sample problem #1: adding fractions

$: Sample Problem #1: Adding fractions$
Sample Gateway Problems:..

Working with Fractions and the Order of OperationsWithout Using a Calculator

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NOTE: Gateway problems 1 & 2 on adding and subtracting fractions can both

be done using the same set of steps.

Adding fractions and subtracting fractions both require finding a least common denominator (LCD), which is most easily done by factoring the denominator (bottom number) of each fraction into a product of prime numbers (a number that can be divided only by itself and 1.)

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Sample Problem #1: Adding fractions

Step 1: Factor the two denominators into prime factors, then write each fraction with its denominator in factored form:

10 = 2 5 ∙ and 35 = 5 7∙ , so 3 + 2 = 3 + 2 . . 10 35 2 5 ∙ 5 7∙Step 2: Find the least common denominator (LCD): LCD = 2 5∙ 7∙

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Sample Problem #1 (continued)Step 3: Multiply the numerator (top)and denominator of each fraction by the factor(s) needed to turn each denominator into the LCD.

LCD = 2 5∙ 7∙ 3 7 ∙ + 2 2 ∙ . 2 5∙ 7∙ 5 7∙ 2∙

Step 4: Multiply each numerator out, leaving the denominators in factored form, then add the two numerators and put them over the common denominator. 21 + 4 = 21 + 4 = 25 (note that 5 7 2 = 2 5 7 ∙ ∙ ∙ ∙

by 2 5 7 5 7 2 2 5 7 2 5 7 ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ the commutative

property)

Step 5: Now factor the numerator, then cancel any common factors that appear in both numerator and denominator. Once you multiply out any remaining factors, the result is your simplified answer.

= 25 = 5 5 ∙ = 5 5 ∙ = 5 = 5 . 2 5 7 2 5 7 2 5 7 2 7 14∙ ∙ ∙ ∙ ∙ ∙ ∙

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Full Solution to Sample Problem #1:

Here is the work we expect to see on your worksheet:

10 = 2 5 and 35 = 5 7, so ∙ ∙ 3 + 2 = 3 + 2 , and LCD = 2 5 7∙ ∙ 10 35 2 5 5 7∙ ∙

3 + 2 = 3 7∙ + 2 2 ∙ = 21 + 4 = 25 = 5 5 ∙ = 5 5 ∙ = 5 = 5 2 5 5 7 2 5 7 5 7 2 2 5 7 2 5 7 2 5 7 2 5 7 ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ 2 5 7 2 7 14∙ ∙ ∙

/ /

5 14

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Sample Problem #2: Subtracting fractions

Step 1: Factor the two denominators into prime factors, then write each fraction with its denominator in factored form:

14 = 2 7 ∙ and 35 = 5 7∙ , so 5 - 2 . 2 7 ∙ 5 7∙

Step 2: Find the least common denominator (LCD): LCD = 2 7∙ 5∙

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Sample Problem #2 (continued)Step 3: Multiply the numerator and denominator of each fraction by the factor(s) needed to turn each denominator into the LCD: form:

LCD = 2 7∙ 5∙ 5 5 ∙ - 2 2 ∙ 2 7∙ 5∙ 5 7∙ 2∙

Step 4: Multiply out the numerators, leaving the denominators in factored form, then add the two numerators and put them over the common denominator. 25 - 4 = 25 - 4 = 21 .

2 5 7 5 7 2 2 5 7 2 5 7∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙

Step 5: Now factor the numerator, then cancel any common factors that appear in both numerator and denominator. Once you multiply out any remaining factors, the result is your simplified answer.

21 = 3 7 ∙ = 3 7 ∙ = 3 = 3 . 2 5 7 2 5 7 2 5 7 2 5 10∙ ∙ ∙ ∙ ∙ ∙ ∙

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14 = 2 7 and 35 = 5 7, so ∙ ∙ 5 - 2 = 5 - 2 , and LCD = 2 5 7∙ ∙ 14 35 2 7 5 7∙ ∙

5 - 2 = 5 5 ∙ - 2 2 ∙ = 25 - 4 = 21 = 3 7 ∙ = 3 7 ∙ = 3 = 3 2 7 5 7 2 7 5 5 7 2 2 5 7 2 5 7 2 5 7 2 5 7 ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ 2 5 7 2 5 10∙ ∙ ∙

3 .

