Question

Solve for x, y and z?

`(5xy)/(x+y)=6`

`(4xz)/(x+z)=3`

`(3yz)/(y+z)=2`

Answer 1

`x=3`, `y=2`, `z=1`

Explanation:

Given:

`{ ((5xy)/(x+y) = 6), ((4xz)/(x+z) = 3), ((3yz)/(y+z) = 2) :}`

Multiplying both sides of the first equation by `(x+y)/(xy)`, the second equation by `2(x+z)/(xz)` and the third by `3(y+z)/(yz)` we get:

`{ (5 = 6(1/x)+6(1/y)), (8 = 6(1/x)+6(1/z)), (9 = 6(1/y)+6(1/z)) :}`

Replacing the last two equations with the result of subtracting the third equation from the second we get:

`{ (5 = 6(1/x)+6(1/y)), (-1 = 6(1/x)-6(1/y)) :}`

Then adding these two equations, we get:

`4 = 12(1/x)`

Hence `x=3`

Then:

`6(1/y) = 5-6(1/x) = 5-2 = 3`

Hence `y=2`

Then:

`6(1/z) = 9-6(1/y) = 9-3 = 6`

Hence `z=1`

Answer 2

See below.

Explanation:

Making `y = lambda x` and `z = mu x`

`(5xy)/(x+y)=6 rArr (lambda x)/(1+lambda) = 6/5`

`(4xz)/(x+z)=3 rArr (mu x)/(1+mu) = 3/4`

`(3yz)/(y+z)=2 rArr (mu lambda x)/(mu+lambda) = 2/3`

and eliminating `x`

`{(mu (1 + lambda)/(mu + lambda) = 5/9),(lambda (1 + mu)/(mu + lambda) = 8/9):}`

and solving for `mu, lambda` we obtain

`mu = 1/3` and `lambda = 2/3` and then

`x = 3`
`y = 2`
`z=1`

Answer 3

`(x,y,z)=(3,2,1)`.

Explanation:

We have, `(5xy)/(x+y)=6`.

`:. (x+y)/(xy)=5/6, or, x/(xy)+y/(xy)=5/6, i.e.,`

`1/y+1/x=5/6................<1>`.

Similarly, `(4xz)/(x+z)=3 rArr 1/z+1/x=4/3=8/6......<2>`, and,

`(3yz)/(y+z)=2 rArr 1/z+1/y=3/2=9/6.............<3>`.

`<<1>>+<<2>>+<<3>>rArr 2(1/x+1/y+1/z)=(5+8+9)/6=22/6`

`rArr 1/x+1/y+1/z=11/6............<4>`.

Then, `<4> - <1> rArr 1/z=(11-5)/6=1 rArr z=1`,

`<4> - <2> rArr 1/y=3/6=1/2 rArr y=2," and, finally, "`

`<4> - <3> rArr 1/x=(11-9)/6=1/3 rArr x=3`.

Altogether, `(x,y,z)=(3,2,1)`.

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