From Förberedande kurs i matematik 1

Revision as of 08:15, 9 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 1.2:1d moved to Solution 1.2:1d: Robot: moved page) ← Previous diff		Current revision (07:36, 19 September 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	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,
-	<center> [[Image:1_2_1d.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{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>.}}
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		+	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>.}}

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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}\,.