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

Answer:

#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

Answer:

#((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])))#