4.3: Equivalent Fractions, Reducing Fractions to Lowest Terms, and Raising Fractions to Higher Terms
- be able to recognize equivalent fractions
- be able to reduce a fraction to lowest terms
- be able to raise a fraction to higher terms
Equivalent Fractions
Let's examine the following two diagrams.
Notice that both \(\dfrac{2}{3}\) and \(\dfrac{4}{6}\) represent the same part of the whole, that is, they represent the same number.
Fractions that have the same value are called equivalent fractions . Equivalent fractions may look different, but they are still the same point on the number line.
There is an interesting property that equivalent fractions satisfy.
A Test for Equivalent Fractions Using the Cross Product
These pairs of products are called cross products .
If the cross products are equal, the fractions are equivalent. If the cross products are not equal, the fractions are not equivalent.
Thus, \(\dfrac{2}{3}\) and \(\dfrac{4}{6}\) are equivalent, that is, \(\dfrac{2}{3} = \dfrac{4}{6}\).
Determine if the following pairs of fractions are equivalent.
\(\dfrac{3}{4}\) and \(\dfrac{6}{8}\). Test for equality of the cross products.
Solution
The cross products are equals.
The fractions \(\dfrac{3}{4}\) and \(\dfrac{6}{8}\) are equivalent, so \(\dfrac{3}{4} = \dfrac{6}{8}\).
\(\dfrac{3}{8}\) and \(\dfrac{9}{16}\). Test for equality of the cross products.
Solution
The cross products are equals.
The fractions \(\dfrac{3}{8}\) and \(\dfrac{9}{16}\) are not equivalent.
Practice Set A
Determine if the pairs of fractions are equivalent.
\(\dfrac{1}{2}\), \(\dfrac{3}{6}\)
- Answer
-
, yes
Practice Set A
\(\dfrac{4}{5}\), \(\dfrac{12}{15}\)
- Answer
-
, yes
Practice Set A
\(\dfrac{2}{3}\), \(\dfrac{8}{15}\)
- Answer
-
\(30 \ne 24\), no
Practice Set A
\(\dfrac{1}{8}\), \(\dfrac{4}{50}\)
- Answer
-
, yes
Practice Set A
\(\dfrac{3}{12}\), \(\dfrac{1}{4}\)
- Answer
-
, yes
Reducing Fractions to Lowest Terms
It is often very useful to conver t one fraction to an equivalent fraction that has reduced values in the numerator and denominator. We can suggest a method for doing so by considering the equivalent fractions \(\dfrac{9}{15}\) and \(\dfrac{3}{5}\). First, divide both the numerator and denominator of \(\dfrac{9}{15}\) by 3. The fraction \(\dfrac{9}{15}\) and \(\dfrac{3}{5}\) are equivalent.
(Can you prove this?) So, \(\dfrac{9}{15} = \dfrac{3}{5}\). We wish to convert \(\dfrac{9}{15}\) to \(\dfrac{3}{5}\). Now divide the numerator and denominator of \(\dfrac{9}{15}\) by 3, and see what happens.
\(\dfrac{9 \div 3}{15 \div 3} = \dfrac{3}{5}\)
The fraction \(\dfrac{9}{15}\) is converted to \(\dfrac{3}{5}\).
A natural question is "Why did we choose to divide by 3?" Notice that
\(\dfrac{9}{15} = \dfrac{3 \cdot 3}{5 \cdot 3}\)
We can see that the factor 3 is common to both the numerator and denominator.
From these observations we can suggest the following method for converting one fraction to an equivalent fraction that has reduced values in the numerator and denominator. The method is called reducing a fraction .
A fraction can be reduced by dividing both the numerator and denominator by the same nonzero whole number.
