LOOKING SIMPLE BUT NOT THAT.

               

Question
Here is the two simple equation but their solution is soo difficult ..
I hope you you enjoy it.

Solution :

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.there is many way to find this problem .

1st method:

Let's started to solve this question.

First here is the equation

x^1/2+y=11 => y=11 -x^1/2………(1)

x+ y^1/2=7 => y^1/2=7-x , squaring both sides

y=49–14x+x^2……………….(2)

from eq.(1) &(2)

49–14x+x^2=11-x^1/2

x^2–14x+x^1/2+38=0

Put x=4 in above equation then  , 

Result=16–56+2+38=0 , therefore (x-4) is a factor.

=x(x-4)+4x-14x+x^1/2+38=0

=x(x-4)-10x+x^1/2+38=0

=x(x-4)-10(x-4)-40+x^1/2+38=0

=x(x-4)-10x(x-4)+(x^1/2–2)=0

=>(x^1/2–2)[x(x^1/2+2)-10x(x^1/2+2)+1]=0

Either x^1/2–2=0=> x=4

Put x=4 in eq.(1)

(4)^1/2+y=11

2+y=11=>y=11–2=9

x=4 ,y =9

2nd Method:

Answer:

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x+√y=7(1)

√x+y=11 ....(2)

from eq 1...

√y=7-x

y=(7-x)^2

from eq 2...

√x=11-y

substitute (7-x)^2 for y

√x=11-(7-x)^2

square both sides

x=121-22(7-x)^2+(7-x)^4

(7-x)^4=(x-7)^2*(x-7)^2=(x^2-14x+49)(x^2-14x-49)=

x^4-28x^3+294x^2-1372x+2401

x=121-22x^2+308x-1078+x^4-28x^3+294x^2-1372x+2401

0=-x+121-22x^2+308x-1078+x^4-28x^3+294x^2-1372x+2401

x^4-28x^3+272x^2-1065x+1444=0

from the rational root theorem, possible roots are factors of 1444

4 is a factor of 1444 and a root of the equation...

256-1792+4352-4260+1444=0

-1536+4352-4260+1444=0

2816-4260+1444=0

-1444+1444=0

0=0

x=4

x+√y=7

4+√y=7

√y=7-4

√y=3

square both sides

y=9

x=4 and y=9

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x−−√+

y4−44y3+712y26

=0