A Quick and Easy Linear Diophantine System

This video is about solving a Linear Diophantine System
If you need to post a picture of your solution or idea:
#NumberTheoryProblems #DiophantineEquations

EXPLORE:

[echandler]

[1] 8x + 5y + z = 100
[2] x + y + z = 20

[1]-5[2] gives
3x - 4z = 0

hence x = 4k, z = 3k

substituting in [2] gives
y = 20 - 7k

For x, y, z in N:
k = 1, 2

thus desired solutions are:
(4, 13, 3) (8, 6, 6)

[randomcubing]

Learning math: Khan Academy/ The Organic Chemistry Tutor
Easier math: SyberMath
Moderate math: Mind Your Decisions
Difficult math: Blackpenredpen
Math memes: *papa flammy*

[scottleung]

I just plugged in some numbers after doing some elementary row operations and got an idea of what the pattern was. From there, I just generalized to account for all solutions in the family.

[DonRedmond-jkhj]

You could also use the Euclidean Algoritm that you might use to find the gcd and work backwards.

More generally, if we are looking for all integer solutions, and not just positive, there are infinitely many solutions....
but they can be written in the parametric form in terms of an integer, say 'k' as follows:

The solution set is
x = 4k, z = 3k, y = 20 - 7k where k is any integer!

It's easy to see y is positive only for k = 1, 2 which gives required solutions!

[georgesbv]

Reduce Z by subtracting: (8-1)x+(5-1)y = 100-20
7x + 4y = 80
7x = 80 - 4y = 4(20-y)
Solutions are: (4, 13, 3) ; (-4, 27, -3)

[DrQuatsch]

Subtract equation 2 from equation 1. This gives 7x + 4y = 80. I will put the 7x on the other side and divide by 4 to get an equation for y: y = (80 - 7x)/4 = 20 - 7x/4. y is an integer, so 7x/4 has to be an integer. Therefore x has to be a multiple of 4. So I will introduce a parameter n which is an integer. x = 4n. This results in y = 20 - 7 * (4n) / 4 = 20 - 7n. Now use the second equation to get a formula for z: z = 20 - x - y = 20 - 4n - (20 - 7n) = 20 - 4n - 20 + 7n = 3n. So the solution set for this diophantine system is: {4n, 20 - 7n, 3n} for all integer n. This gives us all solutions including positive and negative ones. If x, y, z >= 0, then clearly n>=0, but also 20 - 7n >= 0, which means that 7n <= 20, so n <= 20/7, which just leaves n = 0, 1, 2.
Plugging these numbers in gives: (0, 20, 0), (4, 13, 3), (8, 6, 6). If x, y, z > 0 then only (4, 13, 3) and (8, 6, 6) are solutions.

[chillfill]

Instead, get x and y in terms of z. X=4z/3, y = 20- 7z/3. For integer solutions you need z a multiple of 3.

[boolakee]

I solved it through this method :

8x + 5y + z  =  100      Eqn (1)
  x  +  y  + z  =    20      Eqn (2)

Eqn (1) - Eqn (2) :
7x  +  4y  =  80
One solution is : (x, y) = (0, 20)
Hence,
A general set of solutions (x, y) is :
.            (x, y) = [4k, (20-7k)]      where k is an integer
Replacing (x, y) in Eqn (2) and re-arranging :
.               z  =  20 - 4k - (20-7k)  = 3k
Hence, the general set of solutions (x, y, z) is :
.    (x, y, z) = [(4k), (20-7k), (3k)]      where k is an integer.
Positive integer occurs when
7k < 20, i.e. when k =< 2.
Hence the only solutions are when k = 1 or 2 :
For k = 1 : (x, y, z) = (4, 13, 3)
For k = 2 : (x, y, z) = (8, 6, 6)

[mxsjncv]

We have here two plaines and the parametric equation of the common line is (0, 20, 0) +t (4, -7, 3)

[MohammadSami-ul-lah]

Sir please tell me how can find easily the positive integer solution of equations like 6xy + x -- y = n where n is any natural number (say 1, 2, 3, 4, 5, 6 so on)

[hugooliveira]

What about (0, 20, 0)
Isnt that a solution too?

[-basicmaths]

7 x+4 y=80
X is even.x. is not 2, 4 or 6
X is 8.
So y is6, and z= 6