Consider the collection of equivalent fractions
\(\dfrac{5}{20}, \dfrac{4}{16}, \dfrac{3}{12}, \dfrac{2}{8}, \dfrac{1}{4}\)
Reduced to Lowest Terms
Notice that each of the first four fractions can be reduced to the last fraction, \(\dfrac{1}{4}\), by dividing both the numerator and denominator by, respectively, 5, 4, 3, and 2. When a fraction is converted to the fraction that has the smallest numerator and denominator in its collection of equivalent fractions, it is said to be reduced to lowest terms . The fractions \(\dfrac{1}{4}\), \(\dfrac{3}{8}\), \(\dfrac{2}{5}\), and \(\dfrac{7}{10}\) are all reduced to lowest terms.
Observe a very important property of a fraction that has been reduced to lowest terms. The only whole number that divides both the numerator and denominator without a remainder is the number 1. When 1 is the only whole number that divides two whole numbers, the two whole numbers are said to be relatively prime .
Relatively Prime
A fraction is reduced to lowest terms if its numerator and denominator are
relatively prime
.
Methods of Reducing Fractions to Lowest Terms
Method 1: Dividing Out Common Primes
- Write the numerator and denominator as a product of primes.
- Divide the numerator and denominator by each of the common prime factors. We often indicate this division by drawing a slanted line through each divided out factor. This process is also called cancelling common factors .
- The product of the remaining factors in the numerator and the product of remaining factors of the denominator are relatively prime, and this fraction is reduced to lowest terms.
\(\dfrac{6}{18} = \dfrac{\begin{array} {c} {^1} \\ {\cancel{2}} \end{array} \cdot \begin{array} {c} {^1} \\ {\cancel{3}} \end{array}}{\begin{array} {c} {\cancel{2}} \\ {^1} \end{array} \cdot \begin{array} {c} {\cancel{3}} \\ {^1} \end{array} \cdot 3} = \dfrac{1}{3}\) 1 and 3 are relatively prime.
\(\dfrac{16}{20} = \dfrac{\begin{array} {c} {^1} \\ {\cancel{2}} \end{array} \cdot \begin{array} {c} {^1} \\ {\cancel{2}} \end{array} \cdot 2 \cdot 2}{\begin{array} {c} {\cancel{2}} \\ {^1} \end{array} \cdot \begin{array} {c} {\cancel{2}} \\ {^1} \end{array} \cdot 5} = \dfrac{4}{5}\) 4 and 5 are relatively prime.
\(\dfrac{56}{104} = \dfrac{\begin{array} {c} {^1} \\ {\cancel{2}} \end{array} \cdot \begin{array} {c} {^1} \\ {\cancel{3}} \end{array} \begin{array} {c} {^1} \\ {\cancel{2}} \end{array} \cdot 7}{\begin{array} {c} {\cancel{2}} \\ {^1} \end{array} \cdot \begin{array} {c} {\cancel{2}} \\ {^1} \end{array} \cdot \begin{array} {c} {\cancel{2}} \\ {^1} \end{array} \cdot 13} = \dfrac{7}{13}\) 7 and 13 are relatively prime (and also truly prime)
\(\dfrac{315}{336} = \dfrac{\begin{array} {c} {^1} \\ {\cancel{3}} \end{array} \cdot 3 \cdot 5 \cdot \begin{array} {c} {^1} \\ {\cancel{7}} \end{array}}{2 \cdot 2 \cdot 2 \cdot 2 \cdot \begin{array} {c} {\cancel{3}} \\ {^1} \end{array} \cdot \begin{array} {c} {\cancel{7}} \\ {^1} \end{array}} = \dfrac{15}{16}\) 15 and 16 are relatively prime.
Sample Set B
\(\dfrac{8}{15} = \dfrac{2 \cdot 2 \cdot 2}{3 \cdot 5}\) No common prime factors, so 8 and 15 are relatively prime.
The fraction \(\dfrac{8}{15}\) is reduced to lowest terms.
Practice Set B
Reduce each fraction to lowest terms.
