eqn

solve for x,y,z

xy+x+y=23

xz+x+z=41

yz+y+z=27

6 Answers

62
Lokesh Verma ·

this 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..

1
Ricky ·

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 .

341
Hari Shankar ·

Simon's back!

Ricky (5,3,6) is a solution

1
ABHI ·

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

1
avik ·

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

1
ABHI ·

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

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