- In this case, to change the mixed number to an improper fraction, you need to divide 52 by 6. Doing so gives you 8 with 4 remaining. So 8 is the whole number and you place the remainder (4) over the denominator (6) for an answer of 8 4/6. 8 4/6 can be reduced further because 4 and 6 are divisible by 2.
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(ii) Find 2/5 of 15. To find 2/5 of 15, we multiply the numerator 2 by the given whole number 15 and then divide the product 30 by the denominator 5. Solve word problems involving division of whole numbers leading to answers in the form of fractions or mixed numbers, e.g., by using visual fraction models or equations to represent the problem. For example, interpret 3/4 as the result of dividing 3 by 4, noting that 3/4 multiplied by 4 equals 3, and that when 3 wholes are shared equally among.
Dividing fractions is as easy as pie! In fact, after you change mixed numbers to improper fractions, dividing fractions is just like multiplying fractions, except that you flip the numerator and the denominator of the second fraction.
Example: You bought 6 1/2 pounds of sirloin steak, and you want to cut it into pieces that weigh 3/4 pound each. How many pieces will you have?
First, change the word problem into an equation, which looks like this:
Change all mixed numbers to improper fractions.
The only mixed number is 6 1/2, so multiply the denominator by the whole number (2 × 6 = 12), and then add that result to the numerator (12 + 1 = 13). Now the problem looks like this:
Invert (flip) the second fraction, placing the bottom number on top and the top number on the bottom; then change the division symbol to a multiplication symbol.
Multiply the fractions, and reduce if necessary.
Multiply the numerators, straight across the top of the fraction line (13 × 4 = 52). And then multiply the denominators straight across the bottom (2 × 3 = 6).
Always change your answers to mixed numbers and reduce them, if possible.
In this case, to change the mixed number to an improper fraction, you need to divide 52 by 6. Doing so gives you 8 with 4 remaining. So 8 is the whole number and you place the remainder (4) over the denominator (6) for an answer of 8 4/6. 8 4/6 can be reduced further because 4 and 6 are divisible by 2. After you divide the numerator and denominator by 2, your answer becomes 8 2/3.
Here’s how it looks:
You end up with 8 hefty steaks that weigh 3/4 pound each (or 12 ounces), and one steak that weighs 8 ounces (2/3 of 12).
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23
THERE IS ONE RULE for adding or subtracting fractions: The denominators must be the same -- just as in arithmetic.
Add the numerators, and place their sum
over the common denominator.
| Example 1. | 6x + 3 5 | + | 4x − 1 5 | = | 10x + 2 5 |
The denominators are the same. Add the numerators as like terms.
| Example 2. | 6x + 3 5 | − | 4x − 1 5 |
To subtract, change the signs of the subtrahend, and add.
| 6x + 3 5 | − | 4x − 1 5 | = | 6x + 3 − 4x + 1 5 | = | 2x + 4 5 |
Problem 1.
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Do the problem yourself first!
| a) | x 3 | + | y 3 | = | x + y 3 | b) | 5 x | − | 2 x | = | 3 x |
| c) | x x − 1 | + | x + 1 x − 1 | = | 2x + 1 x − 1 | d) | 3x − 4 x − 4 | + | x − 5 x − 4 | = | 4x − 9 x − 4 |
| e) | 6x + 1 x − 3 | − | 4x + 5 x − 3 | = | 6x + 1 − 4x − 5 x − 3 | = | 2x − 4 x − 3 |
| f) | 2x − 3 x − 2 | − | x − 4 x − 2 | = | 2x − 3 − x + 4 x − 2 | = | x + 1 x − 2 |
Different denominators -- The LCM
To add fractions with different denominators, we must learn how to construct the Lowest Common Multiple of a series of terms.
The Lowest Common Multiple (LCM) of a series of terms
is the smallest product that contains every factor of each term.
For example, consider this series of three terms:
pqprps
We will now construct their LCM -- factor by factor.
To begin, it will have the factors of the first term:
LCM = pq
Moving on to the second term, the LCM must have the factors pr. But it already has the factor p -- therefore, we need add only the factor r:
LCM = pqr
Finally, moving on to the last term, the LCM must contain the factors ps. But again it has the factor p, so we need add only the factor s:
LCM = pqrs.
That product is the Lowest Common Multiple of pq, pr, ps. It is the smallest product that contains each of them as factors.
Example 3. Construct the LCM of these three terms: x, x2, x3.
Solution. The LCM must have the factor x.
LCM = x
But it also must have the factors of x2 -- which are x ·x. Therefore, we must add one more factor of x :
LCM = x2
Finally, the LCM must have the factors of x3, which are x·x·x. Therefore,
LCM = x3.
x3 is the smallest product that contains x, x2, and x3 as factors.
We see that when the terms are powers of a variable -- x, x2, x3 -- then their LCM is the highest power.
Problem 2. Construct the LCM of each series of terms.
| a) | ab, bc, cd. abcd | b) | pqr, qrs, rst. pqrst |
| c) | a, a2, a3, a4. a4 | d) | a2b, ab2. a2b2 |
e) ab, cd. abcd
We will now see what this has to do with adding fractions.
| Example 4. Add: | 3 ab | + | 4 bc | + | 5 cd |
Solution. To add fractions, the denominators must be the same. Therefore, as a common denominator choose the LCM of the original denominators. Choose abcd. Then, convert each fraction to an equivalent fraction with denominator abcd.