\(\dfrac{4}{8}\)
- Answer
-
\(\dfrac{1}{2}\)
Practice Set B
\(\dfrac{6}{15}\)
- Answer
-
\(\dfrac{2}{5}\)
Practice Set B
\(\dfrac{6}{48}\)
- Answer
-
\(\dfrac{1}{8}\)
Practice Set B
\(\dfrac{21}{48}\)
- Answer
-
\(\dfrac{7}{16}\)
Practice Set B
\(\dfrac{72}{42}\)
- Answer
-
\(\dfrac{12}{7}\)
Practice Set B
\(\dfrac{135}{243}\)
- Answer
-
\(\dfrac{5}{9}\)
Method 2: Dividing Out Common Factors
- Mentally divide the numerator and the denominator by a factor that is common to each. Write the quotient above the original number.
- Continue this process until the numerator and denominator are relatively prime.
Reduce each fraction to lowest terms.
\(\dfrac{25}{30}\). 5 divides into both 25 and 30.
\(\dfrac{\begin{array} {c} {^5} \\ {\cancel{25}} \end{array}}{\begin{array} {c} {\cancel{30}} \\ {^6} \end{array}} = \dfrac{5}{6}\) 5 and 6 are relatively prime.
\(\dfrac{18}{24}\). Both numbers are even so we can divide by 2.
\(\dfrac{\begin{array} {c} {^9} \\ {\cancel{18}} \end{array}}{\begin{array} {c} {\cancel{24}} \\ {^{12}} \end{array}}\) Now, both 9 and 12 are divisible by 3.
\(\dfrac{\begin{array} {c} {^{^3}} \\ {^{\cancel{9}}} \\ {\cancel{18}} \end{array}}{\begin{array} {c} {\cancel{24}} \\ {^{\cancel{12}}} \\ {^{^4}} \end{array}} = \dfrac{3}{4}\) 3 and 4 are relatively prime.
\(\dfrac{\begin{array} {c} {^{^7}} \\ {^{\cancel{21}}} \\ {\cancel{210}} \end{array}}{\begin{array} {c} {\cancel{150}} \\ {^{\cancel{15}}} \\ {^{^5}} \end{array}} = \dfrac{7}{5}\) 7 and 5 are relatively prime.
\(\dfrac{36}{96} = \dfrac{18}{48} = \dfrac{9}{24} = \dfrac{3}{8}\). 3 and 8 are relatively prime.
Practice Set C
Reduce each fraction to lowest terms.
\(\dfrac{12}{16}\)
- Answer
-
\(\dfrac{3}{4}\)
Practice Set C
\(\dfrac{9}{24}\)
- Answer
-
\(\dfrac{3}{8}\)
Practice Set C
\(\dfrac{21}{84}\)
- Answer
-
\(\dfrac{1}{4}\)
Practice Set C
\(\dfrac{48}{64}\)
- Answer
-
\(\dfrac{3}{4}\)
Practice Set C
\(\dfrac{63}{81}\)
- Answer
-
\(\dfrac{7}{9}\)
Practice Set C
\(\dfrac{150}{240}\)
- Answer
-
\(\dfrac{5}{8}\)
Raising Fractions to Higher Terms
Equally as important as reducing fractions is raising fractions to higher terms. Raising a fraction to higher terms is the process of constructing an equivalent fraction that has higher values in the numerator and denominator than the original fraction.
The fractions \(\dfrac{3}{5}\) and \(\dfrac{9}{15}\) are equivalent, that is, \(\dfrac{3}{5} = \dfrac{9}{15}\). Notice also,
\(\dfrac{3 \cdot 3}{5 \cdot 3} = \dfrac{9}{15}\)
Notice that \(\dfrac{3}{3} = 1\) and that \(\dfrac{3}{5} \cdot 1 = \dfrac{3}{5}\). We are not changing the value of \(\dfrac{3}{5}\).
From these observations we can suggest the following method for converting one fraction to an equivalent fraction that has higher values in the numerator and denominator. This method is called raising a fraction to higher terms .
