FRAME 2-25.
Solution to
The solution to Frame 2-24 is shown below (not sufficient space in
Frame 2-24.
column to right). The problem is worked two ways. The first shows the
problem worked with the common denominator being the product of all of
633
or
211
the denominators (2x3x5x7x9 = 1890). The second shows the problem
1890
630
being worked with a lower common denominator (2x5x7x9 = 630). Did
you notice that the denominator "3" divides evenly into the denominator
"9"?
1 _ 1
1 _ 1
1=
+
+
2
3
5
7
9
1x3x5x7x9 _ 1x2x5x7x9
1x2x3x7x9 _ 1x2x3x5x9
1x2x3x5x7=
+
+
2x3x5x7x9
3x2x5x7x9
5x2x3x7x9
7x2x3x5x9
9x2x3x5x7
945 _ 630
378 _ 270
210
(945-630)+(378-270)+210 =
+
+
=
1890
1890
1890
1890
1890
1890
315+108+210
633
=
1890
1890
NOTE: There are several ways of adding and subtracting the
numerators. One way is shown above. Another (and usually better) way
is to add all of the pluses (positives) together, add all of the minuses
(negatives) together, and subtract as shown below.
(945+378+210) (630+270) = 1533 900 = 633
Using 630 (2x5x7x9) as the common denominator
1x5x7x9 _ 1x2x5x7x3
1x2x7x9 _ 1x2x5x9
1x2x5x7 =
+
+
2x5x7x9
3x2x5x7x3
5x2x7x9
7x2x5x9
9x2x5x7
315 _ 210
126 _ 90
70 = (315-210)+(126-90)+70 =
+
+
630
630
630
630
630
630
105+36+70
211
=
630
630
The converted fractions can also be added and subtracted as below:
(315+126+70) (210+90)
511 300
211
=
=
630
630
630
___________________________________________________________________
MD0900
2-10