominator become larger, this strategy can become cumbersome. That is when the value of the following strategy becomes evident.

Factors of 36 Factors of 45

32 22 5

Multiply all factors in diagram for the LCM: 2 22 3 5 180⋅ ⋅ = .


Strategy 4: Simplifying/Reducing Method. Write the two numbers as a single fraction; then reduce and find the cross products. The product is the LCM.


18 3 ;cross products are 18 4 24 3 or 72. The LCM is 72.24 4

= ⋅ = ⋅

When adding or subtracting fractions, LCM is referred to as the Least Common Denominator (LCD). We have several ways to find a common denominator.

Methods of Finding a Common Denominator

1. Multiply the denominators.

2. List multiples of each denominator, use a common multiple.

3. Find the prime factorization of the denominators, and find the Least Common Multiple.

4. Use the Simplifying/Reducing Method.

Use this method when…

1. the denominators are prime numbers or relatively prime.

2. the denominators are small numbers.

3. the denominators are small numbers; some will advise to never or seldom use this method.

4. the denominators are composite numbers/ large numbers.

Let’s add 1 1 to 3 4

. Will it be 27

? Why not? If we did, the fraction 27

would indicate that we

have two equal size pieces and that 7 of these equal size pieces made one whole unit. That’s just not true. Let’s draw a picture to represent this:

14

13

+

Notice the pieces are not the same size.


1523

+

1 34 12

1 43 12

712

=

+ =

Making the same cuts in each rectangle will result in equally sized pieces. That will allow us to add the pieces together. Each rectangle now has 12 equally sized pieces. Mathematically, we say that 12 is the common denominator. Now let’s count the number of shaded pieces.

From the drawing we can see that 13

is the same as 412

and 14

has the same value as 312

.

Adding the numerators, a total of 7 equally sized pieces are shaded and 12 pieces make one unit. If we did a number of these problems, we would be able to find a way of adding and subtracting fractions without drawing the picture.

Using the algorithm, let’s try one.

Example: Multiply the denominators to find the least common denominator, 5 3 15⋅ = . Now make equivalent fractions and add the numerators.

1 35 152 103 15

1315

=

+ =

Algorithm for Adding/Subtracting Fractions

1. Find a common denominator. 2. Make equivalent fractions. 3. Add/Subtract the numerators. 4. Simplify (reduce), if possible.


3415

+

3 154 201 45 20

1920

=

+ =

These problems can also be written horizontally.

1 2 3 10 135 3 15 15 15

.+ = + =

Let’s try a few. Using the algorithm, first find the common denominator, and then make equal fractions. Once you complete that, you add the numerators and place that result over the common denominator and simplify, if possible.

Remember, the reason you are finding a common denominator is so you have equally sized pieces. To find a common denominator, use one of the strategies shown. Since the denominators are relatively prime, use the “multiply the denominators” method.

Example:

Example:

Writing these problems with variables does not change the strategy.

1114

58

−

To find the common denominator, use the Simplifying/Reducing Method, 8 4 LCD 8 7 56

14 7;= = ⋅ =

11 4414 56

5 358 56

956

=

− =


Example: Simplify the expression. 23 5d d+

The LCD is 15. Making equivalent fractions, we have: If the denominators are larger composite numbers, using the reducing method to find the common denominator may make the work easier.

Example: Simplify the expression. 5 718 24

c c− −

Using the Simplifying/Reducing method: 18 324 4

= , 4 18 72⋅ = , so the LCD is 72.

Another nice feature of using the Simplifying/Reducing Method is that you do not need to compute what 18 72⋅�= or 24 72⋅�= because we can see the number in the cross products. That is, we can identify 18 times 4 is 72, so we multiply −5c by 4 to obtain the new numerator ( 5 4 20c c− ⋅ = − ). Likewise, since 24 times 3 is 72, we determine the other numerator as 7 3 21c c− ⋅ = − .

