Further Fractions 2 (Classes 2.1 and 2.2)
Adding and Subtracting Fractions Class 3.1 (Final Fractions Class)
(Fractions Test for when Classes are completed)
More Fractions – Adding and Subtracting Fractions – Class 3.1
Fractions Class 3.1
Adding and Subtracting Fractions
As stated in Further Fractions Classes, when adding or subtracting two fractions with the same denominator, we simply add or subtract the numerator, and leave the denominator the same, as in the two following examples:
$$\frac{9}{10}-\frac{3}{10}=\frac{6}{10}\;or\;simplified\;\frac{3}{5}$$
$$\frac{2}{11}+\frac{4}{11}=\frac{6}{11}$$
In the first example, above, since both terms of the subtraction equation have the same denominator, 10, the answer’s numerator was found by subtracting the second numerator, 3, from the first numerator, 9. Thus the answer was 6 over the same denominator, 10. In the second example, above, since the denominator 11 was the same in both terms of the addition, the answer’s numerator was found by adding the second numerator, 4, to the first numerator, 2, giving us the answer of 6 over 11, or six elevenths.
Adding and Subtracting Fractions Where the Denominators are Different
How difficult do you think it would be to add one third to one quarter when they do not have the same denominator, as in the illustration below?
When adding or subtracting fractions where the denominators are different, the key to finding the result is in finding a way of expressing these fractions with the same denominator. Fractions with the same denominator are known as homogeneous, while fractions with different denominators are known as heterogeneous.
If we reconstitute the fractions above to have the same denominator of 12, we will more easily be able to add the two fractions together,
The new denominator was found by multiplying the denominators of the original fractions. The numerators must also be changed so that each term properly expresses the original ratio of the two fractions, as shown here,
step 1: $$\frac{1}{3}+\frac{1}{4}=$$
step 2:$$\frac{1×4}{3}+\frac{1×3}{4}=$$
step 3:$$\frac{1×4}{3×4}+\frac{1×3}{4×3}=$$
step 4:$$\frac{4}{12}+\frac{3}{12}=$$
step 5:$$\frac{4}{12}+\frac{3}{12}=\frac{7}{12}$$
At step 1, above, the initial sum, is given. At step 2, numerators are recalculated by multiplying them by the alternate terms original denominators in order to maintain the original ratios in later steps. At step 3, a new denominator is given to each of the terms, which is the product of the two original denominators, which in this case is 12. Step 4 shows the completely reconstituted sum, all ratios intact.
At step 5, the final answer is found by adding the terms which now have the same denominator. The numerators of the terms are added together, while keeping the denominator the same.
To do a subtraction which involves fractions with different denominators, the process is the same as it is for addition, right up to step 4 where you subtract the second numerator from the first, instead of adding it, to get the answer. Here is an example ,
step 1: $$\frac{3}{4}-\frac{2}{5}=$$
step 2:$$\frac{3×5}{4}-\frac{2×4}{5}=$$
step 3:$$\frac{3×5}{4×5}-\frac{2×4}{5×4}=$$
step 4:$$\frac{15}{20}-\frac{8}{20}=$$
step 5:$$\frac{15}{20}-\frac{8}{20}=\frac{7}{20}$$
There are two ways to add or subtract mixed numbers. The usual method is to convert them into improper fractions first, then to find a common denominator and add or subtract them. Or you can add/subtract the whole numbers and fractional parts separately.
Try the following exercises to complete this Class.
$$a)\;20 $$ $$b)\;16$$ $$c)\;12$$ $$d)\;15 $$
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$$a)\;20 $$
$$a)\;12 $$ $$b)\;24$$ $$c)\;21 $$ $$d)\;20 $$
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$$c)\;21 $$
$$a) \;\frac{49}{30} $$ $$b)\;\frac{39}{30} $$ $$c)\;\frac{30}{25} $$ $$d)\;\frac{17}{16} $$
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$$a)\; \frac{49}{30} $$
$$a) \;\frac{4}{5} $$ $$b)\;\frac{6}{11} $$ $$c)\;\frac{22}{27} $$ $$d)\;\frac{20}{21} $$
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$$d)\;\frac{20}{21} $$
$$a)\; \frac{3}{8} $$ $$b)\;\frac{1}{4} $$ $$c)\;\frac{1}{6} $$ $$d)\;\frac{1}{12} $$
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$$c)\;\frac{1}{6} $$
$$a)\; \frac{25}{32} $$ $$b)\;\frac{28}{41} $$ $$c)\; \frac{27}{40} $$ $$d)\;\frac{17}{40} $$
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$$c)\; \frac{27}{40} $$
$$a)\; \frac{13}{27} $$ $$b)\;\frac{8}{11} $$ $$c)\; \frac{9}{16} $$ $$d)\; \frac{7}{12} $$
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$$d)\; \frac{7}{12} $$
$$a)\;\frac{26}{6}\;or\;4\frac{2}{6}$$ $$b)\;\frac{59}{12}\;or\;4\frac{11}{12}$$ $$c)\;\frac{48}{15}\;or\;3\frac{3}{15}$$ $$d)\;\frac{53}{12}\;or\;4\frac{5}{12}$$
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$$b)\;\frac{59}{12}\;or\;4\frac{11}{12}$$
$$a) \;\frac{11}{12} $$ $$b)\;\frac{61}{60} $$ $$c)\;\frac{29}{30} $$ $$d)\;\frac{11}{16} $$
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$$c)\;\frac{29}{30} $$
$$a) \;\frac{41}{12}\;or\;3\frac{5}{12}$$ $$b)\;\frac{28}{7}or\;3\frac{1}{7} $$ $$c)\; \frac{73}{24}\;or\;3\frac{1}{24}$$ $$d)\; \frac{52}{15}\;or\;3\frac{7}{15}$$
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$$a) \;\frac{41}{12}\;or\;3\frac{5}{12}$$