Question

How to solve this system of equations?

Real numbers x and y satisfy the system of equations
`x+y+x/y=10`
`x/y*(x+y)=20`
Determine the sum of all possible values of the expression x+y.

Answer 1

`x=(20(6+sqrt5))/31=5.314` and
`y=(35+11sqrt5)/31=1.922`
or
`x=(20(6-sqrt5))/31=2.428` and
`y=(35-11sqrt5)/31=0.336`

Explanation:

In the system of equations `x+y+x/y=10` and `x/y(x+y)=20`, let us assume `x+y=u` and `x/y=v`, then the equations become

`u+v=10` and `uv=20`. The latter gives `v=20/u` and substituting it in former, we get

`u+20/u=10` or `u^2-10u+20=0`

and using quadratic formula `(-b+-sqrt(b^2-4ac))/(2a)` we get

`u=(10+-sqrt(100-4xx1xx20))/2=(10+-sqrt20)/2=5+-sqrt5`

As `u+v=10`, if `u=5+sqrt5`, `v=5-sqrt5`

and if `u=5-sqrt5`, `v=5+sqrt5`

(1) In first case we have `x+y=5+sqrt5` and `x/y=5-sqrt5`. The latter means `y=x/(5-sqrt5)` and hence

`x+x/(5-sqrt5)=5+sqrt5` or `x(6-sqrt5)=20` or

`x=20/(6-sqrt5)=(20(6+sqrt5))/31=5.314` and

`y=(20(6+sqrt5))/(31(5-sqrt5))=(20(6+sqrt5)(5+sqrt5))/(31*20)`

= `(35+11sqrt5)/31=1.922`

(2) In other case we have `x+y=5-sqrt5` and `x/y=5+sqrt5`. The latter means `y=x/(5+sqrt5)` and hence

`x+x/(5+sqrt5)=5-sqrt5` or `x(6+sqrt5)=20` or

`x=20/(6+sqrt5)=(20(6-sqrt5))/31=2.428` and

`y=(20(6-sqrt5))/(31(5+sqrt5))=(20(6-sqrt5)(5-sqrt5))/(31*20)`

= `(35-11sqrt5)/31=0.336`
graph{(x+y+x/y-10)(x/y(x+y)-20)=0 [-10, 10, -5, 5]}

Answer 2

`((x,y),(20/31 (6 + sqrt[5]),1/31 (35 + 11 sqrt[5])),(20/31 (6 - sqrt[5]),1/31 (35 - 11 sqrt[5])))`

Explanation:

Calling `x/y=a` and `x+y=b` we have

`{(a+b=10),(a b=20):}`

solving we have

`a=5-sqrt(5), b= 5+sqrt(5)` and
`a=5+sqrt(5), b= 5-sqrt(5)`

or

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

and

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

but

`{(x/y=a),(x+y=b):}`

for `x,y` gives

`{(x=(ab)/(1+a)),(y=b-(ab)/(1+a)):}`

so we have finally

`((x,y),(20/31 (6 + sqrt[5]),1/31 (35 + 11 sqrt[5])),(20/31 (6 - sqrt[5]),1/31 (35 - 11 sqrt[5])))`