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How to solve a system of equations x+y+z=7/2 , xyz=1 , 1/x +1/y + 1/z=7/2
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You should rewrite the third equation such that:
`1/x + 1/y + 1/z = (xy+yz+xz)/(xyz)`
Since the problem provides the value of xyz, you need to substitute 1 for xyz such that:
Since `1/x + 1/y + 1/z = 7/2 => xy+yz+xz = 7/2`
Notice that you may solve this system of equations using Vieta's relations, considering x,y,z as the roots of an equation `t^3 + (x+y+z)t^2 + (xy+yz+xz)t + xyz = 0` such that:
`t^3 + (7/2)t^2 + (7/2)t +1 = 0 `
`2t^3 + 7t^2 + 7t +2 = 0`
Notice that t=-1 is a root of this equation such that:
`(t+1)(at^2+bt+c) = 0`
You need to find the polynomial `at^2+bt+c` such that:
`2t^3 + 7t^2 + 7t + 2 = (t+1)(at^2+bt+c)`
`2t^3 + 7t^2 + 7t + 2 = at^3 + bt^2 + ct + at^2 + bt + c`
`2t^3 + 7t^2 + 7t + 2 = at^3 + t^2(a+b) + t(c+b) + c`
Equating coefficients of like powers yields:
`a = 2`
`a+b = 7 => b = 7-2 => b = 5`
`c = 2`
Hence, evaluating the coefficients a,b,c yields:
`at^2+bt+c = 2t^2+5t+2`
You need to use quadratic formula to find the zeroes of `2t^2+5t+2 = 0` such that:
`t_(1,2) = (-5+-sqrt(25 - 16))/4 => t_(1,2) = (-5+-sqrt9)/4`
`t_(1,2) = (-5+-3)/4 => t_1 = -1/2 ; t_2 = -2`
Notice that x, y, z represents the solutions to equation `2t^3 + 7t^2 + 7t + 2 = 0 .`
Hence, evaluating the solutions to the given system of equations yields `x = -1 , y = -1/2 , z = -2` .
Posted by sciencesolve on October 2, 2012 at 4:56 PM (Answer #1)
Here must be 5 answers...
Posted by anuket on October 2, 2012 at 6:06 PM (Answer #3)
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