Posted by kayla on Friday, October 12, 2007 at 12:43am.
What is the difference between an equation with two variables and an equation with three variables?

Math  bobpursley, Friday, October 12, 2007 at 5:34am
one has two, as x and y
one has three, as x, y and z 
Math  tchrwill, Friday, October 12, 2007 at 2:04pm
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
25N + 10D + 25Q = 855
3D = N + Q + 7
4Substituting, 5N + 10N + 10Q + 70 + 25Q = 855.
5Collecting terms, 15N + 35Q = 785 or 3N + 7Q = 157, an equation with 2 variables.
6Dividing 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
8Substituting back into (5) yields 3N + 21k + 7 = 157 or N = 50  7k
9k can be any value from 0 through 7
10k....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
11Therefore, 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 4:56pm
8 divided by 761

Math  Tráº§n Quá»‘c CÆ°á»ng, Wednesday, April 3, 2013 at 4:31am
TÃ´i Ä‘ang cáº§n. Thanks!