Articles

4.14: Add and Subtract Fractions with Different Denominators (Part 2) - Mathematics


Identify and Use Fraction Operations

By now in this chapter, you have practiced multiplying, dividing, adding, and subtracting fractions. The following table summarizes these four fraction operations. Remember: You need a common denominator to add or subtract fractions, but not to multiply or divide fractions.

Summary of Fraction Operations

Fraction multiplication: Multiply the numerators and multiply the denominators.

[dfrac{a}{b} cdot dfrac{c}{d} = dfrac{ac}{bd}]

Fraction division: Multiply the first fraction by the reciprocal of the second.

[dfrac{a}{b} div dfrac{c}{d} = dfrac{a}{b} cdot dfrac{d}{c}]

Fraction addition: Add the numerators and place the sum over the common denominator. If the fractions have different denominators, first convert them to equivalent forms with the LCD.

[dfrac{a}{c} + dfrac{b}{c} = dfrac{a + b}{c}]

Fraction subtraction: Subtract the numerators and place the difference over the common denominator. If the fractions have different denominators, first convert them to equivalent forms with the LCD.

[dfrac{a}{c} - dfrac{a}{c} = dfrac{a - b}{c}]

Example (PageIndex{11}): simplify

Simplify:

  1. (− dfrac{1}{4} + dfrac{1}{6})
  2. (− dfrac{1}{4} div dfrac{1}{6})

Solution

First we ask ourselves, “What is the operation?”

  1. The operation is addition. Do the fractions have a common denominator? No.
Find the LCD.
Rewrite each fraction as an equivalent fraction with the LCD.(- dfrac{1 cdot extcolor{red}{3}}{4 cdot extcolor{red}{3}} + dfrac{1 cdot extcolor{red}{2}}{6 cdot extcolor{red}{2}} )
Simplify the numerators and denominators.(- dfrac{3}{12} + dfrac{2}{12} )
Add the numerators and place the sum over the common denominator.(- dfrac{1}{12} )
Check to see if the answer can be simplified. It cannot.
  1. The operation is division. We do not need a common denominator.
To divide fractions, multiply the first fraction by the reciprocal of the second.(- dfrac{1}{4} cdot dfrac{6}{1})
Multiply.(- dfrac{6}{4})
Simplify.(- dfrac{3}{2} )

Exercise (PageIndex{21})

Simplify:

  1. (− dfrac{3}{4} - dfrac{1}{6})
  2. (− dfrac{3}{4} cdot dfrac{1}{6})
Answer a

(-dfrac{11}{12})

Answer b

(-dfrac{1}{8})

Exercise (PageIndex{22})

Simplify:

  1. (dfrac{5}{6} div left(- dfrac{1}{4} ight))
  2. (dfrac{5}{6} - left(- dfrac{1}{4} ight))
Answer a

(-dfrac{10}{3})

Answer b

(dfrac{13}{12})

Example (PageIndex{12}): simplify

Simplify:

  1. (dfrac{5x}{6} - dfrac{3}{10})
  2. (dfrac{5x}{6} cdot dfrac{3}{10})

Solution

  1. The operation is subtraction. The fractions do not have a common denominator.
Rewrite each fraction as an equivalent fraction with the LCD, 30.(dfrac{5x cdot extcolor{red}{5}}{6 cdot extcolor{red}{5}} - dfrac{3 cdot extcolor{red}{3}}{10 cdot extcolor{red}{3}} )
(dfrac{25x}{30} - dfrac{9}{30} )
Subtract the numerators and place the difference over the common denominator.(dfrac{25x - 9}{30} )
  1. The operation is multiplication; no need for a common denominator.
To multiply fractions, multiply the numerators and multiply the denominators.(dfrac{5x cdot 3}{ 6 cdot 10} )
Rewrite, showing common factors.(dfrac{cancel{5} cdot x cdot cancel{3}}{2 cdot cancel{3} cdot 2 cdot cancel{5}} )
Remove common factors to simplify.(dfrac{x}{4} )

Exercise (PageIndex{23})

Simplify:

  1. (dfrac{3a}{4} - dfrac{8}{9})
  2. (dfrac{3a}{4} cdot dfrac{8}{9})
Answer a

(dfrac{27}{a})

Answer b

(dfrac{2}{a})

Exercise (PageIndex{24})

Simplify:

  1. (dfrac{4k}{5} + dfrac{5}{6})
  2. (dfrac{4k}{5} div dfrac{5}{6})
Answer a

(dfrac{24}{k})

Answer b

(dfrac{24}{k})

Use the Order of Operations to Simplify Complex Fractions

In Multiply and Divide Mixed Numbers and Complex Fractions, we saw that a complex fraction is a fraction in which the numerator or denominator contains a fraction. We simplified complex fractions by rewriting them as division problems. For example,

[dfrac{dfrac{3}{4}}{dfrac{5}{8}} = dfrac{3}{4} div dfrac{5}{8} onumber ]

Now we will look at complex fractions in which the numerator or denominator can be simplified. To follow the order of operations, we simplify the numerator and denominator separately first. Then we divide the numerator by the denominator.

HOW TO: SIMPLIFY COMPLEX FRACTIONS

Step 1. Simplify the numerator.

Step 2. Simplify the denominator.

Step 3. Divide the numerator by the denominator.

Step 4. Simplify if possible.

Example (PageIndex{13}): simplify

Simplify: (dfrac{left(dfrac{1}{2} ight)^{2}}{4 + 3^{2}}).