Raising a Fraction to Higher
Terms
A fraction can be raised to an equivalent fraction that has higher terms in the numerator and denominator by multiplying both the numerator and denominator by the same nonzero whole number.
The fraction \(\dfrac{3}{4}\) can be raised to \(\dfrac{24}{32}\) by multiplying both the numerator and denominator by 8.
Most often, we will want to convert a given fraction to an equivalent fraction with a higher specified denominator. For example, we may wish to convert \(\dfrac{5}{8}\) to an equivalent fraction that has denominator 32, that is,
\(\dfrac{5}{8} = \dfrac{?}{32}\)
This is possible to do because we know the process. We must multiply both the numerator and denominator of \(\dfrac{5}{8}\) by the same nonzero whole number in order to 8 obtain an equivalent fraction.
We have some information. The denominator 8 was raised to 32 by multiplying it by some nonzero whole number. Division will give us the proper factor. Divide the original denominator into the new denominator.
\(32 \div 8 = 4\)
Now, multiply the numerator 5 by 4.
\(5 \cdot 4 = 20\)
Thus,
\(\dfrac{5}{8} = \dfrac{5 \cdot 4}{8 \cdot 4} = \dfrac{20}{32}\)
So,
\(\dfrac{5}{8} = \dfrac{20}{32}\)
Determine the missing numerator or denominator.
\(\dfrac{3}{7} = \dfrac{?}{35}\). Divide the original denominator into the new denominator.
\(35 \div 7 = 5\). The quotient is 5. Multiply the original numerator by 5.
\(\dfrac{3}{7} = \dfrac{3 \cdot 5}{7 \cdot 5} = \dfrac{15}{35}\) The missing numerator is 15.
\(\dfrac{5}{6} = \dfrac{45}{?}\). Divide the original numerator into the new numerator.
\(45 \div 5 = 9\). The quotient is 9. Multiply the original denominator by 9.
\(\dfrac{5}{6} = \dfrac{5 \cdot 9}{6 \cdot 9} = \dfrac{45}{54}\) The missing denominator is 45.
Practice Set D
Determine the missing numerator or denominator.
\(\dfrac{4}{5} = \dfrac{?}{40}\)
- Answer
-
32
Practice Set D
\(\dfrac{3}{7} = \dfrac{?}{28}\)
- Answer
-
12
Practice Set D
\(\dfrac{1}{6} = \dfrac{?}{24}\)
- Answer
-
4
Practice Set D
\(\dfrac{3}{10} = \dfrac{45}{?}\)
- Answer
-
150
Practice Set D
\(\dfrac{8}{15} = \dfrac{?}{165}\)
- Answer
-
88
Exercises
For the following problems, determine if the pairs of fractions are equivalent.
Exercise \(\PageIndex{1}\)
\(\dfrac{1}{2}, \dfrac{5}{10}\)
- Answer
-
equivalent
Exercise \(\PageIndex{2}\)
\(\dfrac{2}{3}, \dfrac{8}{12}\)
Exercise \(\PageIndex{3}\)
\(\dfrac{5}{12}, \dfrac{10}{24}\)
- Answer
-
equivalent
Exercise \(\PageIndex{4}\)
\(\dfrac{1}{2}, \dfrac{3}{6}\)
Exercise \(\PageIndex{5}\)
\(\dfrac{3}{5}, \dfrac{12}{15}\)
- Answer
-
not equivalent
Exercise \(\PageIndex{6}\)
\(\dfrac{1}{6}, \dfrac{7}{42}\)
Exercise \(\PageIndex{7}\)
\(\dfrac{16}{25}, \dfrac{49}{75}\)
- Answer
-
not equivalent
Exercise \(\PageIndex{8}\)
\(\dfrac{5}{28}, \dfrac{20}{112}\)
Exercise \(\PageIndex{9}\)
\(\dfrac{3}{10}, \dfrac{36}{110}\)
- Answer
-
not equivalent
Exercise \(\PageIndex{10}\)
\(\dfrac{6}{10}, \dfrac{18}{32}\)
Exercise \(\PageIndex{11}\)
\(\dfrac{5}{8}, \dfrac{15}{24}\)
- Answer
-
equivalent
Exercise \(\PageIndex{12}\)
\(\dfrac{10}{16}, \dfrac{15}{24}\)
Exercise \(\PageIndex{13}\)
\(\dfrac{4}{5}, \dfrac{3}{4}\)
- Answer
-
not equivalent
Exercise \(\PageIndex{14}\)
\(\dfrac{5}{7}, \dfrac{15}{21}\)
Exercise \(\PageIndex{15}\)
\(\dfrac{9}{11}, \dfrac{11}{9}\)
- Answer
-
not equivalent
For the following problems, determine the missing numerator or denominator.
