eqn

Multiply 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 .

Simon's back!

Ricky (5,3,6) is a solution

Add 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

but there is 1 more solution - (-7,-5,-8)

ya 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