Processing Math: Done
To print higher-resolution math symbols, click the
Hi-Res Fonts for Printing button on the jsMath control panel.
jsMath

Solution 1.2:1d

From Förberedande kurs i matematik 1

(Difference between revisions)
Jump to: navigation, search
m (Lösning 1.2:1d moved to Solution 1.2:1d: Robot: moved page)
Line 1: Line 1:
-
{{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}}
+
 
 +
<math>\frac{1\centerdot 4\centerdot 5}{3\centerdot 4\centerdot 5}+\frac{1\centerdot 3\centerdot 5}{4\centerdot 3\centerdot 5}+\frac{1\centerdot 3\centerdot 4}{5\centerdot 3\centerdot 4}=\frac{20}{60}+\frac{15}{60}+\frac{12}{60}</math>
 +
 
 +
When all three fractions have a common denominator, they can easily be added up
 +
 
 +
 
 +
<math>\frac{20}{60}+\frac{15}{60}+\frac{12}{60}=\frac{20+15+12}{60}=\frac{47}{60}</math>

Revision as of 12:05, 11 September 2008

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,


345145+435135+534134=6020+6015+6012

When all three fractions have a common denominator, they can easily be added up


6020+6015+6012=6020+15+12=6047

Retrieved from "http://wiki.math.se/wikis/2008/bridgecourse1-ImperialCollege/index.php/Solution_1.2:1d"