Exercise \(\PageIndex{16}\)
\(\dfrac{1}{3} = \dfrac{?}{12}\)
Exercise \(\PageIndex{17}\)
\(\dfrac{1}{5} = \dfrac{?}{30}\)
- Answer
-
6
Exercise \(\PageIndex{18}\)
\(\dfrac{2}{3} = \dfrac{?}{9}\)
Exercise \(\PageIndex{19}\)
\(\dfrac{3}{4} = \dfrac{?}{16}\)
- Answer
-
12
Exercise \(\PageIndex{20}\)
\(\dfrac{5}{6} = \dfrac{?}{18}\)
Exercise \(\PageIndex{21}\)
\(\dfrac{4}{5} = \dfrac{?}{25}\)
- Answer
-
20
Exercise \(\PageIndex{22}\)
\(\dfrac{1}{2} = \dfrac{4}{?}\)
Exercise \(\PageIndex{23}\)
\(\dfrac{9}{25} = \dfrac{27}{?}\)
- Answer
-
75
Exercise \(\PageIndex{24}\)
\(\dfrac{3}{2} = \dfrac{18}{?}\)
Exercise \(\PageIndex{25}\)
\(\dfrac{5}{3} = \dfrac{80}{?}\)
- Answer
-
48
Exercise \(\PageIndex{26}\)
\(\dfrac{1}{8} = \dfrac{3}{?}\)
Exercise \(\PageIndex{27}\)
\(\dfrac{4}{5} = \dfrac{?}{100}\)
- Answer
-
80
Exercise \(\PageIndex{28}\)
\(\dfrac{1}{2} = \dfrac{25}{?}\)
Exercise \(\PageIndex{29}\)
\(\dfrac{3}{16} = \dfrac{?}{96}\)
- Answer
-
18
Exercise \(\PageIndex{30}\)
\(\dfrac{15}{16} = \dfrac{225}{?}\)
Exercise \(\PageIndex{31}\)
\(\dfrac{11}{12} = \dfrac{?}{168}\)
- Answer
-
154
Exercise \(\PageIndex{32}\)
\(\dfrac{9}{13} = \dfrac{?}{286}\)
Exercise \(\PageIndex{33}\)
\(\dfrac{32}{33} = \dfrac{?}{1518}\)
- Answer
-
1,472
Exercise \(\PageIndex{34}\)
\(\dfrac{19}{20} = \dfrac{1045}{?}\)
Exercise \(\PageIndex{35}\)
\(\dfrac{37}{50} = \dfrac{1369}{?}\)
- Answer
-
1,850
For the following problems, reduce, if possible, each of the fractions to lowest terms.
