# Inequation problem

1. Feb 25, 2005

### mohlam12

hey eveybody, i have a problem for findin the answer of this, we have to solve for x and y.
$$sqrt(x-1) + sqrt(y-2) = sqrt5$$

thx

Last edited: Feb 25, 2005
2. Feb 25, 2005

### HallsofIvy

Staff Emeritus
Okay, what is your problem? What have you tried? You understand, of course, that you CAN'T generally solve one equation for two unknowns? You can get a quadratic equation for x and y by squaring twice.

3. Feb 25, 2005

### mohlam12

ok, here is what i did but im not sure if it is right.
we have 6>=X>=1 and 7>=y>=2 (because sqrt(x-1)=sqrt5 - sqrt(y-2) and sqrt(y-2)=sqrt5 - sqrt(x-1) )
now we have that x could be {1,2,3,4,5,6} and y could be {2,3,4,5,6,7}
therefore,
if x=1 --> y=7
if x= 2 --> y=8-2sqrt5 (not with the intervalle i gave)
if x=3 --> y/= the intervalle
if x=4 --> y/= the intervalle
if x=5 --> y/= the intervalle
if x=6 --> y=2

finally, the solution would be : (1,7) and (6,2)

is it right? i dun know

4. Feb 25, 2005

### HallsofIvy

Staff Emeritus
If you ask us for help, don't hide part of the problem! You did not say before that x and y had to be positive integers! If that is, in fact, the case, then yes, the simplest way to do this problem is to try each value and find that the only solutions are (1, 7) and (6,2). If x and y can be any real numbers, then there are an infinite number of solutions.

5. Feb 25, 2005

### dextercioby

Okay,so which is the "inequation"...?(v.title)

Daniel.

6. Feb 25, 2005

### mohlam12

im sorry, well im an exchange student so i dun know how to call that :) thx anyways!

7. Feb 26, 2005

### dextercioby

Call it what it is:AN EQUATION...

Daniel.