5 7 20 2118 24 72 72

41 =72

c c c c

c

− − = − −

−

5 2018 72

7 2124 72

4172

c c

c c

c

− = −

− = −

= −

or

2 5 63 5 15 15

11 =15

d d d d

d

+ = +5

3 15

2 65 15

1115

d d

d d

d

=

+ =

=

It is customary to write these problems in a horizontal format like this →


Example: Evaluate the expression.

Regrouping To Subtract Mixed Numbers

The concept of borrowing when subtracting with fractions has been typically a difficult area for kids to master. For example, when subtracting 51

6 612 4− , students usually answer 468 if they

subtract this problem incorrectly. In order to ease the borrowing concept for fraction, it would be a good idea to go back and review borrowing concepts that kids are familiar with. Example: Take away 3 hours 47 minutes from 5 hours 16 minutes. 5 hrs 16 min − 3 hrs 47 min ????????? Subtracting the hours is not a problem but students will see that 47 minutes cannot be subtracted from 16 minutes. In this case, students will see that 1 hour must be borrowed from 5 hrs and added to 16 minutes: 5 hrs 16 min − 3 hrs 47 min ????????? Now the subtraction problem can be rewritten as: 4 hrs 76 min 4 hrs 76 min − 3 hrs 47 min − 3 hrs 47 min ??????????? 1 hr 29 min

3 7 25 10 15x x x

− − +

3 7 25 10 15

18 21 4 35 730 30 30 30 6

x x x

x x x x xor

− − + =

− − + = − −

3 185 30

7 2110 30

2 415 30

35 7 or 30 6

x x

x x

x x

x x

− = −

− = −

+ = +

= − −

or

4hrs

16min 1 16min 60min 76minhr+ = + =


If students can understand the borrowing concept from the previous example, the same concept can be linked to borrowing with mixed numbers. Lets go back to the first example: 51

6 612 4− . It may be easier to link the borrowing concept if the problem

is rewritten vertically:

1 1265 46

−

???????

Example: Subtract Step 1. Find a common denominator:

The common denominator is 10. Step 2. Make Equivalent fractions using 10 as the denominator.

Step 3. It is not possible to subtract the numerators. You cannot take 5 from 4!! Use the concept of borrowing as described in the above examples to re-write this problem. Borrow from 1

from 13 and add 1 ( 1010

) to 410

.

Example: Catherine has a canister filled with 152

cups of flour. She used 314

cups of flour to

bake a cake. How much flour is left in the canister?

Subtract 152−

314

.

Step 1. Find a common denominator: The common denominator between 2 and 4 is 4.

11 1 1 6 716 6 6 6+ = + = 7 11

65 46

−

2 17 76 3=

2 135172

−

4 131057

10−

4 13 1057

10−

121010

+14 121057

10−

9510


334

Step 2. Make equivalent fractions using 4 as the common denominator.

2 35 14 4−

Step 3. When subtracting the numerators, it is not possible to take 3 from 2, therefore borrow. It may be easier to follow the borrowing if the problem is rewritten

vertically . There are cups of flour left in the canister.

Multiplying Fractions and Mixed Numbers Multiplying fractions is pretty straight-forward. So, we’ll just write the algorithm for it, give an example and move on.

Algorithm for Multiplying Fractions and Mixed Numbers

1. Make sure you have proper or improper fractions. 2. Cancel, if possible. 3. Multiply numerators. 4. Multiply denominators. 5. Simplify (reduce), if possible.

44

+2 5 431 4

−

6 4 431 4

−

4

334


Example: 1 43 2 5

Since 132

is not a fraction, we convert it to 7 .2

1 4 7 432 5 2 5

=

7 42 5

can be written as 7 4

2 5

Now what I’m about to say is important and will make your life a lot easier. We know how to reduce fractions, so what we want to do now is cancel with fractions. That’s nothing more than reducing using the commutative and associative properties.

Using the commutative property, we can rewrite this as 4 7 .2 5

Using the associative property, we can rewrite this as 4 7 .2 5

Simplify 4 22 1= .

