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 
. 
