solve: x+y^1/2=11 x^1/2+y=7?

Question

solve mathematically & with proper reasoning

this is a challenge for all the brainy guys if u dare bcoz none has answered properly

Answer 1

y= (11-x)^2
y=7-x^1/2
121+x^2-22x=7-x^1/2
x^3-22x^2+x= -114
I could not solve it for now .Will try it later .
I ve made a mistake in the fourth step

Answer 2

x + y^1/2 = 11 x^1/2 + y = 7
x = 11 - √y
y = 7 - √x
x = 11 - √y
Rather than finding the one usable root of the resulting 4th order polynomial,
If x = 11, y = 3.68337521, x = 9.080787867
If x = 10, y = 3.83772234, x = 9.040989449
If x = 9, y = 4, x = 9

Answer 3

x+y¹/²=11
x¹/²+y=7

(y¹/²)² = (√y)² = (11 - x)²
(x¹/²)² = (√x)² = (7 - y)²

y = 11² - 2.11.x + x²
y = 121 -22x + x²
d = -22² - 4.1.121
d = 484 - 484
d = 0

y = (-22 +/- √0) : 2
y' = y" = -22 : 2 = -11

x = 7² - 2.7.y + y²
x = 49 - 14y + y²
d = -14² - 4.1.49
d = 196 - 196
d = 0
x' = x" = -7
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Answer 4

let x = a^2 and y = b^2

a^2+b = 11 ...1
a + b^2 = 7...2

from 2

a = 7-b^2

so (7-b^2)^2 + b = 11

or 49+b^4-14b^2+b = 11
or b^4-14b^2+b + 36 = 0 say f(b)

if b is integer factors are factors of 36
by checking one by one
f(2) = 0
so b =2 and a = 3
or x=9 and y = 4

Answer 5

x+y^1/2=11.........1st eq
x^1/2+y=7...........2nd eq

from 2nd eq: y=7-x^1/2........... *

substituting * in 1st eq, we get
x+(7-x^1/2)^1/2=11
(7-x^1/2)^1/2=11-x
7-x^1/2=(11-x)^2
7-x^1/2=121-22x+x^2
x^2-22x+121-7+x^1/2=0
x^2-22x+x^1/2+114=0

let x^1/2 =a
then
a^4-22a^2+a+114=0
let f(a)=a^4-22a^2+a+114

by trial and error
f(3)=0
therefore (a-3) is a factor
solve the rest
dnt hav time

Answer 6

21 because they show up alot in the problem. thank me later, no there's no charge.