How to solve linear Diophantine equation with 3 variables?

Question

I worked on this problem several times, but for some reason, I can't seem to get it. Here is the problem:

$$6x + 15y + 10z = 53$$

These are my attempts:

1. Let $w = 3y + 2z$. So our equations are: $6x + 5w = 53\ (1)$ and $ 3y + 2z = w\ (2)$. For $(1)$, after using the Euclidean Algorithm, I got $x = 53 + 5n$ and $w = -53 - 6n$. Now substituting into $(2)$, we get $3y + 2z = -53 -6n$. This is where I'm stuck.

2. Let $w = 2x + 5y$. So our equations are: $2x + 5y = w\ (1)$ and $ 3w + 10z = 53\ (2)$. For $(2)$, after using the Euclidean Algorithm, I got $w = -159 + 10n$ and $w = 53 - 3n$. Now substituting into $(1)$, we get $2x + 5y = -159 +10n$. Again, I'm stuck!

Also, are we allowed to have rational solutions? Any help is greatly appreciated. Thank you in advance!

Answer 1

$x=-2,y=1,z=5$ gives $6(-2) + 15(1) + 10(5) = 53$, by inspection. I'm trying to think of the general rule.

Answer 2

Evaluating equation mod 5 we get x $\equiv$ 3 (mod 5) Evaluating equation mod 3 we get z $\equiv$ 2 (mod 3) Setting x=5m+3, z=3n+2 in equation we get y=1-2m-2n Integer values for m & n will generate Integer solutions for x,y,z

Answer 3

I was able to figure it out. So I thought I would post it in case someone else searched for help on a similar problem!

Solve $6x+15y+10z = 53 \rightarrow 6x+5(3y+2z)=53.$ Let $w=3y+2z.$ Solve:

$$3y+2z=w\ (1) \\6x+5w=53\ (2)$$

Solution to $(2):\ gcd(6,5) = 1.$ So forming a linear combo, get $6(1)+5(-1)=1$. Multiply by 53, get $6(53)+5(-53)=53$. So, $$x=x_0+(b/d)n = 53+5n$$ $$w=w_0-(a/d)n = -53-6n.$$

Solution to $(1):$ First, solve $3y+2z=1.$ So, $gcd(3,2)=1$; form linear combo, get $3(1)+2(-1)=1.$ Multiply by $w=-53-6n$ to get

$$3(w)+2(-w)=w \rightarrow 3(-53-6n)+2(53+6n)=-53-6n$$

Then, $y_0=-53-6n\ and\ z_0=53+6n$. So,

$$y=y_0+(b/d)m =-53-6n+2m$$ $$z=z_0 -(a/d)m = 53+6n-3m $$

Similarly, you can let $w=2x+5y$ and you'll get the same answer. Hope this process makes some sense!

Answer 4

Here’s how I did it.

Reducing the given equation (mod 6) gives us 4y + 3z congruent to 5 mod 6, which is equivalent to 4y + 3z = 5 + 6k. Solving for y gives (5 + 6k - 3z)/4. Repeat the process on the given equation, this time reducing (mod 10) and solving for x. You should get x = (3 + 10m - 5z)/6.

Substitute these expressions for x and y into the original equation and simplify to obtain an expression for z. You should get that z = 15 - 4m - 6k. Now substitute this expression for z into the equation that defines y ((5 + 6k - 3z)/4), and simplify. You should get y = 6k + 3m - 10. Then substitute the expression for z into the equation that defines x ((3 + 10m - 5z)/6) and simplify. You should get 5k + 5m - 12.

These are your parameterized solutions to the general equation: (x, y, z) = (5k + 5m - 12, 6k + 3m - 10, 15 - 4m - 6k), where k and m represent any positive or negative integer. For example, if (k,m) is taken to be (1,3), the solutions of (x,y,z) are (8,5,-3). For (-3,-2) they are (-37,-34,41), and so on.

Does this help?

Answer 5

$6x+15y+10z=53\Rightarrow \begin{cases}x\equiv3\pmod5\Rightarrow x=5s+3\\y\equiv1\pmod2\Rightarrow y=2t+1\end{cases}$

where $s,t\in\mathbb Z$. This gives $$\boxed{(x,y,z)=(5s+3,2t+1,-3(s+t)+2)}$$