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u/New-Ant-2315 13d ago
You can’t solve it. 3(x+y+z)=837 makes x+y+z=279 but then there are infinite solutions to that.
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u/ThatsNotAZombieBite 13d ago
This is one linear equation with three variables.
The solution is a plane in 3-dimensional space. There are infinite point solutions that lie in that plane.
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u/Every-Maths1931 13d ago
Thank you. How would you post it in order to solve it?
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u/CaptainMatticus 13d ago
x + y + z - 279 = 0
That's the equation of the plane. Every point on that plane is a solution.
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u/Glum_Battle6008 13d ago edited 13d ago
Arbitrary and pointless. x = 837/3, y = n, z = -n.
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u/Niels_vdk 13d ago
z=0 as well for your solution.
but yes a single equation with 3 unknowns has infinite possible solutions.
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u/Every-Maths1931 13d ago
What’s a way to rearrange the equation to solve it?
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u/Niels_vdk 13d ago
you could simplify it to x+y+z=279 but thats about it. after that you can pick any number for 2 of the variables and adjust the third accordingly.
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u/CaptainMatticus 13d ago
I wouldn't. You have 3 unknowns and 1 equation. There are an infinite number of solutions.
3 * (x + y + z) = 3 * 279
x + y + z = 279
Even if we said that we had 3 distinct positive integers, we wouldn't have an infinite number of solutions, but we'd still have a lot of solutions.
1 + 2 + 276 , 1 + 3 + 275 , 1 + 4 + 274 , 1 + 5 + 273 , 1 + 6 + 272 , 1 + 7 + 271 , 1 + 8 + 270 , 1+ 9 + 269 , ... , 1 + 135 + 143 , 1 + 136 + 142 , 1 + 137 + 141 , 1 + 138 + 140
And each of those can be shifted around 6 different ways. Now to the next set:
2 + 3 + 274 , 2 + 4 + 273 , 2 + 5 + 272 , 2 + 6 + 271 , ... + 2 + 135 + 142 , 2 + 136 + 141 , 2 + 137 + 140 , 2 + 138 + 139
Then
3 + 4 + 272 , 3 + 5 + 271 , 3 + 6 + 270 , .... , 3 + 133 + 143 , 3 + 134 + 142 , 3 + 135 + 141 , 3 + 136 + 140 , 3 + 137 + 139
And then
4 + 5 + 270 , 4 + 6 + 269 , 4 + 7 + 268 , ... , 4 + 130 + 145 , 4 + 131 + 144 , 4 + 132 + 143 , 4 + 133 + 142 , 4 + 134 + 141 , 4 + 135 + 140 , 4 + 136 + 139 , 4 + 137 + 138
5 + 6 + 268 , .... , 5 + 136 + 138
6 + 7 + 266 , ... , 6 + 136 + 137
7 + 8 + 264 , ... , 7 + 135 + 137
8 + 9 + 262 , ... , 8 + 135 + 136
9 + 10 + 260 , ...
10 + 11 + 258 , ...
11 + 12 + 256 , ...
12 + 13 + 254 , ...
...
90 + 91 + 98 , 90 + 92 + 97 , 90 + 93 + 96 , 90 + 94 + 95
91 + 92 + 96 , 91 + 93 + 95
92 + 93 + 94
So we start off with 137 triplets, then 136, then 134, then 133, then 131, then 130, then 128, then 127, then 125, then 124, and so on. So we need a way to sum all of that.
1 + 2 + 4 + 5 + 7 + 8 + .... + 136 + 137
1 + 2 + 3 + 4 + 5 + 6 + .... + 137 - 3 - 6 - 9 - 12 - .... - 135
1 + 2 + 3 + .... + 137 - 3 * (1 + 2 + 3 + ... + 45)
(137/2) * (137 + 1) - 3 * (45/2) * (45 + 1)
(137/2) * 138 - (135/2) * 46
137 * 69 - 135 * 23
23 * (137 * 3 - 135)
23 * 3 * (137 - 45)
69 * 92 =>
92 * (70 - 1) =>
6440 - 92 =>
6348
Multiply that by 6, because there are 6 permutations to each solution set.
6 * 6348 =>
6 * 6350 - 6 * 2 =>
3 * 12700 - 12 =>
38100 - 12 =>
38088
So even among just distinct positive integer solution sets, we almost have 40,000 solutions. It only gets worse from there. We need more equations. If you have n-unknowns, you need n-distinct equations to get a unique solution.