Find all x,y,x&gt;0 such that3(x+1x)=4(y+1y)=5(z+1z)xy+yz+zx=1The only solution should be x=13,y=12,z=1. There is a way to do it with x=tan⁡α.

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Find all x,y,x&amp;gt;0 such that3(x+1x)=4(y+1y)=5(z+1z)xy+yz+zx=1The only solution should be x=13,y=12,z=1. There is a way to do it with x=tan⁡α.

Yaritza Phillips · Accepted Answer

Rewrite the second equation&amp;nbsp;xy+yz+zx=1&amp;nbsp;as&amp;nbsp;z=1−xyx+y&amp;nbsp;and evaluatez+1z=(x2+1)(y2+1)(x+y)(1−xy)Substitute above&amp;nbsp;z+1z=z2+1z&amp;nbsp;into the first equation3(x2+1)x=4(y2+1)y=5(z2+1)z&amp;nbsp;&amp;nbsp;(1)to get&amp;nbsp;&amp;nbsp;(x2+1)(35x−y2+1(x+y)(1−xy))=0&amp;nbsp;(y2+1)(45y−x2+1(x+y)(1−xy))=0which reduce to&amp;nbsp;3x2y+8xy2+2x−3y=0(2)Note that&amp;nbsp;3×(2)&amp;nbsp;and&amp;nbsp;2×(3)−(2)&amp;nbsp;simplify the equations to&amp;nbsp;3x2y+8xy2+2x−3y=0(2)and, furthermore, with&amp;nbsp;3x×(4)−2y×(5)3x2+xy−2y2=(3x−2y)(x+y)=0Substitute the resulting&amp;nbsp;y=32x&amp;nbsp;and&amp;nbsp;y=−x&amp;nbsp;into (1) to obtain the real solutions(x,y,z)=±(13,12,1)

Marcos Boyer · Accepted Answer

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

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

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