find all the positive integer solution to the diophantine equation?

Question

12x+57y=423



help please

Answer 1

To solve this, all you have to do is use the Euclidean algorithm on 12 and 57.

57 = 4(12) + 9
12 = 1(9) + 3
9 = 3(3)

This tells us that the greatest common factor of 12 and 57 is 3. You have to divide everything of the original equation by 3.

12x + 57y = 423, so dividing by 3, we have

4x + 19y = 141

This gives us a new (but equivalent) equation to solve for.

Now, we use the Euclidean algorithm on 19 and 4, the coefficients of x and y.

19 = 4(4) + 3
4 = 1(3) + 1

Now, we work backwards from the Euclidean algorithm.

1 = 4 + (-1)(3)

But, as per the first equation, 3 = 19 + (-4)(4), so substituting this,

1 = 4 + (-1)(19 + (-4)(4))

Distributing the (-1),

1 = 4 + (-1)(19) + 4(4)

And now, grouping the like terms (Note: 4 is the same as 1(4)):

1 = 5(4) + (-1)(19)

Now, we want to multiply both sides of this by 141. Our goal is to retain the 4 and the 19, so when multiplying both sides by 141, we're going to do so with the 5 and the -1.

141 = (705)(4) + (-141)(19)

Compare this to the equation we're solving for:

4x + 19y = 141

Looking back at the other equation, let's rewrite it slightly.

4(705) + 19(-141) = 141

Therefore, one solution of the Linear Diophantine equation is:
x = 705, y = -141

However, since we want positive integer solutions, remember that as soon as we get one solution for the linear diophantine equation, we can get others using the following formula:

For the equation ax + by = c, and for solutions a1 and b1

x = a1 + bk
y = b1 - ak

Therefore, the general solution of the linear diophantine equation is:

x = 705 + 19k
y = -141 - 4k

We want positive integer solutions, so x > 0 and y > 0. This means

705 + 19k > 0
19k > -705
k > -705/19, and -705/19 is approximately equal to -37.105

-141 - 4k > 0
-4k > 141
k < -141/4, and -141/4 is approximately equal to -35.25

Therefore, all the positive integer solutions occur at the integers between -37.105 and -35.25; these are

k = -37, -36

Remember that

x = 705 + 19k
y = -141 - 4k

So we plug in each of k = -37, -36 obtaining the following solutions:

k = -37: x = 2, y = 7
k = -36: x = 21, y = 3

As you can see, x and y are both positive.

Answer 2

First, divide both sides of the equation by 3 and simplify it to
4x + 19y = 141.
Then 4x = 141 - 19y Notice that if y>=8 then 19 y >= 152 and so
4x < 0. Since x and y must both be positive, then 1 <= y <= 7.

x = (141 - 19y) / 4 = 141 / 4 - 19y / 4 = (35 + 1/4) - (5y - y / 4)
= (35 - 5y) + (1 + y) / 4

Now 35 - 5y is an integer when y is an integer. When is (1 + y) / 4 an integer if 1<= y <= 7? It is easy to see this is true only if y = 3 or y = 7.

If y = 3 then x = (35 - (5)(3)) + (1 + 3) / 4 = 35 - 15 + 4/4 = 20 + 1
= 21
If y = 7 then x = (35 - (5)(7)) + (1 + 7) /4 = 35 - 35 + 8 / 4 = 0 + 2
= 2

x = 21 and y = 3: (12)(21) + (57)(3) = 252 + 171 = 423
x = 2 and y = 7: (12)(2) + (57)(7) = 24 + 399 = 423

Answer 3

thrilling. in the past i glance on the links some observations: Set y=p/q lowest phrases, x=r/s lowest phrases, then p^2s^3 = r^3q^2 - 2q^2s^3 provides s^3 | q^2, q^2 | s^3 and s^3 = q^2 = t^6, giving a diophantine equation (*)............. p^2 = r^3 - 2t^6. For which you have got here across p=383, r=129, t= 10. The values for the denominators are many times q=t^3 and s=t^2 ; t=a million provides p=5,r=3. At this factor i do no longer understand some million

Answer 4

It's really not that hard since y has to be less than 7.

This also 4x + 19 y =141. y has to be odd. Look at 1 3 5 and 7.

1 and 5 won't work because 141 - 19 y is not a multiple of 4. It remains 3 and 7 and the x's are 21 and 2. That's it (2,31) and (2,7)

Answer 5

I don't feel like doing the problem but I'd recommend prime factoring all the coefficients.

4+2+3 = 9 which indicates that 423 is divisible by 3

57 is also divisible by 3, as is 12

I cheated and put this into my TI-83, set the table so that it step by integers, and got:

x,y =
2,7
21,3

Answer 6

12x+57y=423
12(2)+57(7)
24 + 399 = 423

x=2 y= 7

Answer 7

stop cheating. Look in your book and maybe try to understand how to do homework.