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

Integrated Publishing, Inc. |