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.

2 Answers
Jan 12, 2017

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]}

Jan 12, 2017

((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])))