Solution

Simplify the numerator.(dfrac{dfrac{1}{4}}{4 + 3^{2}})
Simplify the term with the exponent in the denominator.(dfrac{dfrac{1}{4}}{4 + 9} )
Add the terms in the denominator.(dfrac{dfrac{1}{4}}{13} )
Divide the numerator by the denominator.(dfrac{1}{4} div 13 )
Rewrite as multiplication by the reciprocal.(dfrac{1}{4} cdot dfrac{1}{13} )
Multiply.(dfrac{1}{52})

Exercise (PageIndex{25})

Simplify: (dfrac{left(dfrac{1}{3} ight)^{2}}{2^{3} + 2}).

Answer

(dfrac{1}{90})

Exercise (PageIndex{26})

Simplify: (dfrac{1 + 4^{2}}{left(dfrac{1}{4} ight)^{2}}).

Answer

(272)

Example (PageIndex{14}): simplify

Simplify: (dfrac{dfrac{1}{2} + dfrac{2}{3}}{dfrac{3}{4} - dfrac{1}{6}}).

Solution

Rewrite numerator with the LCD of 6 and denominator with LCD of 12.(dfrac{dfrac{3}{6} + dfrac{4}{6}}{dfrac{9}{12} - dfrac{2}{12}} )
Add in the numerator. Subtract in the denominator.(dfrac{dfrac{7}{6}}{dfrac{7}{12}} )
Divide the numerator by the denominator.(dfrac{7}{6} div dfrac{7}{12})
Rewrite as multiplication by the reciprocal.(dfrac{7}{6} cdot dfrac{12}{7} )
Rewrite, showing common factors.(dfrac{cancel{7} cdot cancel{6} cdot 2}{cancel{6} cancel{7} cdot 1} )
Simplify.(2 )

Exercise (PageIndex{27})

Simplify: (dfrac{dfrac{1}{3} + dfrac{1}{2}}{dfrac{3}{4} - dfrac{1}{3}}).

Answer

(2)

Exercise (PageIndex{28})

Simplify: (dfrac{dfrac{2}{3} - dfrac{1}{2}}{dfrac{1}{4} + dfrac{1}{3}}).

Answer

(dfrac{2}{7})

Evaluate Variable Expressions with Fractions

We have evaluated expressions before, but now we can also evaluate expressions with fractions. Remember, to evaluate an expression, we substitute the value of the variable into the expression and then simplify.

Example (PageIndex{15}): evaluate

Evaluate (x + dfrac{1}{3}) when

  1. (x = - dfrac{1}{3})
  2. (x = - dfrac{3}{4})

Solution

  1. To evaluate (x + dfrac{1}{3}) when (x = − dfrac{1}{3}), substitute (− dfrac{1}{3}) for (x) in the expression.
Substitute ( extcolor{red}{- dfrac{1}{3}}) for x.( extcolor{red}{- dfrac{1}{3}} + dfrac{1}{3} )
Simplify.(0 )
  1. To evaluate (x + dfrac{1}{3}) when (x = − dfrac{3}{4}), we substitute (− dfrac{3}{4}) for (x) in the expression.
Substitute ( extcolor{red}{- dfrac{3}{4}}) for x.( extcolor{red}{- dfrac{1}{3}} + dfrac{1}{3})
Rewrite as equivalent fractions with the LCD, 12.(- dfrac{3 cdot 3}{4 cdot 3} + dfrac{1 cdot 4}{3 cdot 4} )
Simplify the numerators and denominators.(- dfrac{9}{12} + dfrac{4}{12} )
Add.(- dfrac{5}{12} )

Exercise (PageIndex{29})

Evaluate (x + dfrac{3}{4}) when:

  1. (x = - dfrac{7}{4})
  2. (x = - dfrac{5}{4})
Answer a

(-1)

Answer b

(-dfrac{1}{2})

Exercise (PageIndex{30})

Evaluate (y + dfrac{1}{2}) when:

  1. (y = dfrac{2}{3})
  2. (y = - dfrac{3}{4})
Answer a

(dfrac{7}{6})

Answer b

(-dfrac{1}{4})

Example (PageIndex{16}): evaluate

Evaluate (y − dfrac{5}{6}) when (y = - dfrac{2}{3}).

Solution

We substitute (− dfrac{2}{3}) for (y) in the expression.

Substitute ( extcolor{red}{- dfrac{2}{3}}) for y.( extcolor{red}{- dfrac{2}{3}} - dfrac{5}{6})
Rewrite as equivalent fractions with the LCD, 6.(- dfrac{4}{6} - dfrac{5}{6} )
Subtract.(- dfrac{9}{6} )
Simplify.(- dfrac{3}{2} )

Exercise (PageIndex{31})

Evaluate (y − dfrac{1}{2}) when (y = - dfrac{1}{4}).

Answer

(-dfrac{3}{4})

Exercise (PageIndex{32})

Evaluate (x − dfrac{3}{8}) when (x = - dfrac{5}{2}).

Answer

(-dfrac{23}{8})

Example (PageIndex{17}):

Evaluate (2x^2y) when (x = dfrac{1}{4}) and (y = − dfrac{2}{3}).

Solution

Substitute the values into the expression. In (2x^2y), the exponent applies only to (x).

Substitute ( extcolor{red}{dfrac{1}{4}}) for x and ( extcolor{blue}{- dfrac{2}{3}}) for y.(2 left( extcolor{red}{dfrac{1}{4}} ight)^{2} left( extcolor{blue}{- dfrac{2}{3}} ight) )
Simplify exponents first.(2 left(dfrac{1}{16} ight) left(- dfrac{2}{3} ight))
Multiply. The product will be negative.(- dfrac{2}{1} cdot dfrac{1}{16} cdot dfrac{2}{3} )
Simplify.(- dfrac{4}{48} )
Remove the common factors.(- dfrac{1 cdot cancel{4}}{cancel{4} cdot 12} )
Simplify.(- dfrac{1}{12} )

Exercise (PageIndex{33})

Evaluate: (3ab^2) when (a = − dfrac{2}{3}) and (b = − dfrac{1}{2}).