10

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NOTE: Gateway problems 3 & 5 on multiplying and dividing fractions can both

be done using the similar steps.

Neither multiplying fractions nor dividing fractions requires finding an LCD. These kinds of problems can be most easily done by factoring both the numerator (top number) and denominator of both fractions into a product of prime numbers, and then canceling any common factors (numbers that appear on both the top and the bottom.)9

Sample Problem #3: Multiplying fractions

Step 1: Factor both the numerators and denominators into prime factors, then write each fraction in factored form:

First fraction: 39= 3 13 ∙ and 50 = 2 5 5∙ ∙ Second fraction: 15= 3 5 ∙ and 26 = 2 13∙

So you can write 39 • 15 as 3 13 ∙ • 3 5∙ . 50 26 2 5 5∙ ∙ 2 13∙NOTE: You do NOT need an LCD when multiplying fractions.10

Sample Problem #3 (continued)

Step 2: Now just cancel any common factors that appear in both numerator and denominator. Once you multiply out any remaining factors, the result is your simplified answer.

3 13∙ • 3 5∙ = 3 3∙ = 9 .

2 5 5 2 13 2 5 2 20∙ ∙ ∙ ∙ ∙

/ / / /

NOTE: It is much easier to factor first and then cancel, rather than multiplying out the numerators and denominators and then trying to simplify the answer (especially if you aren’t using a calculator!) If you multiplied first, you’d have gotten 585 , which would be nasty to simplify by hand…1300

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39 • 15 = 3 13 ∙ • 3 5∙ = 3 13 ∙ • 3 5∙ = 3 3 ∙ = 9 . 50 26 2 5 5 2 13 2 5 5 2 13∙ ∙ ∙ ∙ ∙ ∙ 2 5 2 ∙ ∙20

/ / / /

9 .

20

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Sample Problem #5: Dividing fractions

Step 1: Multiply the first fraction by the 45 ÷ 21 = 45 • 26reciprocal of the second fraction. 13 26 14 21 (i.e. flip the second fraction upside down and change ÷ to • .)


First fraction: 45 = 3 3 5 ∙ ∙ and 13 = 13 (prime) Second fraction: 26 = 2 13 ∙ and 21 = 3 7 ∙

So you can write 45 • 26 as 3 3 5∙ ∙ • 2 13∙ . 13 21 13 3 7∙ 13


NOTE: You do NOT need an LCD when dividing fractions.

Step 3: Now just cancel any common factors that appear in both numerator and denominator. Once you multiply out any remaining factors, the result is your simplified answer.

3 3 5∙ ∙ • 2 13∙ = 3 5 2∙ ∙ = 30 13 3 7 7 7∙

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NOTE: Once again, it is much easier to factor first and then cancel, rather than multiplying out the numerators and denominators and then trying to simplify the answer (especially if you aren’t using a calculator!) If you multiplied first, you’d have gotten 1170 , which would be pretty hard to simplify by hand. 273 14



45 ÷ 21 = 45 • 26 = 3 3 5 ∙ ∙ • 2 13∙ = 3 3 5 ∙ ∙ • 2 13∙ 13 26 13 21 13 3 7 13 ∙3 7 ∙ .

= 3 5 2 ∙ ∙ = 30 1 7 7∙

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9 .

20

30 .

7

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NOTE: Gateway problems 4 & 6 using mixed numbers both start

with the same step.

A mixed number consists of an integer part and a fraction part. We want to covert the mixed number into an improper fraction, This is done by multiplying the integer part by the denominator of the fraction part, then adding that product to the numerator of the fraction and putting that sum over the original denominator. 16

Sample Problem #4: Multiplying mixed numbers

Step 1: Convert the mixed number into an improper fraction: (Note that ) .

52 2 23 3 1 35 5

235

5 5 15 172 2 21 3 1 3 33 3 3

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So becomes , which we can thensolve the same way we did problem #3.

623 75 17 6

3 717



First fraction: 17 and 3 are both prime Second fraction: 6 = 2 3∙ and 7 is prime

So you can write 17 ∙ 6 as 17 ∙ 2 3∙ . 3 7 3 7Step 3: Now just cancel any common factors that appear in both numerator and denominator. Once you multiply out any remaining factors, the result is your simplified answer.

