Find $x$ and $y$: $\sqrt {x+1}+\sqrt {y+1}=3$, $xy-x-y+15=0$

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SUMMARY

The discussion focuses on solving the system of equations defined by $\sqrt{x+1}+\sqrt{y+1}=3$ and $xy-x-y+15=0$. The first equation can be manipulated to express $y$ in terms of $x$, leading to a quadratic equation when substituted into the second equation. The solutions for $x$ and $y$ are derived as $(x, y) = (4, -1)$ and $(x, y) = (-1, 4)$, confirming the correctness of the answers provided by participants.

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Albert1
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$\sqrt {x+1}+\sqrt {y+1}=3$

$xy-x-y+15=0$

find :$x,y$
 
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Albert said:
$\sqrt {x+1}+\sqrt {y+1}=3$

$xy-x-y+15=0$

find :$x,y$

I copy the 1st equation
$\sqrt{x+1} + \sqrt{y+1} = 3... (1) $
Though x,y are integer is not specified both $\sqrt{x+1}$ and $\sqrt{y+1}$ have to be integers.
The second equation given

$xy – x – y + 1 + 14 = 0$
Or
$(x-1)(y-1) = - 14... (2)$

One one of (x-1) and (y-1) is –ve
As it is symmetric in xand y we can assume x to be so
So x – 1 = - 14 or -2 or – 1
From the 1st equation x + 1 >= 0 so x >= -1
So x = -1 , 0
X = 0 => y- 1 = 15 so $\sqrt{y+1} = 4$ does not satisfy (1)
X= -1 => y = 8 and it saitsfiles (1)
So x = -1 , y = 8 or x = 8, y = -1
 
Last edited:
kaliprasad said:
I copy the 1st equation
$\sqrt{x+1} + \sqrt{y+1} = 3... (1) $
Though x,y are integer is not specified both $\sqrt{x+1}$ and $\sqrt{y+1}$ have to be integers.
The second equation given

$xy – x – y + 1 + 14 = 0$
Or
$(x-1)(y-1) = - 14... (2)$

One one of (x-1) and (y-1) is –ve
As it is symmetric in xand y we can assume x to be so
So x – 1 = - 14 or -2 or – 1
From the 1st equation x + 1 >= 0 so x >= -1
So x = -1 , 0
X = 0 => y- 1 = 15 so $\sqrt{y+1} = 4$ does not satisfy (1)
X= -1 => y = 8 and it saitsfiles (1)
So x = -1 , y = 8 or x = 8, y = -1
your answer is correct:)
 

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