Answer

(-dfrac{1}{2})

Exercise (PageIndex{34})

Evaluate: (4c^3d) when (c = − dfrac{1}{2}) and (d = − dfrac{4}{3}).

Answer

(dfrac{2}{3})

Example (PageIndex{18}): evaluate

Evaluate: (dfrac{p + q}{r}) when (p = −4), (q = −2), and (r = 8).

Solution

We substitute the values into the expression and simplify.

Substitute ( extcolor{red}{-4}) for p, ( extcolor{blue}{-2}) for q and ( extcolor{magenta}{8}) for r.(dfrac{ extcolor{red}{-4} + extcolor{blue}{(-2)}}{ extcolor{magenta}{8}} )
Add in the numerator first.(- dfrac{6}{8})
Simplify.(- dfrac{3}{4})

Exercise (PageIndex{35})

Evaluate: (dfrac{a + b}{c}) when (a = −8), (b = −7), and (c = 6).

Answer

(-dfrac{5}{2})

Exercise (PageIndex{36})

Evaluate: (dfrac{x + y}{z}) when (x = 9), (y = −18), and (z =- 6).

Answer

(dfrac{3}{2})

Practice Makes Perfect

Find the Least Common Denominator (LCD)

In the following exercises, find the least common denominator (LCD) for each set of fractions.

  1. (dfrac{2}{3}) and (dfrac{3}{4})
  2. (dfrac{3}{4}) and (dfrac{2}{5})
  3. (dfrac{7}{12}) and (dfrac{5}{8})
  4. (dfrac{9}{16}) and (dfrac{7}{12})
  5. (dfrac{13}{30}) and (dfrac{25}{42})
  6. (dfrac{23}{30}) and (dfrac{5}{48})
  7. (dfrac{21}{35}) and (dfrac{39}{56})
  8. (dfrac{18}{35}) and (dfrac{33}{49})
  9. (dfrac{2}{3}, dfrac{1}{6}) and (dfrac{3}{4})
  10. (dfrac{2}{3}, dfrac{1}{4}) and (dfrac{3}{5})

Convert Fractions to Equivalent Fractions with the LCD

In the following exercises, convert to equivalent fractions using the LCD.

  1. (dfrac{1}{3}) and (dfrac{1}{4}), LCD = 12
  2. (dfrac{1}{4}) and (dfrac{1}{5}), LCD = 20
  3. (dfrac{5}{12}) and (dfrac{7}{8}), LCD = 24
  4. (dfrac{7}{12}) and (dfrac{5}{8}), LCD = 24
  5. (dfrac{13}{16}) and (- dfrac{11}{12}), LCD = 48
  6. (dfrac{11}{16}) and (- dfrac{5}{12}), LCD = 48
  7. (dfrac{1}{3}, dfrac{5}{6}), and (dfrac{3}{4}), LCD = 12
  8. (dfrac{1}{3}, dfrac{3}{4}), and (dfrac{3}{5}), LCD = 60

Add and Subtract Fractions with Different Denominators

In the following exercises, add or subtract. Write the result in simplified form.

  1. (dfrac{1}{3} + dfrac{1}{5})
  2. (dfrac{1}{4} + dfrac{1}{5})
  3. (dfrac{1}{2} + dfrac{1}{7})
  4. (dfrac{1}{3} + dfrac{1}{8})
  5. (dfrac{1}{3} - left(- dfrac{1}{9} ight))
  6. (dfrac{1}{4} - left(- dfrac{1}{8} ight))
  7. (dfrac{1}{5} - left(- dfrac{1}{10} ight))
  8. (dfrac{1}{2} - left(- dfrac{1}{6} ight))
  9. (dfrac{2}{3} + dfrac{3}{4})
  10. (dfrac{3}{4} + dfrac{2}{5})
  11. (dfrac{7}{12} + dfrac{5}{8})
  12. (dfrac{5}{12} + dfrac{3}{8})
  13. (dfrac{7}{12} - dfrac{9}{16})
  14. (dfrac{7}{16} - dfrac{5}{12})
  15. (dfrac{11}{12} - dfrac{3}{8})
  16. (dfrac{5}{8} - dfrac{7}{12})
  17. (dfrac{2}{3} - dfrac{3}{8})
  18. (dfrac{5}{6} - dfrac{3}{4})
  19. (− dfrac{11}{30} + dfrac{27}{40})
  20. (− dfrac{9}{20} + dfrac{17}{30})
  21. (− dfrac{13}{30} + dfrac{25}{42})
  22. (− dfrac{23}{30} + dfrac{5}{48})
  23. (− dfrac{39}{56} - dfrac{22}{35})
  24. (− dfrac{33}{49} - dfrac{18}{35})
  25. (- dfrac{2}{3} - left(- dfrac{3}{4} ight))
  26. (- dfrac{3}{4} - left(- dfrac{4}{5} ight))
  27. (- dfrac{9}{16} - left(- dfrac{4}{5} ight))
  28. (- dfrac{7}{20} - left(- dfrac{5}{8} ight))
  29. 1 + (dfrac{7}{8})
  30. 1 + (dfrac{5}{6})
  31. 1 − (dfrac{5}{9})
  32. 1 − (dfrac{3}{10})
  33. (dfrac{x}{3} + dfrac{1}{4})
  34. (dfrac{y}{2} + dfrac{2}{3})
  35. (dfrac{y}{4} - dfrac{3}{5})
  36. (dfrac{x}{5} - dfrac{1}{4})

Identify and Use Fraction Operations

In the following exercises, perform the indicated operations. Write your answers in simplified form.

