Solving sqrt(x-1) + sqrt(y-2) = sqrt(5) for x and y

[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

 

[HallsofIvy]

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.

 

[mohlam12]

ok, here is what i did but I am 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

 

[HallsofIvy]

ok, here is what i did but I am 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

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.

 

[dextercioby]

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

Daniel.

 

[mohlam12]

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

 

[dextercioby]

Call it what it is:AN EQUATION...

Daniel.