I dunno

[sand pawn]

1. Ask your question and show the work you've done so far. If you've posted a screenshot of a question, specify which part you need help with.
2. Wait patiently for a helper to come along.
3. Once someone helps you, say thank you and close the thread with:

   +close
   

4. Feel free to nominate the person for helper of the week in #helper-nominations
5. Do not ping the mods, unless someone is breaking the rules.
6. If you're happy with the help you got here, and the server overall, you can contribute financially as well:

[tired breach]

I got

$[x+(y^2+1)^1/2][y+(y^2+1)^1/2]= 1+ (x-y)[(y^2+1)1/2 - (x^2+1)1/2]$

[silent herald]

anonymoushuman69

[tired breach]

$[x+(y^2+1)^(1/2)][y+(y^2+1)^(1/2)]= 1+ (x-y)[(y^2+1)^(1/2) - (x^2+1)^(1/2)]$

[silent herald]

anonymoushuman69

[tired breach]

$[x+(y^2+1)^1/2][y+(y^2+1)^1/2]= 1+ (x-y)[(y^2+1)^1/2 - (x^2+1)^1/2]$

[silent herald]

anonymoushuman69

[tired breach]

$[x+(y^2+1)^{1/2}][y+(y^2+1)^{1/2}]= 1+ (x-y)[(y^2+1)^{1/2} - (x^2+1)^{1/2}]$

[silent herald]

anonymoushuman69

[tired breach]

how do you do after

[dense tangle]

@tired breach multiply em and substitute xy in the other equation

[tired breach]

i got this though

[dense tangle]

Ahh that works too

[tired breach]

what then?

[viscid flame]

@tired breach substitute x=ty

[tired breach]

what do u mean ty?

[viscid flame]

"ty"

[tired breach]

thank you?

[viscid flame]

no

instead of x write ty

[tired breach]

oh t*y

[viscid flame]

either way it's obvious that x= something * y

[tired breach]

oh ok