  1. (a) (dfrac{3}{4} + dfrac{1}{6}) (b) (dfrac{3}{4} div dfrac{1}{6})
  2. (a) (dfrac{2}{3} + dfrac{1}{6}) (b) (dfrac{2}{3} div dfrac{1}{6})
  3. (a) (- dfrac{2}{5} - dfrac{1}{8}) (b) (- dfrac{2}{5} cdot dfrac{1}{8})
  4. (a) (- dfrac{4}{5} - dfrac{1}{8}) (b) (- dfrac{4}{5} cdot dfrac{1}{8})
  5. (a) (dfrac{5n}{6} div dfrac{8}{15}) (b) (dfrac{5n}{6} - dfrac{8}{15})
  6. (a) (dfrac{3a}{8} div dfrac{7}{12}) (b) (dfrac{3a}{8} - dfrac{7}{12})
  7. (a) (dfrac{9}{10} cdot left(− dfrac{11d}{12} ight)) (b) (dfrac{9}{10} + left(− dfrac{11d}{12} ight))
  8. (a) (dfrac{4}{15} cdot left(− dfrac{5}{q} ight)) (b) (dfrac{4}{15} + left(− dfrac{5}{q} ight))
  9. (- dfrac{3}{8} div left(- dfrac{3}{10} ight))
  10. (- dfrac{5}{12} div left(- dfrac{5}{9} ight))
  11. (- dfrac{3}{8} + dfrac{5}{12})
  12. (- dfrac{1}{8} + dfrac{7}{12})
  13. (dfrac{5}{6} − dfrac{1}{9})
  14. (dfrac{5}{9} − dfrac{1}{6})
  15. (dfrac{3}{8} cdot left(− dfrac{10}{21} ight))
  16. (dfrac{7}{12} cdot left(− dfrac{8}{35} ight))
  17. (− dfrac{7}{15} - dfrac{y}{4})
  18. (− dfrac{3}{8} - dfrac{x}{11})
  19. (dfrac{11}{12a} cdot dfrac{9a}{16})
  20. (dfrac{10y}{13} cdot dfrac{8}{15y})

Use the Order of Operations to Simplify Complex Fractions

In the following exercises, simplify.

  1. (dfrac{left(dfrac{1}{5} ight)^{2}}{2 + 3^{2}})
  2. (dfrac{left(dfrac{1}{3} ight)^{2}}{5 + 2^{2}})
  3. (dfrac{2^{3} + 4^{2}}{left(dfrac{2}{3} ight)^{2}})
  4. (dfrac{3^{3} - 3^{2}}{left(dfrac{3}{4} ight)^{2}})
  5. (dfrac{left(dfrac{3}{5} ight)^{2}}{left(dfrac{3}{7} ight)^{2}})
  6. (dfrac{left(dfrac{3}{4} ight)^{2}}{left(dfrac{5}{8} ight)^{2}})
  7. (dfrac{2}{dfrac{1}{3} + dfrac{1}{5}})
  8. (dfrac{5}{dfrac{1}{4} + dfrac{1}{3}})
  9. (dfrac{dfrac{2}{3} + dfrac{1}{2}}{dfrac{3}{4} - dfrac{2}{3}})
  10. (dfrac{dfrac{3}{4} + dfrac{1}{2}}{dfrac{5}{6} - dfrac{2}{3}})
  11. (dfrac{dfrac{7}{8} - dfrac{2}{3}}{dfrac{1}{2} + dfrac{3}{8}})
  12. (dfrac{dfrac{3}{4} - dfrac{3}{5}}{dfrac{1}{4} + dfrac{2}{5}})

Mixed Practice

In the following exercises, simplify.

  1. (dfrac{1}{2} + dfrac{2}{3} cdot dfrac{5}{12})
  2. (dfrac{1}{3} + dfrac{2}{5} cdot dfrac{3}{4})
  3. 1 − (dfrac{3}{5} div dfrac{1}{10})
  4. 1 − (dfrac{5}{6} div dfrac{1}{12})
  5. (dfrac{2}{3} + dfrac{1}{6} + dfrac{3}{4})
  6. (dfrac{2}{3} + dfrac{1}{4} + dfrac{3}{5})
  7. (dfrac{3}{8} - dfrac{1}{6} + dfrac{3}{4})
  8. (dfrac{2}{5} + dfrac{5}{8} - dfrac{3}{4})
  9. 12(left(dfrac{9}{20} − dfrac{4}{15} ight))
  10. 8(left(dfrac{15}{16} − dfrac{5}{6} ight))
  11. (dfrac{dfrac{5}{8} + dfrac{1}{6}}{dfrac{19}{24}})
  12. (dfrac{dfrac{1}{6} + dfrac{3}{10}}{dfrac{14}{30}})
  13. (left(dfrac{5}{9} + dfrac{1}{6} ight) div left(dfrac{2}{3} − dfrac{1}{2} ight))
  14. (left(dfrac{3}{4} + dfrac{1}{6} ight) div left(dfrac{5}{8} − dfrac{1}{3} ight))

In the following exercises, evaluate the given expression. Express your answers in simplified form, using improper fractions if necessary.

