Katie from New Zealand

[katiegrey]

Hi Everyone

I've got a question. I'm first year physics student and I can't work this one out.

z = x*​y+​(x-​1)*​(9*​y-​1)

I know what Z is.. is there a way to get x or y?
Katie xx

 

[katiegrey]

Actually..
it's

z = x*​y+​(x+1)*​(9*​y-​1)

here's an example. Normally we only get the z data from our experiment but we need a way to calculate the x and y.


18587681 = 2717 * 6157 + 2718 * 684
18587681 = 6156 * 302 + 6157 * 2717

I hope this helps explain my dilemma. Me and the girls would really appreciate any help. xx

 

[Deleted member 4993]

Hi Everyone

I've got a question. I'm first year physics student and I can't work this one out.

z = x*​y+​(x-​1)*​(9*​y-​1)

I know what Z is.. is there a way to get x or y?
Katie xx

In the absence of any other information, you cannot solve for 2 unknowns (x &y) from 1 equation.

 

[Bob Brown MSEE]

I know what Z is.. is there a way to get x or y?
Katie xx

Are you desiring all INTEGER x and y, when Z = 18587681 ?

 

[katiegrey]

Are you desiring all INTEGER x and y, when Z = 18587681 ?

Yes if possible. If there is a way to do that. Oh.. also.. see my other reply.Thanks for taking interest.

 

[katiegrey]

In the absence of any other information, you cannot solve for 2 unknowns (x &y) from 1 equation.

OK.. would this count as a second equation?

543227 = 464 * 1054 + 463 * 117


4889042 = 1053 * 464 + 1054 * 4175

Thanks so much for your interest.

 

[katiegrey]

OK this is an update. Z is the data we receive. We know that
z1 = x*​y+​(x-​1)*​(9*​y-​1)
z2 = x*​y+​(x-​1)*​(9*​y-​1)

z1 = 543227 = 464 * 1054 + 463 * 117
AND
z1 = 543227 = 118 * 464 + 117 * 4175
AND
z2 = 4889042 = 4175 * 1054 + 4176 * 117
AND
z2 = 4889042 = 1053 * 464 + 1054 * 4175

Where z2 = (9 * z1) + 1

I hope this extra data somehow helps.
Katie

 

[mmm4444bot]

this is an update

We know that

z1 = x*​y+​(x-​1)*​(9*​y-​1)

z2 = x*​y+​(x-​1)*​(9*​y-​1)

z1 = 543227

z2 = 4889042

I'm thinking that you also need subscripted variables for x and y because the expressions highlighted in red are equal (i.e., z1=z2). Like this:

z₁ = x₁*​y₁+​(x₁-​1)*​(9*​y₁-​1)


z₂ = x₂*​y₂+​(x₂-​1)*​(9*​y₂-​1)


Or, for ease of typing, you could use upper- and lower-case symbols, to discuss a specific pair of solutions {xyz,XYZ} to the original equation. Like this:

z = x​y + ​(x - ​1)​(9​y - ​1)


Z = X​Y + ​(X - ​1)​(9​Y - ​1)


z = 543227


Z = 4889042


Earlier, you posted:

Actually..

it's

z = x*​y+​(x+1)*​(9*​y-​1)

So, I had thought that you changed the binomial x-1 in your original equation to x+1 (highlighted in red above). Looking at your most recent post, however, it appears that you've changed it back to x-1.

Please confirm the equation.

Cheers :cool:

 

[katiegrey]

I'm thinking that you also need subscripted variables for x and y because the expressions highlighted in red are equal (i.e., z1=z2). Like this:

z₁ = x₁*​y₁+​(x₁-​1)*​(9*​y₁-​1)


z₂ = x₂*​y₂+​(x₂-​1)*​(9*​y₂-​1)


Or, for ease of typing, you could use upper- and lower-case symbols, to discuss a specific pair of solutions {xyz,XYZ} to the original equation. Like this:

z = x​y + ​(x - ​1)​(9​y - ​1)


Z = X​Y + ​(X - ​1)​(9​Y - ​1)


z = 543227


Z = 4889042


Earlier, you posted:



I had thought that you changed the binomial (highlighted in red) in your original equation. Looking at your most recent post, it appears that you've changed it back to x-1.

Please confirm the equation.

Cheers :cool:

Thanks so much.

Yes the proper equation is : z = x*​y+​(x+1)*​(9*​y-​1) (I can refer this via the sample data)

The z1 and z2 values are correct as well. (In the example data I supplied)


I really appreciate you sharing your intelligence with me on this one.
Katie

 

[Quaid]

Are you desiring all INTEGER x and y, when Z = 18587681 ?

Yes if possible. If there is a way to do that.

One way would be to program a computer. Here's a quick one, in justBASIC:

[start]
  input "Enter z: "; z
  for x = 0 to z
    y = (z+x+1)/(10*x+9)
    if y - int(y) = 0 then print "(";x;", ";y;")"
  next x
  print "Done"
[end]

Here are the printouts.

Enter z: 18587681
(0, 2065298)
(301, 6157)
(6156, 302)
(2065297, 1)
Done


Enter z: 543227
(1, 28591)
(3, 13929)
(13928, 4)
(28590, 2)
Done


Enter z: 4889042
(0, 543227)
(463, 1054)
(1053, 464)
(543226, 1)
Done


Interesting pattern. If you know an x in one solution, add 1 to get a y in another solution. If you know a y in one solution, then subtract 1 to get an x in another solution.

:cool:

 

[Info]

Hello
If you have one individual system of two equations as:
543227=x*y+(x+1)*(9y-1)
4889042=x*y+(x+1)*(9y-1)
Then i should say that your problem is impossible to have any answer. Because as we know, two quantities equal to another quantity are equal to each other.
but 543227 is not equal to 4889042 so the problem is impossible to have any answer.

 

[Quaid]

If you have These two equations as:

543227=x*y+(x+1)*(9y-1)

4889042=x*y+(x+1)*(9y-1)

Hi Info:

I think Katie is discussing sets of (x_n, y_n, z_n) solutions to the equation

z_n = x_n*​y_n + ​(x_n​ + 1)*​(9​y_n - ​1)


So, she ought to have written something like

543227 = x*y + (x + 1)*(9y - 1)

4889042 = X*Y + (X + 1)*(9Y - 1)

for solutions (x, y) and (X, Y)

:cool:

 

[katiegrey]

Hello
If you have one individual system of two equations as:
543227=x*y+(x+1)*(9y-1)
4889042=x*y+(x+1)*(9y-1)
Then i should say that your problem is impossible to have any answer. Because as we know, two quantities equal to another quantity are equal to each other.
but 543227 is not equal to 4889042 so the problem is impossible to have any answer.

Well I know that (543227 * 9) + 1 = 4889042

 

[katiegrey]

Hi Info:

I think Katie is discussing sets of (x_n, y_n, z_n) solutions to the equation

z_n = x_n*​y_n + ​(x_n​ + 1)*​(9​y_n - ​1)


So, she ought to have written something like

543227 = x*y + (x + 1)*(9y - 1)

4889042 = X*Y + (X + 1)*(9Y - 1)

for solutions (x, y) and (X, Y)

:cool:

Yes!!!! Quaid.. I think you understand my problem the best

X = 9x + 1
Thanks Quaid... will this help you?