MHB Solving Diophantine Equation: 97x+35y=13

[Petrus]

Hello MHB,
I got problem with this http://staff.www.ltu.se/~larserik/applmath/chap10/part4.html (It's 'Exempel 1' and yes it's on Swedish), I follow my book method and get,
$$x=169-35K, \ y= -468-97K$$
I get the 'homogen soloution' that $$x=-35k, \ y=97k$$
'Solve equation $$97x+35y=13$$'

Regards,

 

[I like Serena]

Hello MHB,
I got problem with this http://staff.www.ltu.se/~larserik/applmath/chap10/part4.html (It's 'Exemple 1' and yes it's on Swedish), I follow my book method and get,
$$x=169-35K, \ y= -468-97K$$
I get the 'homogen soloution' that $$x=-35k, \ y=97k$$
'Solve equation $$97x+35y=13$$'

Regards,

Erm... what is your question?

 

[Petrus]

Solve equation $$97x+35y=13$$
ehmm. I am suposed to find one soultion with gcd, eucliders algoritmen then find all other soloution, cause it got infinity soloution, for x and y. Do it make sense now?

Regards,

 

[I like Serena]

Solve equation $$97x+35y=13$$
ehmm. I am suposed to find one soultion with gcd, eucliders algoritmen then find all other soloution, cause it got infinity soloution, for x and y. Do it make sense now?

Regards,

Ah, okay.
So do you know how to apply Euclid's algorithm?
(Btw, isn't this university-level stuff instead of pre-algebra?)

 

[MarkFL]

...
(Btw, isn't this university-level stuff instead of pre-algebra?)

Good call...moved to the Number Theory sub-forum.

 

[Petrus]

Ah, okay.
So do you know how to apply Euclid's algorithm?
(Btw, isn't this university-level stuff instead of pre-algebra?)

Yes, I get x=169 and y=-468

 

[I like Serena]

Yes, I get x=169 and y=-468

That looks about right.
Wolfram says x=29, y=-80, which is effectively the same answer.
(Since you can add a multiple of 35 to x if you also subtract the same multiple of 97 from y.)
So you're all good?

 

[Petrus]

That looks about right.
Wolfram says x=29, y=-80, which is effectively the same answer.
(Since you can add a multiple of 35 to x if you also subtract the same multiple of 97 from y.)
So you're all good?

so this answer is correct?
$$x=169-35K, \ y= -468-97K$$
cause that website says the answer shall be
http://staff.www.ltu.se/~larserik/applmath/chap10/eq409.gif

 

[Opalg]

so this answer is correct?
$$x=169-35K, \ y= -468-97K$$
cause that website says the answer shall be
http://staff.www.ltu.se/~larserik/applmath/chap10/eq409.gif

The website answer is correct. If you add a multiple of 35 to $x$ then you have to subtract that multiple of 97 from $y$.

 

[Petrus]

The website answer is correct. If you add a multiple of 35 to $x$ then you have to subtract that multiple of 97 from $y$.

I am kinda confused cause I follow the method from my book.
* find a soloution $$x=x_0 , y=t_0$$ (particulate solution) that is what we find we euclidmens algorithmen.
*Find all soloution to homogeneous equation $$ax'+by'=0$$ and that is the one i do it wrong then...

 

[I like Serena]

*Find all soloution to homogeneous equation $$ax'+by'=0$$ and that is the one i do it wrong then...

The solution is $x'=kb,\ y'=-ka$, for integers k (note the minus sign).
Substitute to verify...

 

[Petrus]

The solution is $x'=kb,\ y'=-ka$, for integers k (note the minus sign).
Substitute to verify...

that is the part I strugle, my book say $$x'=-kb, \ y'= ka$$

 

[I like Serena]

that is the part I strugle, my book say $$x'=-kb, \ y'= ka$$

That works as well.
Can you substitute that solution in the equation?

 

[Petrus]

That works as well.
Can you substitute that solution in the equation?

$$x' = -35k , \ y'= 97k$$

 

[I like Serena]

$$x' = -35k , \ y'= 97k$$

Yes.
Can you substitute that in the equation $ax′+by′=0$?

 

[Petrus]

Yes.
Can you substitute that in the equation $ax′+by′=0$?

$$35(-35k)+97*97k=0$$ ?

 

[I like Serena]

$$35(-35k)+97*97k=0$$ ?

I kind of lost track of the equation. :)
But the equation was $97x'+35y'=0$.

Can you redo the substitution with this equation?

 

[Petrus]

I kind of lost track of the equation. :)
But the equation was $97x'+35y'=0$.

Can you redo the substitution with this equation?

$$97(-35k)+35*97k$$
my book says $$x=x_0+x' , y=y_0+y'$$
and our $$x_0=169$$ and $$y_0=-468$$

 

[I like Serena]

$$97(-35k)+35*97k$$

Do you see that this is zero?

my book says $$x=x_0+x' , y=y_0+y'$$
and our $$x_0=169$$ and $$y_0=-468$$

Yes?

 

[Petrus]

Do you see that this is zero?

Yes,

Yes?

That means the answer is $$x=169-35K, \ y= -468-97K$$

 

[I like Serena]

$$97(-35k)+35*97k$$
my book says $$x=x_0+x' , y=y_0+y'$$
and our $$x_0=169$$ and $$y_0=-468$$

That means the answer is $$x=169-35K, \ y= -468-97K$$

Erm.

You have $x'=-35k,\ y'=97k$ and $$x_0=169$$, $$y_0=-468$$

If I substitute that into $$x=x_0+x' , y=y_0+y'$$, I get
$$\qquad x=169+(-35k) , y=-468+(97k)$$.

See the difference?

 

[Petrus]

Erm.

You have $x'=-35k,\ y'=97k$ and $$x_0=169$$, $$y_0=-468$$

If I substitute that into $$x=x_0+x' , y=y_0+y'$$, I get
$$\qquad x=169+(-35k) , y=-468+(97k)$$.

See the difference?

Finally, I did find my mistake...! I somehow did have $$-97k$$ in my paper... and confused myself..! Can you confirm that this is correct answer?

 

[I like Serena]

Finally, I did find my mistake...! I somehow did have $$-97k$$ in my paper... and confused myself..! Can you confirm that this is correct answer?

Erm, yes. It is the correct answer.
It's what the website solution said...

 

[Petrus]

Erm, yes. It is the correct answer.
It's what the website solution said...

Hello I like Serena,
I am sorry I just got confused, Thanks for taking your time and now I understand diophantine equation a lot better! Hey look at the bright side, you gained some 'thanks' ;)

Regards,

 

[I like Serena]

Yep. Thanks! ;)