<< Chapter < Page | Chapter >> Page > |
To add two or more fractions that have the same denominators, add the numerators and place the resulting sum over the common denominator . Reduce, if necessary.
Find the following sums.
$\frac{3}{7}$ + $\frac{2}{7}$
The denominators are the same.
Add the numerators and place the sum over the common denominator, 7.
$\frac{3}{7}$ + $\frac{2}{7}$ = $\frac{3+2}{7}$ = $\frac{5}{7}$
When necessary, reduce the result.
$\frac{1}{8}$ + $\frac{3}{8}$ = $\frac{1+3}{8}$ = $\frac{4}{8}$ = $\frac{1}{2}$
To see what happens if we mistakenly add the denominators as well as the numerators, let’s add
$\frac{1}{2}$ and $\frac{1}{2}$ .
Adding the numerators and mistakenly adding the denominators produces:
$\frac{1}{2}$ + $\frac{1}{2}$ = $\frac{1+1}{2+2}$ = $\frac{2}{4}$ = $\frac{1}{2}$
This means that $\frac{1}{2}$ + $\frac{1}{2}$ is the same as $\frac{1}{2}$ , which is preposterous! We do not add denominators .
$\frac{3}{8}$ + $\frac{3}{8}$
$\frac{6}{8}$ = $\frac{3}{4}$
$\frac{7}{\text{11}}$ + $\frac{4}{\text{11}}$
$\frac{\text{11}}{\text{11}}$ = 1
$\frac{\text{15}}{\text{20}}$ + $\frac{1}{\text{20}}$ + $\frac{2}{\text{20}}$
$\frac{\text{18}}{\text{20}}$ = $\frac{9}{\text{10}}$
To subtract two or more fractions that have the same denominators, subtract the numerators and place the resulting difference over the common denominator . Reduce, if necessary.
Find the following differences.
$\frac{3}{5}$ - $\frac{1}{5}$
The denominators are the same.
Subtract the numerators and place the difference over the common denominator, 5.
$\frac{3}{5}$ - $\frac{1}{5}$ = $\frac{3-1}{5}$ = $\frac{2}{5}$
When necessary, reduce the result.
$\frac{8}{6}$ - $\frac{2}{6}$ = $\frac{6}{6}$ = 1
To see what happens if we mistakenly subtract the denominators as well as the numerators, let’s subtract
$\frac{7}{\text{15}}$ - $\frac{4}{\text{15}}$ .
Subtracting the numerators and mistakenly subtracting the denominators produces:
$\frac{7}{\text{15}}$ - $\frac{4}{\text{15}}$ = $\frac{7-4}{\text{15}-\text{15}}$ = $\frac{3}{0}$
We end up dividing by zero, which is undefined. We do not subtract denominators.
$\frac{5}{\text{12}}$ - $\frac{1}{\text{12}}$
$\frac{4}{\text{12}}$ = $\frac{1}{3}$
$\frac{3}{\text{16}}$ - $\frac{3}{\text{16}}$
Result is 0
$\frac{\text{16}}{5}$ - $\frac{1}{5}$ - $\frac{2}{5}$
Result is $\frac{\text{13}}{5}$
Basic Rule: Fractions can only be added or subtracted conveniently if they have like denominators.
To see why this rule makes sense, let’s consider the problem of adding a quarter and a dime.
A quarter is $\frac{1}{4}$ of a dollar.
A dime is $\frac{1}{\text{10}}$ of a dollar.
We know that 1 quarter + 1 dime = 35 cents. How do we get to this answer by adding $\frac{1}{4}$ and $\frac{1}{\text{10}}$ ?
We convert them to quantities of the same denomination.
A quarter is equivalent to 25 cents, or $\frac{\text{25}}{\text{100}}$ .
A dime is equivalent to 10 cents, or $\frac{\text{10}}{\text{100}}$ .
By converting them to quantities of the same denomination, we can add them easily:
$\frac{\text{25}}{\text{100}}$ + $\frac{\text{10}}{\text{100}}$ = $\frac{\text{35}}{\text{100}}$ .