17 63 7

//

17 ∙ 2 3∙ = 17 2∙ = 34 . 3 7 7 7

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5 5 3 15 172 2 2 2 23 3 1 3 13 3 3 3 35 5

6 17 6 17 2 3 17 2 3 17 2 3423 7 3 7 3 7 3 7 7 75

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Sample Problem #6: Dividing mixed numbers

Step 1: Convert the mixed numbers into improper fractions:

77 7 49 501 1 1 1 1

7 7 1 7 1 7 7 7 777 7

251 1 12 1 12 22

1 24 12 2 1 2 1 2 2 2 212 12

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First fraction: 50 = 2 5 5 ∙ ∙ and 7 is prime Second fraction: 2 is prime and 25 = 5 5∙So you can write 50 • 2 as 2 5 5∙ ∙ • 2 . 7 25 7 5 5∙Step 3: Now just cancel any common factors that appear in Both numerator and denominator. Once you multiply out anyremaining factors, the result is your simplified answer.

2 5 5∙ ∙ • 2 = 2 2∙ = 4 7 5 5 7 7∙

50 25 501 1 27 2 7 2 7 257 12

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/ / / /

7 7 7 49 501 1 1 1 17 7 1 7 17 7 7 7 77 7

251 1 12 1 12 2 1 24 1

2 2 1 2 1 2 2 2 2 212 12

50 25 501 1 27 2 7 2 7 25

2 5 5 2 5 52 2 2 2 47 5 5 7 5 5 7 7

7 12

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NOTE: Gateway problems 7 & 8 both require using the order of operations.

Order of operations:1) First, calculate expressions within grouping symbols

(parentheses, brackets, braces,absolute values, fraction bars).

If there are nested sets of grouping symbols, start with the innermost ones first and work your way out.

2) Exponential expressions – left to right

3) Multiplication and division – left to right

4) Addition and subtraction – left to right

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Order of operations memory device:

“Please excuse my dear Aunt Sally”

1. Please (Parentheses)

2. Excuse (Exponents)

3. My Dear (Multiply and Divide)

4. Aunt Sally (Add and Subtract)

… or just remember PEMDAS

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Sample Problem # 7: Order of Operations

Strategy: Calculate out the entire top expression and then the entire bottom expression, using the order of operations on each part. Then simplify the resulting fraction, if necessary.

TOP EXPRESSION: 24 – 4(7 + 2)

Step 1: Parentheses: 24 – 4(7 + 2) = 24 – 4(9)

Step 2: Exponents: 24 – 4(9) = 2•2•2•2 – 4(9) = 16 – 4(9) (because 2•2•2•2 = 4•2•2 = 8•2 = 16)

Step 3: Multiply/Divide: 16 – 4(9) = 16 – 4•9 = 16 – 36

Step 4: Add/Subtract: 16 – 36 = -20 25

Now calculate the bottom expression: 2(6+2) + 4

Step 1: Parentheses: 2(6+2) + 4 = 2(8) + 4

Step 2: Exponents: There aren’t any in this part.

Step 3: Multiply/Divide: 2(8) + 4 = 2•8 + 4 = 16 + 4

Step 4: Add/Subtract: 16 + 4 = 20

Now put the top over the bottom and simplify the resulting fraction:

TOP = 24 – 4(7 + 2) = -20 = -1 = -1BOTTOM 2(6+2) + 4 20 1

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24 – 4(7 + 2) = 24 – 4(9) = 16 – 4(9) = 16 – 36 = -20 = -1 = -1 2(6+2) + 4 2(8) + 4 16 + 4 20 20 1

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Sample Problem # 8: Order of Operations

Strategy: Deal with the expressions inside the grouping symbols (parentheses, brackets) first, starting with the innermost set (-3 + 6).

STEP 1: (inside the parentheses) 3[17 + 5(-3 + 6) - 10] = 3[17 + 5(3) - 10] STEP 2: (inside the brackets; multiply first, then add and subtract)

3[17 + 5(3) -10] = 3[17 + 5•3 -10] = 3[17 + 15 - 10] = 3[17 + 15 - 10] = 3[32 - 10] = 3[22] STEP 3: Do the final multiplication: 3[22] = 3•22 = 66 28



3[17 + 5(-3 + 6) - 10] = 3[17 + 5(3) - 10] =

3[17 + 15 - 10] = 3[32 - 10] = 3[22] = 66

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sample problem #1: adding fractions

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