341Simon's back!
Ricky (5,3,6) is a solution
6 Answers
62this one is a very old olympiad quest..
Hint try to think of what is to be a added to make xy+y+x as a product of 2 linear terms..
1Multiply the first eqn. by " z " and the second by " y " .
x y z + x z + y z = 23 z
x y z + x y + y z = 41 y
Subtracting , x = -1 8 or , y = z .
Again , doing the same process , y = - 4 or x = y .
And , z = - 14 or , x = z .
For , x = y or , x = z or z = y , no solution will be there .
For , x = - 18 , y = - 4 , and z = - 14 , again no solution will exist .
1Add 1 to all the 3 equations :
xy + x + y + 1 = 24
→(x+1)(y+1) = 24
Similarly
(x+1)(z+1) = 42
(y+1)(z+1) = 28
So multiplying and taking the square root of that,
(x+1)(y+1)(z+1)=168
Now dividing this by each of the three equations,
So, z+1 = 7 ...z = 6
x+1 = 6....x = 5
y+1 = 4....y = 3
So solns - (5,3,6)
Ds used to come in ur class 9 exams
1ya avik ...sorry i missed dat part while taking square root
(x+1)(y+1)(z+1) = -168
n do d same process....
u will get
x = -7
y= - 5
z = -8
sorry for overlooking d negative terms