  1. x + (dfrac{1}{2}) when
    1. x = (− dfrac{1}{8})
    2. x = (− dfrac{1}{2})
  2. x + (dfrac{2}{3}) when
    1. x = (− dfrac{1}{6})
    2. x = (− dfrac{5}{3})
  3. x + (left(− dfrac{5}{6} ight)) when
    1. x = (dfrac{1}{3})
    2. x = (− dfrac{1}{6})
  4. x + (left(− dfrac{11}{12} ight)) when
    1. x = (dfrac{11}{12})
    2. x = (dfrac{3}{4})
  5. x − (dfrac{2}{5}) when
    1. x = (dfrac{3}{5})
    2. x = (- dfrac{3}{5})
  6. x − (dfrac{1}{3}) when
    1. x = (dfrac{2}{3})
    2. x = (- dfrac{2}{3})
  7. (dfrac{7}{10}) − w when
    1. w = (dfrac{1}{2})
    2. w = (- dfrac{1}{2})
  8. (dfrac{5}{12}) − w when
    1. w = (dfrac{1}{4})
    2. w = (- dfrac{1}{4})
  9. 4p2q when p = (- dfrac{1}{2}) and q = (dfrac{5}{9})
  10. 5m2n when m = (- dfrac{2}{5}) and n = (dfrac{1}{3})
  11. 2x2y3 when x = (- dfrac{2}{3}) and y = (- dfrac{1}{2})
  12. 8u2v3 when u = (- dfrac{3}{4}) and v = (- dfrac{1}{2})
  13. (dfrac{u + v}{w}) when u = −4, v = −8, w = 2
  14. (dfrac{m + n}{p}) when m = −6, n = −2, p = 4
  15. (dfrac{a + b}{a - b}) when a = −3, b = 8
  16. (dfrac{r - s}{r + s}) when r = 10, s = −5

Everyday Math

  1. Decorating Laronda is making covers for the throw pillows on her sofa. For each pillow cover, she needs (dfrac{3}{16}) yard of print fabric and (dfrac{3}{8}) yard of solid fabric. What is the total amount of fabric Laronda needs for each pillow cover?
  2. Baking Vanessa is baking chocolate chip cookies and oatmeal cookies. She needs (1 dfrac{1}{4}) cups of sugar for the chocolate chip cookies, and (1 dfrac{1}{8}) cups for the oatmeal cookies How much sugar does she need altogether?

Writing Exercises

  1. Explain why it is necessary to have a common denominator to add or subtract fractions.
  2. Explain how to find the LCD of two fractions.

Self Check

(a) After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

(b) After looking at the checklist, do you think you are well prepared for the next section? Why or why not?


The equation means $ad+bc=bd k$. It follows that $b$ divides $ad$, hence also $d$. The rest is for you .

Bezout's Identity says that since $(a,b)=1$ $ ax+by=1 ag <1>$ and that since $(c,d)=1$ $ cu+dv=1 ag <2>$ Multiplying your equation by $bd$ gives $ ad+bc=bdk ag <3>$ Multiply $(1)$ by $d$ to get $ color<#C00000>+bdy=d ag <4>$ Multiply $(3)$ by $x$ to get $ color<#C00000>+bcx=bdkx ag <5>$ Solving $(5)$ for $adx$ and plugging that into $(4)$ yields $ d=b(dy+dkx-cx) ag <6>$ Multiply $(2)$ by $b$ to get $ color<#C00000>+bdv=b ag <7>$ Multiply $(3)$ by $u$ to get $ adu+color<#C00000>=bdku ag <8>$ Solving $(8)$ for $bcu$ and plugging that into $(7)$ yields $ b=d(bku-au+bv) ag <9>$ Equations $(6)$ and $(9)$ should finish things off.


Key Fraction Facts

To understand how to work out fractions, it’s important to get to grips with the fundamentals. First, let’s look at the three different types of fractions:

Fraction Definitions and Examples

Proper fraction – A proper fraction is a fraction in which the numerator is of lesser value than the denominator. 1/2, 10/15, and 85/100 are all examples of proper fractions. The overall value of a proper fraction is always less than one.

Improper fraction – In an improper fraction, the value of the numerator is greater than that of the denominator. 6/3, 25/18 and 50/20 are all examples of improper fractions. The overall value of an improper fraction is always more than one.

Mixed fractions – A mixed fraction is presented as a whole number followed by a fractional number, such as 2⅔, 6⅘ or 25⅝. Mixed fractions are also known as mixed numbers.

Key Terms

Now we know the different types of fractions, let’s look at some other key terms and phrases:

Equivalent fractions – These are fractions that appear different but hold the same value. For example, 2/3 is the same as 4/6.

Simplified fractions – These are fractions reduced to their lowest form. Basically, a lower equivalent of a higher fraction. So, using the example above, 2/3 is a simplified version of 4/6.

Reciprocals – This is where the fraction is reversed by placing the denominator above the numerator. As an example, the reciprocal of 2/3 is 3/2. Reciprocals are used when dividing and multiplying fractions (5 ÷ 1/5 is the same as 5 x 5/1 or 5 x 5).

Fractions can also be presented as decimals and percentages. We’ll look at how to convert fractions in the example equations below.


Adding and subtracting unlike fractions

This fifth grade lesson teaches how to add and subtract unlike fractions (fractions with different denominators). First, we use visual models to learn that the fractions need converted into like fractions, using equivalent fractions. Students do several exercises using visual models, and try to look for a pattern in the common denominators. The next lesson concentrates on how we find the common denominator.

The video below outlines a lesson plan for teaching adding unlike fractions (which I consider to be the most difficult topic in fraction arithmetic). In the video, I first go through exercises that have a visual model and the common denominator is given. Then, we work exercises without a visual model where the common denominator is still given. Lastly, we study the rule about finding the common denominator. I also have another lesson that concentrates on the common denominator.

Cover the page below the black line. Then try to figure out the addition problems below.