Then multiply and simplify, as a mixed number. 2 7 14 421 5 5 5

= =

Rather than going through all those steps, we could take a shortcut and cancel. Now rather than going through all those steps, using the commutative and associative properties, we could have taken a shortcut and cancelled. To cancel, we would look for common factors in the numerator and the denominator and divide them out. In our problem, there is a common factor of 2. By dividing out a 2, the problem looks like this:

7 2 14 421 5 5 5

or=

Let’s look at another one.

72

4⋅

5 2

1


82 8 8 22

÷

826242220

−

−

−

−

Example: 3 23 25 9

18 205 9 Rewrite as improper fractions.

Cancel 18 and 9 by common factor of 9. Cancel 20 and 5 by common factor of 5.

2 4 8 81 1 1

= = Multiply numerators, multiply denominators, simplify.

When variables are added to these problems, the strategy remains the same.

Example: Simplify the expression. 2 23 2

4 7n n

⋅

Cancel the 2 and 4 by common factor of 2. Multiply the numerators and denominators.

Dividing Fractions and Mixed Numbers Before we learn how to divide fractions, let’s revisit the concept of division using whole numbers. When I ask, how many 2’s are there in 8, I can write that mathematically three ways. To find out how many 2’s there are in 8, we will use the subtraction model:

Now, how many times did we subtract 2? Count them: there are 4 subtractions. So there are 4 twos in eight. Mathematically, we say 8 ÷ 2 = 4. The good news is, division has been defined as repeated subtraction That won’t change because we are using a different number set. In other words, to divide fractions, I could also do repeated subtraction.

2 4

1 1

185

20

9

1

2

2 2 43 2 34 7 14n n n

⋅ =


Example: 1 112 4÷

Another way to look at this problem is using your experiences with money. How many quarters are there in $1.50? Using repeated subtraction we have:

How many times did we subtract 14

? Six. Therefore, 1 11 62 4÷ = . But this took

a lot of time and space.

A visual representation of division of fractions would look like the following.

Example: 1 12 8÷ =

We have 1 .2

Representing that would be

Since the question we need to answer is how many 1 '8

s are there in 12

, we

need to cut this entire diagram into eighths. Then count each of the shaded one-eighths.

As you can see there are four. So 1 1 42 8÷ = .

1 21 12 4

14114141

=

−

−

41414341424

=

−

−

241414140

−

−


Example: 5 16 3÷ =

We have 56

. Representing that would be

Since the question we need to answer is how many 1 '3

s are there in 56

, we need to use the cuts

for thirds only. Then count each of the one-thirds.

1 2 12

As you can see that are 122

. So 5 1 12 .6 3 2÷ =

Be careful to choose division examples that are easy to represent in visual form. Well, because some enjoy playing with numbers, they found a quick way of dividing fractions. They did this by looking at fractions that were to be divided and they noticed a pattern. And here is what they noticed.

Algorithm for Dividing Fractions and Mixed Numbers

1. Make sure you have proper or improper

fractions. 2. Invert the divisor (2nd number). 3. Cancel, if possible. 4. Multiply numerators. 5. Multiply denominators. 6. Simplify (reduce), if possible.


The very simple reason we tip the divisor upside-down, then multiply for division of fractions is because it works. And it works faster than if we did repeated subtractions, not to mention it takes less time and less space.

Example: 3 24 5÷ 3 5

4 2

15 718 8=

(Invert the divisor.)

Example: 1 433 9÷ 10 4

3 9÷ = Make sure you have proper or improper fractions.

10 93 4

=

Invert the divisor.

10 93 4

= Cancel 10 and 4 by 2, and cancel 9 and 3 by 3.

Computing with Fractions and Signed Numbers

Syllabus Objective: (3.9) The student will perform operations with positive and negative fractions and mixed numbers. The rules for adding, subtracting, multiplying and dividing fractions with signed numbers are the same as before, the only difference is you integrate the rules for integers.

Example: 3 24 7

− ÷ Invert divisor

3 74 2

−

21 52

8 8−

= −

Example:

9 2510 12

− −

9 25

10 12 − −

3 5 15 712 4 8 8

− − = + =

Multiply numerators and denominators, and simplify.

Multiply numerators and denominators, and simplify.

3 5

2 4

5 3

1 2

5 31 2

152172

=

=

Multiply numerators and denominators. Simplify.


