Adding Mixed Numbers © Math As A Second Language All Rights Reserved next #7 Taking the Fear out of Math 1313 1 1212 2 +

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Adding Mixed Numbers © Math As A Second Language All Rights Reserved next #7 Taking the Fear out of Math

Students have probably added mixed numbers many times and not even realized that they were doing it. © Math As A Second Language All Rights Reserved A Non-Technical Introduction next For example, suppose you rent a machine at a cost of $12 per hour and you have to pay for it to the nearest minute. Suppose further that you used the machine for 6 hours and 40 minutes one day and for 3 hours and 45 minutes the next day, and you want too compute how much it cost you to rent the machine during the two days.

© Math As A Second Language All Rights Reserved next One way to proceed would be to start by computing the total time you used the machine, namely… 6 hours + 40 minutes + 3 hours + 45 minutes Without thinking about it consciously you would probably use the properties of arithmetic to conclude that… 6 hours + 40 minutes + 3 hours + 45 minutes = 6 hours + 3 hours + 40 minutes + 45 minutes = 9 hours + 85 minutes

Recognizing that 85 minutes is more than 1 hour, the sum could be rewritten as… © Math As A Second Language All Rights Reserved next 9 hours + 60 minutes + 25 minutes = 9 hours + 1 hour + 25 minutes = 10 hours + 25 minutes

Since 10 hours and 25 minutes is more than 10 hours but less than 11 hours, we see that at a cost of $12 per hour, the total cost is more than 10 × $12 or $120 but less than 11 × $12 or $132. © Math As A Second Language All Rights Reserved next To find the exact answer, we might next observe that… 25 minutes = of an hour = 5 12 of an hour …and therefore that the additional cost is 5 / 12 of $12 or $5.

© Math As A Second Language All Rights Reserved In summary, the cost for 10 hours is $120 and the cost for the additional 25 minutes is $5. Therefore the total cost is $120 + $5 or $125. If we suffered from “fraction aversion”, we might have decided to work solely with whole numbers. In that case, we would have converted the times into minutes rather than hours. next

© Math As A Second Language All Rights Reserved Since there are 60 minutes in an hour, 10 hours would equal 600 minutes, and… Using the fact that 100 cents equals one dollar, the total cost is × 20 cents = 12,500 cents = $ This checks with the answer, we obtained previously. next 10 hours + 25 minutes = 600 minutes + 25 minutes = 625 minutes

Although the arithmetic appears to be more abstract, adding mixed numbers is no more complicated than what we did above. © Math As A Second Language All Rights Reserved Adding Mixed Numbers next As we saw in our previous presentation, if 38 corn breads are divided equally among 7 people the number of corn breads each person gets is / 7.

The fact that the plus sign separates the whole number from the fraction makes it easy for one to assume that only the fraction is modifying “corn breads” 1. © Math As A Second Language All Rights Reserved next To avoid the possibility of misinterpretation, we should use parentheses and write (5 + 3 / 7 ) corn breads. note 1 This type of misinterpretation happens frequently to beginning students in algebra. For example, when they are trying to simplify an expression such as (a + b)x they rewrite the expression as a + bx rather than as ax + bx.

However, because it’s cumbersome to write the answer in this form, we agree to omit the plus sign and write the fractional part immediately to the right of the whole number; that is, 5 3 / 7 means (5 + 3 / 7 ). © Math As A Second Language All Rights Reserved next In any event, the associative and commutative properties of addition now show us that to add two mixed numbers… We simply have to add the two whole numbers to get the whole number part of the sum and the two fractions to get the fraction part of the sum. next

© Math As A Second Language All Rights Reserved For example, let’s find the sum of 5 2 / 7 and 6 4 / 7 as a mixed number. First, we may rewrite the problem in the form… next (5 + 2 / 7 ) + (6 + 4 / 7 ) Then using the associative and commutative properties of addition, we may rewrite the above expression in the form… (5 + 6) + ( 2 / / 7 ) …from which we see, the sum is / 7 or in more standard form, 11 6 / 7.

