Express X in terms of x and z, Y in terms of y and z if












0












$begingroup$


Given

$$X+Y=x+y$$
$$X-Y=z(x-y)$$
where z is a positive constant.



Attempts so far



Its easy when $z=1$. I have tried shuffling the variables around trying to eliminate either $x$ or $y$ but I'm unsuccessful.I also tried to assume that $X=ax+bz+czx$,$Y=py+qz+rzy$ and tried to solve for $a,b,c$ but I couldn't.






algebra-precalculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I see that its easy when z=1. I have tried shuffling the variables around trying to eliminate either x or y but I'm unsuccessful.
    $endgroup$
    – user276750
    Nov 29 '15 at 9:01






  • 1




    $begingroup$
    I also tried to assume that X=ax+bz+czx and tried to solve for a,b,c but I couldn't
    $endgroup$
    – user276750
    Nov 29 '15 at 9:02












  • $begingroup$
    add your last comments to the question under "My attempt so far:"!
    $endgroup$
    – Jesse P Francis
    Nov 29 '15 at 9:03


















0












$begingroup$


Given

$$X+Y=x+y$$
$$X-Y=z(x-y)$$
where z is a positive constant.



Attempts so far



Its easy when $z=1$. I have tried shuffling the variables around trying to eliminate either $x$ or $y$ but I'm unsuccessful.I also tried to assume that $X=ax+bz+czx$,$Y=py+qz+rzy$ and tried to solve for $a,b,c$ but I couldn't.






algebra-precalculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I see that its easy when z=1. I have tried shuffling the variables around trying to eliminate either x or y but I'm unsuccessful.
    $endgroup$
    – user276750
    Nov 29 '15 at 9:01






  • 1




    $begingroup$
    I also tried to assume that X=ax+bz+czx and tried to solve for a,b,c but I couldn't
    $endgroup$
    – user276750
    Nov 29 '15 at 9:02












  • $begingroup$
    add your last comments to the question under "My attempt so far:"!
    $endgroup$
    – Jesse P Francis
    Nov 29 '15 at 9:03
















0












0








0





$begingroup$


Given

$$X+Y=x+y$$
$$X-Y=z(x-y)$$
where z is a positive constant.



Attempts so far



Its easy when $z=1$. I have tried shuffling the variables around trying to eliminate either $x$ or $y$ but I'm unsuccessful.I also tried to assume that $X=ax+bz+czx$,$Y=py+qz+rzy$ and tried to solve for $a,b,c$ but I couldn't.






algebra-precalculus







share|cite|improve this question











$endgroup$




Given

$$X+Y=x+y$$
$$X-Y=z(x-y)$$
where z is a positive constant.



Attempts so far



Its easy when $z=1$. I have tried shuffling the variables around trying to eliminate either $x$ or $y$ but I'm unsuccessful.I also tried to assume that $X=ax+bz+czx$,$Y=py+qz+rzy$ and tried to solve for $a,b,c$ but I couldn't.






algebra-precalculus




algebra-precalculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 '15 at 22:53









Daniel Robert-Nicoud

20.4k33696




20.4k33696










asked Nov 29 '15 at 8:45









user276750user276750

413




413








  • 1




    $begingroup$
    I see that its easy when z=1. I have tried shuffling the variables around trying to eliminate either x or y but I'm unsuccessful.
    $endgroup$
    – user276750
    Nov 29 '15 at 9:01






  • 1




    $begingroup$
    I also tried to assume that X=ax+bz+czx and tried to solve for a,b,c but I couldn't
    $endgroup$
    – user276750
    Nov 29 '15 at 9:02












  • $begingroup$
    add your last comments to the question under "My attempt so far:"!
    $endgroup$
    – Jesse P Francis
    Nov 29 '15 at 9:03
















  • 1




    $begingroup$
    I see that its easy when z=1. I have tried shuffling the variables around trying to eliminate either x or y but I'm unsuccessful.
    $endgroup$
    – user276750
    Nov 29 '15 at 9:01






  • 1




    $begingroup$
    I also tried to assume that X=ax+bz+czx and tried to solve for a,b,c but I couldn't
    $endgroup$
    – user276750
    Nov 29 '15 at 9:02












  • $begingroup$
    add your last comments to the question under "My attempt so far:"!
    $endgroup$
    – Jesse P Francis
    Nov 29 '15 at 9:03










1




1




$begingroup$
I see that its easy when z=1. I have tried shuffling the variables around trying to eliminate either x or y but I'm unsuccessful.
$endgroup$
– user276750
Nov 29 '15 at 9:01




$begingroup$
I see that its easy when z=1. I have tried shuffling the variables around trying to eliminate either x or y but I'm unsuccessful.
$endgroup$
– user276750
Nov 29 '15 at 9:01