Same denomination $\to $ same denominator
If the denominators are not the same, make them the same by building up the fractions so that they both have a common denominator. A common denominator can always be found by multiplying all the denominators, but it is not necessarily the Least Common Denominator.
The LCD is the smallest number that is evenly divisible by all the denominators.
It is the least common multiple of the denominators.
The LCD is the product of all the prime factors of all the denominators, each factor taken the greatest number of times that it appears in any single denominator.
Find the sum of these unlike fractions.
$\frac{1}{\text{12}}$ + $\frac{4}{\text{15}}$
Factor the denominators:
12 = 2 × 2 × 3
15 = 3 × 5
What is the greatest number of times the prime factor 2 appear in any single denominator? Answer: 2 times. That is the number of times the prime factor 2 will appear as a factor in the LCD.
What is the greatest number of times the prime factor 3 appear in any single denominator? Answer: 1 time. That is the number of times the prime factor 3 will appear as a factor in the LCD.
What is the greatest number of times the prime factor 5 appear in any single denominator? Answer: 1 time. That is the number of times the prime factor 5 will appear as a factor in the LCD.
So we assemble the LCD by multiplying each prime factor by the number of times it appears in a single denominator, or:
2 × 2 × 3 × 5 = 60
60 is the Least Common Denominator (the Least Common Multiple of 12 and 15) .
To create fractions with like denominators, we now multiply the numerators by whatever factors are missing when we compare the original denominator to the new LCD.
In the fraction $\frac{1}{\text{12}}$ , we multiply the denominator 12 by 5 to get the LCD of 60. Therefore we multiply the numerator 1 by the same factor (5).
$\frac{1}{\text{12}}$ × $\frac{5}{5}$ = $\frac{5}{\text{60}}$
Similarly,
$\frac{4}{\text{15}}$ × $\frac{4}{4}$ = $\frac{\text{16}}{\text{60}}$
We can now add the two fractions because they have like denominators:
$\frac{5}{\text{60}}$ + $\frac{\text{16}}{\text{60}}$ = $\frac{\text{21}}{\text{60}}$
Reduce the result: $\frac{\text{21}}{\text{60}}$ = $\frac{7}{\text{20}}$
$\frac{1}{6}$ + $\frac{3}{4}$
Result is $\frac{\text{11}}{\text{12}}$
$\frac{5}{9}$ - $\frac{5}{\text{12}}$
Result is $\frac{5}{\text{36}}$
$\frac{\text{15}}{\text{16}}$ + $\frac{1}{2}$ - $\frac{3}{4}$
Result is $\frac{\text{35}}{\text{16}}$
$\frac{9}{\text{15}}$ + $\frac{4}{\text{15}}$
Result is $\frac{\text{13}}{\text{15}}$
$\frac{7}{\text{10}}$ - $\frac{3}{\text{10}}$ + $\frac{\text{11}}{\text{10}}$
Result is $\frac{\text{15}}{\text{10}}$ (reduce to 1 $\frac{1}{2}$ )
Find the total length of the screw in this diagram:
Total length is $\frac{\text{19}}{\text{32}}$ in.
$\frac{5}{2}$ + $\frac{\text{16}}{2}$ - $\frac{3}{2}$
Result is $\frac{\text{18}}{2}$ (reduce to 9)
$\frac{3}{4}$ + $\frac{1}{3}$
Result is $\frac{\text{13}}{\text{12}}$
Two months ago, a woman paid off $\frac{3}{\text{24}}$ of a loan. One month ago, she paid off $\frac{4}{\text{24}}$ of the loan. This month she will pay off $\frac{5}{\text{24}}$ of the total loan. At the end of this month, how much of her total loan will she have paid off?
She will have paid off $\frac{\text{12}}{\text{24}}$ , or $\frac{1}{2}$ of the total loan.
$\frac{8}{3}$ - $\frac{1}{4}$ + $\frac{7}{\text{36}}$
Result is $\frac{\text{94}}{\text{36}}$ (reduce to $\frac{\text{47}}{\text{18}}$ )
Notification Switch
Would you like to follow the 'Contemporary math applications' conversation and receive update notifications?