© Math As A Second Language All Rights Reserved If we are uncomfortable with mixed numbers but are more comfortable with common fractions, we can translate every mixed number problem into an equivalent (improper) fraction problem. Conversely, if we prefer, we can convert improper fractions into mixed numbers. next

© Math As A Second Language All Rights Reserved Hence, 5 2 / / 7 = 37 / / 7 = 83 / 7 = 11 6 / 7. next 5 2 / 7 =5 + 2 / 7 = 5 / / 7 = 35 / / 7 = 37 / / 7 =6 + 4 / 7 = 6 / / 7 = 42 / / 7 = 46 / 7 Mixed Numbers to Improper Fractions next However, in this case it is much simpler just to add the whole numbers and to add the fractional parts.

For example, if one person had 5 2 / 7 corn breads and another person has 6 4 / 7 corn breads, the total corn breads is 11 6 / 7 corn breads. © Math As A Second Language All Rights Reserved Using the Corn Bread next corn bread 1/71/7 1/71/7 1/71/7 1/71/7 1/71/7 1/71/7 1/71/7 52/752/7 64/764/7 1/71/7 1/71/7 1/71/7 1/71/7 1/71/7 1/71/7 1/71/7 1/71/7 1/71/ / 7 +=

next © Math As A Second Language All Rights Reserved As another example, let’s find the sum of 6 2 / 3 and 3 3 / 4 as a mixed number. First, we may rewrite the problem in the form… next (6 + 2 / 3 ) + (3 + 3 / 4 ) Then using the associative and commutative properties of addition, we may rewrite the above expression in the form… (6 + 3) + ( 2 / / 4 ) A common denominator for 2 / 3 and 3 / 4 is 12.

© Math As A Second Language All Rights Reserved 6 2 / / 4 next Since 2 / 3 equals 8 / 12, and 3 / 4 equals 9 / 12, we can combine the fractions using the common denominator 12… = (6 + 3) + ( 2 / / 4 ) = (6 + 3) + ( 8 / / 12 ) = 9 + ( 17 / 12 ) = 9 + ( 12 / / 12 ) = 9 + (1 + 5 / 12 ) = / 12 = 10 5 / 12

We would not write the answer as 9 17 / 12. Remember that to be a mixed number, the fractional part has to be less than 1. © Math As A Second Language All Rights Reserved Notes next Notice that because the denominators were not the same, we had to find a common denominator before we could add the fractions.

Notice that because the denominator was 12, we traded 12 twelfths for 1 whole rather ten twelfths. © Math As A Second Language All Rights Reserved Notes next Very often students make the mistake of not looking at the denominator of a mixed number and continue to “trade in” by tens no matter what the denominator was.

Once again we could have solved this problem by converting the two mixed numbers to improper fractions and then using our algorithm for adding fractions. © Math As A Second Language All Rights Reserved Notes next However, this process is both more cumbersome and more “mechanical”.

This problem brings us back full circle to our opening discussion. © Math As A Second Language All Rights Reserved next If we assume that 6 2 / 3 and 3 3 / 4 modify “hours”, 40 minutes is 2 / 3 of an hour and 45 minutes is 3 / 4 of an hour. So if you rented a machine for 6 hours and 40 minutes one day and for 3 hours and 45 minutes the next day, the total amount of time that you rented the machine was 10 hours and 25 minutes.

If we convert our minutes to hours we would obtain… © Math As A Second Language All Rights Reserved next 6 hours + 40 minutes + 3 hours + 45 minutes = 6 hours + 3 hours + 40 minutes + 45 minutes = 9 hours + 85 minutes = 9 hours + 60 minutes + 25 minutes = 9 hours + 1 hour + 25 minutes = 10 hours + 25 minutes = 10 hours + 25 / 60 hours = 10 hours + 5 / 12 hours = 10 5 / 12 hours

10 5 / 12 hours is exactly what we obtained earlier in our presentation. © Math As A Second Language All Rights Reserved next Sometimes using such nouns as hours and minutes, as we just did above, allows students to better visualize what is happening when we add mixed numbers.

In our next section we will discuss the process of subtracting one mixed number from another. © Math As A Second Language All Rights Reserved 6 2 / 3 – 3 3 / 4 = ?