1 2 63 5 151 1 5 3 3 15

1115

x

x

+ = − → −

− − → −

= −

25 3

5 2 51 5 3 1

10 1 33 3

x

x

x or

=−

− − = − −

= −

25 3

x=

−

2 4 203

x − ≤

Solving Equations and Inequalities Containing Fractions and Decimals

Syllabus Objective: (3.10) The student will solve equations and inequalities involving fractions and mixed numbers. (2.8) The student will solve multi-step inequalities.

First Strategy for Solving: You solve equations and inequalities containing fractions and decimals the same as you do with whole numbers; the strategy does not change. To solve linear equations or inequalities, put the variable terms on one side of the equal sign, and put the constant (number) terms on the other side. To do this, use opposite (inverse) operations.

Example: Solve: 1 2 .3 5

x + = −

Example: Solve:

Example: Solve:

Undo adding one-third by subtracting one-third from both sides of the equation; make equivalent fractions with a common denominator of 15.

Undo dividing by –5 by multiplying both sides by –5. Cancel. Multiply numerators and denominators, and simplify.


( ) ( )

2 4 203

4 4

2 243

23 3 243

2 72

2 722 2

36

x

x

x

x

x

x

− ≤

+ +

≤

≤

≤

≤

≤

2 5 5 3 0 7

5 3 5 3

2 8 0 7

2 8 0 70 7 0 7

4

. . . x

. .

. . x

. . x. .

x

= +

− = −

− =

−=

− =

We could have saved a little time by recognizing that multiplying by 3 and then dividing by 2

could have been done in one step by multiplying by the reciprocal 32

.

2 4 203

4 4

2 243

3 2 3 242 3 2

36

x

x

x

x

− ≤

+ +

≤

≤

≤

Example: Solve 2 5 5 3 0 7. . . x= +


Second Strategy for Solving: Another way to solve an equation or inequality with fractions is to “clear the fractions” by multiplying both sides of the equation or inequality by the LCD of the fractions. The resulting equation/inequality is equivalent to the original. You can also clear decimals by determining the greatest number of decimal places and multiplying both sides of the equation/inequality by that power of 10.

Example: Solve 2 553 2

x + = .

( )

2 553 2

2 56 5 63 2

2 56 6 5 63 2

4 30 15

30 30

4 15

4 154 4

334

x

x

x

x

x

x

x

+ =

+ =

+ =

+ =

− = −

= −

−=

= −

Example: Solve 5 14 0 8 2 3. . x .+ ≤ .

Original equation. Multiply each side by LCD of 6. Distribute. Simplify. Undo adding 30 by subtracting 30from both sides. Undo multiplying by 4 by dividing by 4. Simplify.

Original inequality. Since greatest number of decimals is 2, multiply by 210 or 100. Distribute and simplify. Undo addition by subtracting 514 from each side. Simplify. Undo multiplication by dividing both sides by 80. Simplify.

( ) ( )

5 14 0 8 2 3

100 5 14 0 8 100 2 3

514 80 230

514 514

80 284

80 28480 80

3 55

. . x .

. . x .

x

x

x

x .

+ ≤

+ ≤

+ ≤

− ≤ −

≤ −

−≤

≤ −


Example: Solve 2 875 9 12 45. .− + ≤ .

( ) ( )

2 875 9 12 45

1000 2 875 9 1000 12 45

2875 9000 12450

9000 9000

2875 3450

2875 34502875 2875

1 2

. x .

. x .

x

x

x

x .

− + ≤

− + ≤

− + ≤

− ≤ −

− ≤

−≥

− −

≥ −

Original inequality. Since greatest number of decimals is 3, multiply by 310 or 1000. Distribute and simplify. Undo addition by subtracting 9000 from each side. Simplify. Undo multiplication by dividing both sides by −2875. Reverse the inequality. Simplify.

Documents

Pre-Algebra Notes – Unit Five: Rational Numbers and Equationsrpdp.net/admin/images/uploads/resource_2635.pdfPre-Algebra Notes – Unit Five: Rational Numbers and Equations