# Math

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What is the difference between an equation with two variables and an equation with three variables?

• Math -

one has two, as x and y
one has three, as x, y and z

• Math -

As 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 -

8 divided by 761

• Math -

TÃ´i Ä‘ang cáº§n. Thanks!

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