We choose a common multiple of the denominators because we change a denominator by multiplying it. Lesson 22.
Solution. The lowest common multiple of 3 and 4 is their product, 12. (Lesson 22, Question
4.)
We will convert each fraction to an equivalent fraction with denominator 12.
2 3 |
+ |
1 4 |
= |
8 12 |
+ |
3 12 |
|
|
|
|
= |
11 12 |
. |
We converted |
2 3 |
to |
8 12 |
by saying, "3 goes into |
(is contained in) 12 four times. Four times 2 is 8."
(In that way, we multiplied both 2 and 3 by the same number, namely 4. See Lesson 22, Question 3.)
We converted |
1 4 |
to |
3 12 |
by saying, "4 goes into 12 three |
times. Three times 1 is 3." (We multiplied both 1 and 4 by 3.)
The fact that we say what we do shows again that arithmetic is a spoken skill.
In practice, it is necessary to write the common denominator only once:
2 3 |
+ |
1 4 |
= |
8 + 3 12 |
= |
11 12 |
. |
Solution. The LCM of 5 and 15 is 15. Therefore,
4 5 |
+ |
2 15 |
= |
12 + 2 15 |
= |
14 15 |
. |
We changed |
4 5 |
to |
12 15 |
by saying, "5 goes into 15 three |
times. Three times 4 is 12."
We did not change |
2 15 |
, because we are not changing the |
denominator 15.
Example 5. |
2 3 |
+ |
1 6 |
+ |
7 12 |
Solution. The LCM of 3, 6, and 12 is 12.
2 3 |
+ |
1 6 |
+ |
7 12 |
= |
8 + 2 + 7 12 |
2 3 |
+ |
1 6 |
+ |
7 12 |
= 1 |
5 12 |
. |
We converted |
2 3 |
to |
8 12 |
by saying, "3 goes into 12 four |
times. Four times 2 is 8."
We converted |
1 6 |
to |
2 12 |
by saying, "6 goes into 12 two |
times. Two times 1 is 2."
We did not change |
7 12 |
, because we are not changing the |
denominator 12.
Finally, we changed the improper fraction |
17 12 |
to 1 |
5 12 | by |
dividing 17 by 12. (Lesson 20.)
"12 goes into 17 one (1) time with remainder 5."
Solution. The LCM of 6 and 9 is 18.
5 6 |
+ |
7 9 |
= |
15 + 14 18 |
= |
29 18 |
= 1 |
11 18 |
. |
We changed |
5 6 |
to |
15 18 |
by multiplying both terms by 3. |
We changed |
7 9 |
to |
14 18 |
by multiplying both terms by 2. |
Example 7. Add mentally |
1 2 |
+ |
1 4 |
. |
Answer. |
1 2 |
is how many |
1 4 |
's? |
Just as 1 is half of 2, so 2 is half of 4. Therefore,
The student should not have to write any problem in which one of
the fractions is |
1 2 |
, and the denominator of the other is even. |
For example,
Example 8. In a recent exam, one eighth of the students got A, two fifths got B, and the rest got C. What fraction got C?
Solution. Let 1 represent the whole number of students. Then the question is:
Now,
1 8 |
+ |
2 5 |
= |
5 + 16 40 |
= |
21 40 |
. |
The rest, the fraction that got C, is the complement of |
21 40 |
. |
|