It is necessary to write the common denominator only once:
| 3 ab | + | 4 bc | + | 5 cd | = | 3cd + 4ad + 5ab abcd |
To change into an equivalent fraction with denominator abcd, simply multiply ab by the factors it is missing, namely cd. Therefore, we must also multiply 3 by cd. That accounts for the first term in the numerator.
To change into an equivalent fraction with denominator abcd, multiply bc by the factors it is missing, namely ad. Therefore, we must also multiply 4 by ad. That accounts for the second term in the numerator.
To change into an equivalent fraction with denominator abcd, multiply cd by the factors it is missing, namely ab. Therefore, we must also multiply 5 by ab. That accounts for the last term in the numerator.
That is how to add fractions with different denominators.
Each factor of the original denominators must be a factor
of the common denominator.
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Problem 3. Add.
| a) | 5 ab | + | 6 ac | = | 5c + 6b abc |
| b) | 2 pq | + | 3 qr | + | 4 rs | = | 2rs + 3ps + 4pq pqrs |
| c) | 7 ab | + | 8 bc | + | 9 abc | = | 7c + 8a + 9 abc |
| d) | 1 a | + | 2 a2 | + | 3 a3 | = | a2 + 2a + 3 a3 |
| e) | 3 a2b | + | 4 ab2 | = | 3b + 4a a2b2 |
| f) | 5 ab | + | 6 cd | = | 5cd + 6ab abcd |
| g) | _2_ x(x + 2) | + | __3__ (x + 2)(x − 3) | = | 2(x − 3) + 3x x(x + 2)(x − 3) |
| = | _ 2x − 6 + 3x_ x(x + 2)(x − 3) | ||||
| = | _5x − 6_ x(x + 2)(x − 3) |
At the 2nd Level we will see a similar problem, but the denominators will not be factored.
| Problem 4. Add: 1 − | 1 a | + | c + 1 ab | . But write the answer as |
1 − a fraction.
| 1 − | 1 a | + | c + 1 ab | = | 1 − ( | 1 a | − | c + 1 ab | ) |
| = | 1 − | b − (c + 1) ab |
| = | 1 − | b − c − 1 ab |
Example 5. Denominators with no common factors.
| a m | + | b n |
When the denominators have no common factors, their LCM is simply their product, mn.
| a m | + | b n | = | an + bm mn |
The numerator then appears as the result of 'cross-multiplying' :
an + bm
However, that technique will work only when adding two fractions, and the denominators have no common factors.
| Example 6. | 2 x − 1 | − | 1 x |
Solution. These denominators have no common factors -- x is not a factor of x − 1. It is a term. Therefore, the LCM of denominators is their product.
| 2 x − 1 | − | 1 x | = | 2x − (x − 1) (x − 1)x | = | 2x − x + 1 (x − 1)x | = | _x + 1_ (x − 1)x |
Note: The entire x − 1 is being subtracted. Therefore, we write it in parentheses -- and its signs change.
Problem 5.
| a) | x a | + | y b | = | xb + ya ab | b) | x 5 | + | 3x 2 | = | 2x + 15x 10 | = | 17x 10 |
| c) | 6 x − 1 | + | 3 x + 1 | = | 6(x + 1) + 3(x − 1) (x + 1)(x − 1) |
| = | 6x + 6 + 3x − 3 (x + 1)(x − 1) | ||||
| = | _9x + 3_ (x + 1)(x − 1) |
| d) | 6 x − 1 | − | 3 x + 1 | = | 6(x + 1) − 3(x − 1) (x + 1)(x − 1) |
| = | 6x + 6 − 3x + 3 (x + 1)(x − 1) | ||||
| = | _3x + 9_ (x + 1)(x − 1) |
| e) | 3 x − 3 | − | 2 x | = | 3x − 2(x − 3) (x − 3)x |
| = | 3x − 2x + 6 (x − 3)x | ||||
| = | x + 6 (x − 3)x |
| f) | 3 x − 3 | − | 1 x | = | 3x − (x − 3) (x − 3)x |
| = | 3x − x + 3 (x − 3)x | ||||
| = | 2x + 3 (x − 3)x |
| g) | 1 x | + | 2 y | + | 3 z | = | yz + 2xz + 3xy xyz |
| Example 7. Add: a + | b c | . |
Solution. We have to express a with denominator c.
| a | = | ac c | (Lesson 20) |
2.3 Changing Form: Fractions And Decimalsmr. Mac's Page Using
Therefore,
| a + | b c | = | ac + b c | . |
Problem 6.
| a) | p q | + r | = | p + qr q | b) | 1 x | − 1 | = | 1 − x x |
| c) x − | 1 x | = | x2 − 1 x | d) 1 − | 1 x2 | = | x2 − 1 x2 |
| e) 1 − | 1 x + 1 | = | x + 1 − 1 x + 1 | = | x x + 1 |
| f) 3 + | 2 x + 1 | = | 3x + 3 + 2 x + 1 | = | 3x + 5 x + 1 |
| Problem 7. Write the reciprocal of | 1 2 | + | 1 3 | . |
| [Hint: Only a single fraction | a b | has a reciprocal; it is | b a | .] |
| 1 2 | + | 1 3 | = | 3 + 2 6 | = | 5 6 |
| Therefore, the reciprocal is | 6 5 | . |
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