A pure equation

Algebra Level 2

x + y + x y = 19 , y + z + y z = 29 , z + x + z x = 23 x + y + xy = 19, \quad y + z + yz = 29, \quad z + x + zx = 23 If x , y , x, y, and z z are positive real numbers satisfying the system above, then find x , y , x, y, and z . z.

7,11,13 1,2,3 can't be determined 3, 4, 5

Tanay Gaurav by tanay gaurav , 24, India

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2 solutions

Tanay Gaurav
Sep 7, 2015

Here x+y+xy=19 or, (x+y)^2=19+xy and, x+y=19-xy so, (19-xy)^2=19+xy Compute xy, yz and zx from these equations and use the above equations to find x, y, z.

if any problem arises, feel free to contact me... -S.B Tanay Gaurav

1 Helpful 0 Interesting 0 Brilliant 0 Confused

There is a much easier way with these problems.

Add 1 1 to both sides and factorise.

We get: ( 1 + x ) ( 1 + y ) = 20 , ( 1 + y ) ( 1 + z ) = 30 and ( 1 + x ) ( 1 + z ) = 24 (1+x)(1+y)=20, \; (1+y)(1+z)=30 \; \text{and} \; (1+x)(1+z)=24

And it can easily be deduced that x = 3 , y = 4 and z = 5 x=3, \: y=4 \; \text{and} \; z=5 .

Isaac Buckley - 5 years, 9 months ago

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Very nice...

tanay gaurav - 5 years, 6 months ago
Vignesh Rao
Dec 23, 2015

@Andrew Ellinor @tanay gaurav It is also possible that x = − 5 ; y = − 6 ; z = − 7 x=-5 ; \ y=-6 ; \ z = -7 . Do correct me if I am wrong in saying so.

0 Helpful 0 Interesting 0 Brilliant 1 Confused

Indeed. I have updated the problem statement accordingly.

Andrew Ellinor - 5 years, 5 months ago

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