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March 30, 2015

March 30, 2015

Posted by **kayla** on Friday, October 12, 2007 at 12:43am.

- Math -
**bobpursley**, Friday, October 12, 2007 at 5:34amone has two, as x and y

one has three, as x, y and z

- Math -
**tchrwill**, Friday, October 12, 2007 at 2:04pmAs bobpursley said one would have variables such as "x" and "y" and the other would have variables of "x", "y' and "z". The one with variables of x and y could be solved for either x or y and substiruted into the one with variables of x, y and z yielding you a new equation with variables of either x and z or y and z which can now be solved by the method of successive reductions or the Euclidian Algorithm.

Example of successive reductions:

Janet has $8.55 in nickels, dimes, and quarters. She has 7 more dimes than

>nickels and quarters combined. How many of each coin does she have?

>

1--.05N + .10D + .25Q = 8.55

2--5N + 10D + 25Q = 855

3--D = N + Q + 7

4--Substituting, 5N + 10N + 10Q + 70 + 25Q = 855.

5--Collecting terms, 15N + 35Q = 785 or 3N + 7Q = 157, an equation with 2 variables.

6--Dividing through by the lowest coefficiet, 3 yields N + 2Q + Q/3 = 52 + 1

7--(Q - 1)/3 must be an integer k making Q = 3k + 1

8--Substituting back into (5) yields 3N + 21k + 7 = 157 or N = 50 - 7k

9--k can be any value from 0 through 7

10--k....0....1....2....3....4....5....6....7

.....N...50...43..36..29..22..15...8....1

.....Q....1....4....7...10..13..16..19..22

.....D...58...54..50..46..42..38..34..30

11--Therefore, there are 8 solutions.

Example of Euclidian Algorithm:

What is the smallest positive integer that leaves a remainder of 1 when divided by 1000 and a remainder of 8 when divided by 761?

This can be expressed by 1000x + 1 = 761y + 8 = N.

Rearranging, 1000x - 761y = 7, an equation of 2 variables.

First find a solution to 1000x - 761y = 1

Using the Euclidian Algorithm:

1000 = 1(761) + 239

761 = 3(239) + 44

239 = 5(44) + 19

44 = 2(19) + 6

19 = 3(6) + 1

Then

1 = 19 - 3(6)

1 = 19 - 3(440 + 6(19) = 7(19) - 3(44)

1 = 7(239) - 35(440 - 3(44) = 7(239) - 38(44)

1 = 7(239) - 38(761) + 114(239) = 121(230) - 38(761)

1 = 121(1000) - 121(761) - 38(761) = 121(1000) - 159(761)1 =

Therefore, x = 121 and y = 159 is one solution to 1000x - 761y = 1

Multiplying by 7 yields x = 847 and y = 1113, as a solution to 1000x - 761y = 7.

The general solution is then x = 847 - 761t and y = 1113 - 1000t.

The smallest solution occurs when t = 1 yielding x = 86 and y = 113.

This then permits the definition of the positive solutions as x = 86 + 761t and y = 113 + 1000t.

- Math -
**Anonymous**, Wednesday, December 8, 2010 at 4:56pm8 divided by 761

- Math -
**Trần Quốc Cường**, Wednesday, April 3, 2013 at 4:31amTôi đang cần. Thanks!

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