Adding Fractions with Unlike Denominators

If the denominators are not the same, then you have to use equivalent fractions which do have a common denominator . To do this, you need to find the least common multiple (LCM) of the two denominators.

To add fractions with unlike denominators, rename the fractions with a common denominator. Then add and simplify.

For example, suppose you want to add:

The LCM of 3 and 11 is 33 . So, we need to find fractions equivalent to 1 11 and 2 3 which have 33 in the denominator. Multiply the numerator and denominator of 1 11 by 3 , and multiply the numerator and denominator of 2 3 by 11 .

Now we have like denominators, and we can add as described above.

Download our free learning tools apps and test prep books

Names of standardized tests are owned by the trademark holders and are not affiliated with Varsity Tutors LLC.

4.9/5.0 Satisfaction Rating over the last 100,000 sessions. As of 4/27/18.

Media outlet trademarks are owned by the respective media outlets and are not affiliated with Varsity Tutors.

Award-Winning claim based on CBS Local and Houston Press awards.

Varsity Tutors does not have affiliation with universities mentioned on its website.

Varsity Tutors connects learners with experts. Instructors are independent contractors who tailor their services to each client, using their own style, methods and materials.


Adding Fractions with Unlike Denominators (A)

Teacher s can use math worksheets as test s, practice assignment s or teaching tool s (for example in group work , for scaffolding or in a learning center ). Parent s can work with their children to give them extra practice , to help them learn a new math skill or to keep their skills fresh over school breaks . Student s can use math worksheets to master a math skill through practice, in a study group or for peer tutoring .

Use the buttons below to print, open, or download the PDF version of the Adding Fractions with Unlike Denominators (A) math worksheet. The size of the PDF file is 27095 bytes . Preview images of the first and second (if there is one) pages are shown. If there are more versions of this worksheet, the other versions will be available below the preview images. For more like this, use the search bar to look for some or all of these keywords: fractions, mathematics, math, adding .

The Print button will initiate your browser's print dialog. The Open button will open the complete PDF file in a new tab of your browser. The Teacher button will initiate a download of the complete PDF file including the questions and answers (if there are any). If a Student button is present, it will initiate a download of only the question page(s). Additional options might be available by right-clicking on a button (or holding a tap on a touch screen). I don't see buttons!

The Adding Fractions with Unlike Denominators (A) Math Worksheet Page 1 The Adding Fractions with Unlike Denominators (A) Math Worksheet Page 2

Solution

The denominators of these fractions are 2 and 14. Since 2 divides evenly into 14, 14 is a multiple of 2, so 14 itself is a common denominator for the fractions $frac<1><2>$ and $frac<1><14>$. The picture below shows this:

Here is a picture showing the fractions when they are both written in terms of fourteenths:

Any multiple of 14 is a common multiple of 2 and 14. So $2 imes14=28$ is common multiple and is a common denominator for the fractions $frac<1><2>$ and $frac<1><14>$. Thus, 14 and 28 are two different common denominators for the fractions $frac<1><2>$ and $frac<1><14>$.

The first common denominator that we identified in part (i) was 14. Here is a picture that represents $frac12 - frac<1><14>$:

This is how we can write the process of finding a common denominator and subtracting using symbols:

The second common denominator that we identified in part (i) was 28. This is how we might use this common denominator to solve the given subtraction problem:

Note that $egin frac<6><14>&=frac<3 imes 2><7 imes 2> &=frac<3><7> &=frac<3 imes 4><7 imes 4> &=frac<12> <28>end$ So we get the same answer using different denominators, as we would expect!

In order to find a solution to this subtraction problem, we must first find a common denominator for the fractions $frac<5><9>$ and $frac<1><6>$. 18 is a common multiple of the denominators 9 and 6 because $9 imes2=18$ and $6 imes3=18$. This means that 18 is a common denominator for the fractions $frac<5><9>$ and $frac<1><6>$. Here is a picture that shows both of these fractions when they are written in terms of eighteenths:

Here is a picture that represents $frac<5><9>-frac<1><6>$ using this common denominator:

Any other common multiple of the denominators 9 and 6 could also be used when solving this subtraction problem. This is how we write the process of finding the common denominator and subtracting using symbols:

The picture shows that after we convert both fractions to eighteenths we can subtract 3 eighteenths from 10 eighteenths and are left with 7 eighteenths, the same answer that we found symbolically.

In order to find a solution to this subtraction problem, we must again first find a common denominator for the fractions $frac<21><10>$ and $frac<24><15>$. 30 is a common multiple of the denominators 10 and 15 because $10 imes3=30$ and $15 imes2=30$. This means that 30 is a common denominator for the fractions $frac<21><10>$ and $frac<24><15>$. This is how we might use the common denominator 30 in order to solve this subtraction problem using symbols:


Statistic modular 2

1-1/3
−8-2/5
14-1/2
We can read these mixed numbers as:

Step 1: Write the fraction as a division problem: 9 divided by 5 .

nine divided by five
Step 2: Solve by dividing numerator 9 by denominator 5 .

solve by dividing nine by five
5 goes into 9 one time with a remainder of 4 .

Step 3: Write the answer using the quotient, 1 , followed by a fraction whose numerator is the remainder, 4 , and whose denominator is the denominator from the original fraction, 5 .

Step 1: Multiply the whole number by the fraction's denominator.
1×5=5
Step 2: Add that product to the numerator of the fraction.

5+4=9
Step 3: The answer from step 2 now becomes the numerator of the improper fraction. Rewrite the answer as an improper fraction.

Step 4: Reduce the fraction.

Cancel, or divide by, all the factors that are common to both the numerator and denominator.

Are you finished yet? Notice that fraction begin. numerator: -4
denominator: 6
fraction end. still has a common factor. You are not finished yet and need to reduce further.