Exercise \(\PageIndex{36}\)
\(\dfrac{6}{8}\)
Exercise \(\PageIndex{37}\)
\(\dfrac{8}{10}\)
- Answer
-
\(\dfrac{4}{5}\)
Exercise \(\PageIndex{38}\)
\(\dfrac{5}{10}\)
Exercise \(\PageIndex{39}\)
\(\dfrac{6}{14}\)
- Answer
-
\(\dfrac{3}{7}\)
Exercise \(\PageIndex{40}\)
\(\dfrac{3}{12}\)
Exercise \(\PageIndex{41}\)
\(\dfrac{4}{14}\)
- Answer
-
\(\dfrac{2}{7}\)
Exercise \(\PageIndex{42}\)
\(\dfrac{1}{6}\)
Exercise \(\PageIndex{43}\)
\(\dfrac{4}{6}\)
- Answer
-
\(\dfrac{2}{3}\)
Exercise \(\PageIndex{44}\)
\(\dfrac{18}{14}\)
Exercise \(\PageIndex{45}\)
\(\dfrac{20}{8}\)
- Answer
-
\(\dfrac{5}{2}\)
Exercise \(\PageIndex{46}\)
\(\dfrac{4}{6}\)
Exercise \(\PageIndex{47}\)
\(\dfrac{10}{6}\)
- Answer
-
\(\dfrac{5}{3}\)
Exercise \(\PageIndex{48}\)
\(\dfrac{6}{14}\)
Exercise \(\PageIndex{49}\)
\(\dfrac{14}{6}\)
- Answer
-
\(\dfrac{7}{3}\)
Exercise \(\PageIndex{50}\)
\(\dfrac{10}{12}\)
Exercise \(\PageIndex{51}\)
\(\dfrac{16}{70}\)
- Answer
-
\(\dfrac{8}{35}\)
Exercise \(\PageIndex{52}\)
\(\dfrac{40}{60}\)
Exercise \(\PageIndex{53}\)
\(\dfrac{20}{12}\)
- Answer
-
\(\dfrac{5}{3}\)
Exercise \(\PageIndex{54}\)
\(\dfrac{32}{28}\)
Exercise \(\PageIndex{55}\)
\(\dfrac{36}{10}\)
- Answer
-
\(\dfrac{18}{5}\)
Exercise \(\PageIndex{56}\)
\(\dfrac{36}{60}\)
Exercise \(\PageIndex{57}\)
\(\dfrac{12}{18}\)
- Answer
-
\(\dfrac{2}{3}\)
Exercise \(\PageIndex{58}\)
\(\dfrac{18}{27}\)
Exercise \(\PageIndex{59}\)
\(\dfrac{18}{24}\)
- Answer
-
\(\dfrac{3}{4}\)
Exercise \(\PageIndex{60}\)
\(\dfrac{32}{40}\)
Exercise \(\PageIndex{61}\)
\(\dfrac{11}{22}\)
- Answer
-
\(\dfrac{1}{2}\)
Exercise \(\PageIndex{62}\)
\(\dfrac{27}{81}\)
Exercise \(\PageIndex{63}\)
\(\dfrac{17}{51}\)
- Answer
-
\(\dfrac{1}{3}\)
Exercise \(\PageIndex{64}\)
\(\dfrac{16}{42}\)
Exercise \(\PageIndex{65}\)
\(\dfrac{39}{13}\)
- Answer
-
3
Exercise \(\PageIndex{66}\)
\(\dfrac{44}{11}\)
Exercise \(\PageIndex{67}\)
\(\dfrac{66}{33}\)
- Answer
-
2
Exercise \(\PageIndex{68}\)
\(\dfrac{15}{1}\)
Exercise \(\PageIndex{69}\)
\(\dfrac{15}{16}\)
- Answer
-
already reduced
Exercise \(\PageIndex{70}\)
\(\dfrac{15}{40}\)
Exercise \(\PageIndex{71}\)
\(\dfrac{36}{100}\)
- Answer
-
\(\dfrac{9}{25}\)
Exercise \(\PageIndex{72}\)
\(\dfrac{45}{32}\)
Exercise \(\PageIndex{73}\)
\(\dfrac{30}{75}\)
- Answer
-
\(\dfrac{2}{5}\)
Exercise \(\PageIndex{74}\)
\(\dfrac{121}{132}\)
Exercise \(\PageIndex{75}\)
\(\dfrac{72}{64}\)
- Answer
-
\(\dfrac{9}{8}\)