1




1




$begingroup$
I also tried to assume that X=ax+bz+czx and tried to solve for a,b,c but I couldn't
$endgroup$
– user276750
Nov 29 '15 at 9:02






$begingroup$
I also tried to assume that X=ax+bz+czx and tried to solve for a,b,c but I couldn't
$endgroup$
– user276750
Nov 29 '15 at 9:02














$begingroup$
add your last comments to the question under "My attempt so far:"!
$endgroup$
– Jesse P Francis
Nov 29 '15 at 9:03






$begingroup$
add your last comments to the question under "My attempt so far:"!
$endgroup$
– Jesse P Francis
Nov 29 '15 at 9:03












1 Answer
1






active

oldest

votes


















0












$begingroup$

Hint:



the solutions of the system are, for generic $z$:
$$
X=frac{x(1+z)+y(1-z)}{2} qquad
Y=frac{x(1-z)+y(1+z)}{2}
$$
so we can have $X=f(z,x)$ and $Y=f(z,y)$ only for $z=1$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So you're saying unless z=1 we cannot express X as X=f(z,x)?
    $endgroup$
    – user276750
    Nov 29 '15 at 9:27










  • $begingroup$
    Yes, if $x$ and $y$ are independent.
    $endgroup$
    – Emilio Novati
    Nov 29 '15 at 9:52











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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

Hint:



the solutions of the system are, for generic $z$:
$$
X=frac{x(1+z)+y(1-z)}{2} qquad
Y=frac{x(1-z)+y(1+z)}{2}
$$
so we can have $X=f(z,x)$ and $Y=f(z,y)$ only for $z=1$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So you're saying unless z=1 we cannot express X as X=f(z,x)?
    $endgroup$
    – user276750
    Nov 29 '15 at 9:27










  • $begingroup$
    Yes, if $x$ and $y$ are independent.
    $endgroup$
    – Emilio Novati
    Nov 29 '15 at 9:52
















0












$begingroup$

Hint:



the solutions of the system are, for generic $z$:
$$
X=frac{x(1+z)+y(1-z)}{2} qquad
Y=frac{x(1-z)+y(1+z)}{2}
$$
so we can have $X=f(z,x)$ and $Y=f(z,y)$ only for $z=1$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So you're saying unless z=1 we cannot express X as X=f(z,x)?
    $endgroup$
    – user276750
    Nov 29 '15 at 9:27










  • $begingroup$
    Yes, if $x$ and $y$ are independent.
    $endgroup$
    – Emilio Novati
    Nov 29 '15 at 9:52














0












0








0





$begingroup$

Hint:



the solutions of the system are, for generic $z$:
$$
X=frac{x(1+z)+y(1-z)}{2} qquad
Y=frac{x(1-z)+y(1+z)}{2}
$$
so we can have $X=f(z,x)$ and $Y=f(z,y)$ only for $z=1$.






share|cite|improve this answer











$endgroup$



Hint:



the solutions of the system are, for generic $z$:
$$
X=frac{x(1+z)+y(1-z)}{2} qquad
Y=frac{x(1-z)+y(1+z)}{2}
$$
so we can have $X=f(z,x)$ and $Y=f(z,y)$ only for $z=1$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 29 '15 at 9:19

























answered Nov 29 '15 at 9:09









Emilio NovatiEmilio Novati

52k43474




52k43474












  • $begingroup$
    So you're saying unless z=1 we cannot express X as X=f(z,x)?
    $endgroup$
    – user276750
    Nov 29 '15 at 9:27










  • $begingroup$
    Yes, if $x$ and $y$ are independent.
    $endgroup$
    – Emilio Novati
    Nov 29 '15 at 9:52


















  • $begingroup$
    So you're saying unless z=1 we cannot express X as X=f(z,x)?
    $endgroup$
    – user276750
    Nov 29 '15 at 9:27










  • $begingroup$
    Yes, if $x$ and $y$ are independent.
    $endgroup$
    – Emilio Novati
    Nov 29 '15 at 9:52
















$begingroup$
So you're saying unless z=1 we cannot express X as X=f(z,x)?
$endgroup$
– user276750
Nov 29 '15 at 9:27




$begingroup$
So you're saying unless z=1 we cannot express X as X=f(z,x)?
$endgroup$
– user276750
Nov 29 '15 at 9:27












$begingroup$
Yes, if $x$ and $y$ are independent.
$endgroup$
– Emilio Novati
Nov 29 '15 at 9:52




$begingroup$
Yes, if $x$ and $y$ are independent.
$endgroup$
– Emilio Novati
Nov 29 '15 at 9:52


















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