3/8 ≟ 15/40
A simple way to determine whether these fractions are equal is to cross-multiply. In this example, cross-multiplying means multiplying 3×40 and 8×15 . If the two products are equal, the fractions are equivalent, or equal. Since 3×40=120 and 8×15=120 , we know that 3/8=15/40

6/17 ≟ 36/101
Now draw "butterfly wings" around the 6 and the 101 , and around the 17 and the 36 , like this:

What is a common multiple of 2 and 3 ?
To find a common multiple, multiply 2×3=6 .
Clearly, both 2 and 3 evenly divide 6 , so 6 is a multiple of both.

The multiples of 8 are: 8 , 16 , 24 , 32 , 40 , 48 , 56 , 64 , 72 .

21÷3=7
Multiply the numerator and denominators by 7 .

(1×7)(3×7)=7/21
Now, let's transform −2/7 into an equivalent fraction with the common denominator
Divide the LCD (the new denominator) by the current denominator.

21÷7=3
Multiply the numerator and denominators by 3 .
(−2×3)(7×3)=−6/21
ur two new fractions with a common denominator are:

(5−1)6=4/6
Step 2: Reduce the fraction (if necessary).

4/6
Notice that the both the numerator, 4 , and the denominator, 6 , are both divisible by 2 . Therefore we can reduce this fraction to:

Step 2: Convert fractions to have like denominators:

Multiplying both the numerator and denominator by 3 is equivalent to dividing each of the old parts into 3 parts. Multiplying the numerator and denominator by 4 is equivalent to dividing each of the old parts into 4 parts. The amount of each fraction (shaded region of the pies) remains the same, but we have the thinner slices in each pie. The point is we have the same thinner slices for each of the pies.

1/4+1/3=(1⋅3)(4⋅3)+(1⋅4)(3⋅4)=3/12+4/12
Now, we can add the like fractions:

Step 2: Convert fractions to have like denominators.

2/3−5/9=(2⋅3)(3⋅3)−59=6/9−5/9
Step 3: Subtract the like fractions.

Here is a reminder of how to change a mixed number to an improper fraction.

1-2/5=7/5
Step 2: Find equivalent fractions with the least common denominator.

Convert the fractions to equivalent fractions with the LCD. The lowest common denominator is 5×2=10 , therefore:

Multiply the numerator and denominator of 7/5 by 2
Multiply the numerator and denominator of 3/2 by 5
75−32=14/10−15/10
Step 3: Subtract Like Fractions

Next, subtract the numerators of the fractions.

Remember to keep the same denominator—do not subtract them!

14/10−15/10=−1/10
Step 4: To complete the problem, convert any improper fractions to lowest terms.

Here is a reminder of how to change a mixed number to an improper fraction.

7-3/4=31/4
Step 2: Find equivalent fractions with the least common denominator.

Convert the fractions to equivalent fractions with the LCD. The lowest common denominator is 3×4=12 , therefore:

Multiply the numerator and denominator of 20/3 by 4
Multiply the numerator and denominator of 31/4 by 3
20/3+31/4=80/12+93/12
Step 3: Add like fractions.

Next, add the fractions. If this were a subtraction problem, you would simply subtract instead of adding.

80/12+93/12=173/12
Step 4: To complete the problem, convert any improper fractions to lowest terms.

One of the most difficult aspects of working with fractions is remembering how to handle the denominators. In multiplication and division, do not find common denominators. Follow the steps below to multiply:

A helpful way to think about the multiplication of fractions is that in solving these types of problems, we are looking for a part OF another. Therefore, it is important to keep in mind that the word "of" can be used to refer to multiplication. For example, what is 12 of 13 .

One of the most difficult aspects of working with fractions is remembering how to handle the denominators. In multiplication and division, do not find common denominators. Follow the steps below to multiply:

A helpful way to think about the multiplication of fractions is that in solving these types of problems, we are looking for a part OF another. Therefore, it is important to keep in mind that the word "of" can be used to refer to multiplication. For example, what is 12 of 13 .

One of the most difficult aspects of working with fractions is remembering how to handle the denominators. In multiplication and division, do not find common denominators. Follow the steps below to multiply:


More word problem worksheets

Explore all of our math word problem worksheets, from kindergarten through grade 5.

K5 Learning offers free worksheets, flashcards and inexpensive workbooks for kids in kindergarten to grade 5. We help your children build good study habits and excel in school.

Download & Print
From only $2.20

K5 Learning offers free worksheets, flashcards and inexpensive workbooks for kids in kindergarten to grade 5. We help your children build good study habits and excel in school.


ADDING ALGEBRAIC FRACTIONS

T HERE 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. 6 x + 3
5
+ 4 x &minus 1
5
= 10 x + 2
5

The denominators are the same. Add the numerators as like terms.

Example 2. 6 x + 3
5
&minus 4 x &minus 1
5

To subtract, change the signs of the subtrahend, and add.

6 x + 3
5
&minus 4 x &minus 1
5
= 6 x + 3 &minus 4 x + 1
5
= 2 x + 4
5

To see the answer, pass your mouse over the colored area.
To cover the answer again, click "Refresh" ("Reload").
Do the problem yourself first!

a) x
3
+ y
3
= x + y
3
b) 5
x
&minus 2
x
= 3
x
c) x
x &minus 1
+ x + 1
x &minus 1
= 2 x + 1
x &minus 1
d) 3 x &minus 4
x &minus 4
+ x &minus 5
x &minus 4
= 4 x &minus 9
x &minus 4
e) 6 x + 1
x &minus 3
&minus 4 x + 5
x &minus 3
= 6 x + 1 &minus 4 x &minus 5
x &minus 3
= 2 x &minus 4
x &minus 3
f) 2 x &minus 3
x &minus 2
&minus x &minus 4
x &minus 2
= 2 x &minus 3 &minus x + 4
x &minus 2
= x + 1
x &minus 2

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:

We will now construct their LCM -- factor by factor.

