Solution 1.2:1d
From Förberedande kurs i matematik 1
(Difference between revisions)
m |
|||
| (One intermediate revision not shown.) | |||
| Line 1: | Line 1: | ||
When we have three fractions involved in an addition, we need to multiply the top and bottom of each fraction by the product of the other fraction's denominators so that all fractions have a common denominator, | When we have three fractions involved in an addition, we need to multiply the top and bottom of each fraction by the product of the other fraction's denominators so that all fractions have a common denominator, | ||
| - | + | {{Displayed math||<math>\frac{1\cdot 4\cdot 5}{3\cdot 4\cdot 5}+\frac{1\cdot 3\cdot 5}{4\cdot 3\cdot 5}+\frac{1\cdot 3\cdot 4}{5\cdot 3\cdot 4}=\frac{20}{60}+\frac{15}{60}+\frac{12}{60}\,</math>.}} | |
| - | <math>\frac{1\ | + | |
When all three fractions have a common denominator, they can easily be added up | When all three fractions have a common denominator, they can easily be added up | ||
| - | + | {{Displayed math||<math>\frac{20}{60}+\frac{15}{60}+\frac{12}{60}=\frac{20+15+12}{60}=\frac{47}{60}\,</math>.}} | |
| - | <math>\frac{20}{60}+\frac{15}{60}+\frac{12}{60}=\frac{20+15+12}{60}=\frac{47}{60}</math> | + | |
Current revision
When we have three fractions involved in an addition, we need to multiply the top and bottom of each fraction by the product of the other fraction's denominators so that all fractions have a common denominator,
| \displaystyle \frac{1\cdot 4\cdot 5}{3\cdot 4\cdot 5}+\frac{1\cdot 3\cdot 5}{4\cdot 3\cdot 5}+\frac{1\cdot 3\cdot 4}{5\cdot 3\cdot 4}=\frac{20}{60}+\frac{15}{60}+\frac{12}{60}\,. |
When all three fractions have a common denominator, they can easily be added up
| \displaystyle \frac{20}{60}+\frac{15}{60}+\frac{12}{60}=\frac{20+15+12}{60}=\frac{47}{60}\,. |
