Find all \(\displaystyle{x},{y},{x}{>}{0}\) such that \(\displaystyle{3}{\left({x}+{\frac{{{1}}}{{{x}}}}\right)}={4}{\left({y}+{\frac{{{1}}}{{{y}}}}\right)}={5}{\left({z}+{\frac{{{1}}}{{{z}}}}\right)}\) \(\displaystyle{x}{y}+{y}{z}+{z}{x}={1}\) The

serpebm2r

serpebm2r

Answered question

2022-04-03

Find all x,y,x>0 such that
3(x+1x)=4(y+1y)=5(z+1z)
xy+yz+zx=1
The only solution should be x=13,y=12,z=1. There is a way to do it with x=tanα.

Answer & Explanation

Yaritza Phillips

Yaritza Phillips

Beginner2022-04-04Added 12 answers

Rewrite the second equation xy+yz+zx=1 as z=1xyx+y and evaluate
z+1z=(x2+1)(y2+1)(x+y)(1xy)
Substitute above z+1z=z2+1z into the first equation
3(x2+1)x=4(y2+1)y=5(z2+1)z  (1)
to get

  (x2+1)(35xy2+1(x+y)(1xy))=0 (y2+1)(45yx2+1(x+y)(1xy))=0
which reduce to
 3x2y+8xy2+2x3y=0(2)
Note that 3×(2) and 2×(3)(2) simplify the equations to
 3x2y+8xy2+2x3y=0(2)
and, furthermore, with 3x×(4)2y×(5)
3x2+xy2y2=(3x2y)(x+y)=0
Substitute the resulting y=32x and y=x into (1) to obtain the real solutions
(x,y,z)=±(13,12,1)

Marcos Boyer

Marcos Boyer

Beginner2022-04-05Added 12 answers

Note that xy+yz+zx=1, then we have:
43=x+1xy+1y=x+xy+yz+zxxy+xy+yz+zxy=(x+y)(x+z)x(y+z)(y+x)y=y(x+z)x(z+y)=yz+yxxy+xz

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?