To begin, it will have the factors of the first term:

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 :

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 :

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 , x 2 , x 3 .

Solution . The LCM must have the factor x .

But it also must have the factors of x 2 -- which are x · x . Therefore, we must add one more factor of x :

Finally, the LCM must have the factors of x 3 , which are x · x · x . Therefore,

x 3 is the smallest product that contains x , x 2 , and x 3 as factors.

We see that when the terms are powers of a variable -- x , x 2 , x 3 -- 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 , a 2 , a 3 , a 4 . a 4 d) a 2 b , a b 2 . a 2 b 2

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
a b
+ 4
b c
+ 5
c d
= 3 c d + 4 a d + 5 a b
a b c d

To change into an equivalent fraction with denominator a b c d , simply multiply a b by the factors it is missing, namely c d . Therefore, we must also multiply 3 by c d . That accounts for the first term in the numerator.

To change into an equivalent fraction with denominator a b c d , multiply b c by the factors it is missing, namely a d . Therefore, we must also multiply 4 by a d . That accounts for the second term in the numerator.

To change into an equivalent fraction with denominator a b c d , multiply c d by the factors it is missing, namely a b . Therefore, we must also multiply 5 by a b . 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.

a) 5
ab
+ 6
ac
= 5 c + 6 b
abc
b) 2
pq
+ 3
qr
+ 4
rs
= 2 rs + 3 ps + 4 pq
pqrs
c) 7
ab
+ 8
bc
+ 9
abc
= 7 c + 8 a + 9
abc
d) 1
a
+ 2
a 2
+ 3
a 3
= a 2 + 2 a + 3
a 3
e) 3
a 2 b
+ 4
a b 2
= 3 b + 4 a
a 2 b 2
f) 5
ab
+ 6
cd
= 5 cd + 6 ab
abcd
g) _2_
x ( x + 2)
+ __3__
( x + 2)( x &minus 3)
= 2( x &minus 3) + 3 x
x ( x + 2)( x &minus 3)
= _ 2 x &minus 6 + 3 x _
x ( x + 2)( x &minus 3)
= _5 x &minus 6_
x ( x + 2)( x &minus 3)

At the 2nd Level we will see a similar problem, but the denominators will not be factored.

Problem 4. Add: 1 &minus 1
a
+ c + 1
ab
. But write the answer as

1 &minus 1
a
+ c + 1
ab
= 1 &minus ( 1
a
&minus c + 1
ab
)
= 1 &minus b &minus ( c + 1)
ab
= 1 &minus b &minus c &minus 1
ab

Example 5. Denominators with no common factors.

When the denominators have no common factors, their LCM is simply their product, mn .

The numerator then appears as the result of "cross-multiplying" :

However, that technique will work only when adding two fractions, and the denominators have no common factors.

Solution . These denominators have no common factors -- x is not a factor of x &minus 1. It is a term. Therefore, the LCM of denominators is their product.

2
x &minus 1
&minus 1
x
= 2 x &minus ( x &minus 1)
( x &minus 1) x
= 2 x &minus x + 1
( x &minus 1) x
= _ x + 1_
( x &minus 1) x

Note: The entire x &minus 1 is being subtracted. Therefore, we write it in parentheses -- and its signs change.

a) x
a
+ y
b
= xb + ya
ab
b) x
5
+ 3 x
2
= 2 x + 15 x
10
= 17 x
10
c) 6
x &minus 1
+ 3
x + 1
= 6( x + 1) + 3( x &minus 1)
( x + 1)( x &minus 1)
= 6 x + 6 + 3 x &minus 3
( x + 1)( x &minus 1)
= _9 x + 3_
( x + 1)( x &minus 1)
d) 6
x &minus 1
&minus 3
x + 1
= 6( x + 1) &minus 3( x &minus 1)
( x + 1)( x &minus 1)
= 6 x + 6 &minus 3 x + 3
( x + 1)( x &minus 1)
= _3 x + 9_
( x + 1)( x &minus 1)
e) 3
x &minus 3
&minus 2
x
= 3 x &minus 2( x &minus 3)
( x &minus 3) x
= 3 x &minus 2 x + 6
( x &minus 3) x
= x + 6
( x &minus 3) x
f) 3
x &minus 3
&minus 1
x
= 3 x &minus ( x &minus 3)
( x &minus 3) x
= 3 x &minus x + 3
( x &minus 3) x
= 2 x + 3
( x &minus 3) x
g) 1
x
+ 2
y
+ 3
z
= yz + 2 xz + 3 xy
xyz
Example 7. Add: a + b
c
.

Solution. We have to express a with denominator c.

a) p
q
+ r = p + qr
q
b) 1
x
&minus 1 = 1 &minus x
x
c) x &minus 1
x
= x 2 &minus 1
x
d) 1 &minus 1
x 2
= x 2 &minus 1
x 2
e) 1 &minus 1
x + 1
= x + 1 &minus 1
x + 1
= x
x + 1
f) 3 + 2
x + 1
= 3 x + 3 + 2
x + 1
= 3 x + 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
.

Please make a donation to keep TheMathPage online.
Even $1 will help.


Watch the video: ΠΛΗ10 ΜΑΘΗΜΑ - ΜΕΡΟΣ 613 - Αφαίρεση στο Δυαδικό Σύστημα Αρίθμησης (November 2021).