Exercise \(\PageIndex{76}\)
\(\dfrac{30}{105}\)
Exercise \(\PageIndex{77}\)
\(\dfrac{46}{60}\)
- Answer
-
\(\dfrac{23}{30}\)
Exercise \(\PageIndex{78}\)
\(\dfrac{75}{45}\)
Exercise \(\PageIndex{79}\)
\(\dfrac{40}{18}\)
- Answer
-
\(\dfrac{20}{9}\)
Exercise \(\PageIndex{80}\)
\(\dfrac{108}{76}\)
Exercise \(\PageIndex{81}\)
\(\dfrac{7}{21}\)
- Answer
-
\(\dfrac{1}{3}\)
Exercise \(\PageIndex{82}\)
\(\dfrac{6}{51}\)
Exercise \(\PageIndex{83}\)
\(\dfrac{51}{12}\)
- Answer
-
\(\dfrac{17}{4}\)
Exercise \(\PageIndex{84}\)
\(\dfrac{8}{100}\)
Exercise \(\PageIndex{85}\)
\(\dfrac{51}{54}\)
- Answer
-
\(\dfrac{17}{18}\)
Exercise \(\PageIndex{86}\)
A ream of paper contains 500 sheets. What fraction of a ream of paper is 200 sheets? Be sure to reduce.
Exercise \(\PageIndex{87}\)
There are 24 hours in a day. What fraction of a day is 14 hours?
- Answer
-
\(\dfrac{7}{12}\)
Exercise \(\PageIndex{88}\)
A full box contains 80 calculators. How many calculators are in \(\dfrac{1}{4}\) of a box?
Exercise \(\PageIndex{89}\)
There are 48 plants per flat. How many plants are there in \(\dfrac{1}{3}\) of a flat?
- Answer
-
16
Exercise \(\PageIndex{90}\)
A person making $18,000 per year must pay $3,960 in income tax. What fraction of this person's yearly salary goes to the IRS?
For the following problems, find the mistake.
Exercise \(\PageIndex{91}\)
\(\dfrac{3}{24} = \dfrac{\cancel{3}}{\cancel{3} \cdot 8} = \dfrac{0}{8} = 0\)
- Answer
-
Should be \(\dfrac{1}{8}\); the cancellation is division, so the numerator should be 1.
Exercise \(\PageIndex{92}\)
\(\dfrac{8}{10} = \dfrac{\cancel{2} + 6}{\cancel{2} + 8} = \dfrac{6}{8} = \dfrac{3}{4}\)
Exercise \(\PageIndex{93}\)
\(\dfrac{7}{15} = \dfrac{\cancel{7}}{\cancel{7} + 8} = \dfrac{1}{8}\)
- Answer
-
Cancel factors only, not addends; \(\dfrac{7}{15}\) is already reduced.
Exercise \(\PageIndex{94}\)
\(\dfrac{6}{7} = \dfrac{\cancel{5} + 1}{\cancel{5} + 2} = \dfrac{1}{2}\)
Exercise \(\PageIndex{95}\)
\(\dfrac{\cancel{9}}{\cancel{9}} = \dfrac{0}{0} = 0\)
- Answer
-
1
Exercises for Review
Exercise \(\PageIndex{96}\)
Round 816 to the nearest thousand.
Exercise \(\PageIndex{97}\)
Perform the division: \(0 \div 6\).
- Answer
-
0
Exercise \(\PageIndex{98}\)
Find all the factors of 24.
Exercise \(\PageIndex{99}\)
Find the greatest common factor of 12 and 18.
- Answer
-
6
Exercise \(\PageIndex{100}\)
Convert \(\dfrac{15}{8}\) to a mixed number.