# Given 3(x+y)^1/2+2(8-x)^1/2+(6-y)^1/2=14 what are x, y real?

sciencesolve | Certified Educator

You should consider the following inequality such that:

`(3sqrt(x + y) + 2sqrt(8 - x) + sqrt(6 - y))^2 =< (3^2 + 2^2 + 1^2)(x + y + 8 - x + 6 - y)`

Reducing like terms yields:

`(3sqrt(x + y) + 2sqrt(8 - x) + sqrt(6 - y))^2 =< 14*14`

`(3sqrt(x + y) + 2sqrt(8 - x) + sqrt(6 - y))^2 =< 14^2`

The problem provides the information `(3sqrt(x + y) + 2sqrt(8 - x) + sqrt(6 - y)) = 14 => (3sqrt(x + y) + 2sqrt(8 - x) + sqrt(6 - y))^2 = 14^2`

Hence, the equality holds for the followings:

`(x + y)/9 = (8 - x)/4 = (6 - y)/1`

Using the properties of fractions yields:

`(x + y)/9 = (8 - x)/4 = (6 - y)/1 = (x + y + 8 - x + 6 - y)/(9+4+1)`

`(x + y)/9 = (8 - x)/4 = (6 - y)/1 = 14/14`

`(x + y)/9 = (8 - x)/4 = (6 - y)/1 = 1 => {((x + y)/9 = 1),((8 - x)/4 = 1),((6 - y)/1 = 1):}`

`x + y = 9`

`8 - x = 4 => -x = 4 - 8 => -x = -4 => x = 4`

`6 - y = 1 => y = 6 - 1 => y = 5`

You should substitute 4 for x and 5 for y in the first equation to check it such that:

`4 + 5 = 9 => 9 = 9`

Hence, evaluating the values for x and y for the equation to hold yields `x = 4, y = 5` .