SOLUTION: Question: how i can solve this equation. x + y^(.5) = 7 y + x^(.5) = 11 i know the answer of these equations but dont know how to solve ? meance in first equation just y plus x

Question 1827: Question: how i can solve this equation.
x + y^(.5) = 7
y + x^(.5) = 11
i know the answer of these equations but dont know how to solve ?
meance in first equation just y plus x to the power .5 and in second equation swap x and y.
answer is x = 4 ...... and y = 9
but i want to know the method for solving this equation :-)
Answer by khwang(438) (Show Source): You can put this solution on YOUR website!
You can use substitution tosolve this system of equations.
Solve x + y^(1/2) = 7,,,(1)
y + x^(1/2) = 11...(2)
Sol: y^(1/2) = 7 - x, y = (7 - x)^2
(7 - x)^2 + x^(1/2) = 11.
Simplify: x^(1/2) + 38 - 14x + x^2 =0,
Let X = x^(1/2),
X^4 - 14X^2 + X + 38 = 0..(**)
Factor: (X-2)(X^3 + 2X^2 -10 X -19) = 0.
So,X = 2 is a root and hence x = X^2 = 4.
Hence, y = (7 - x)^2 = (7 - 4)^2 = 9
Thus,we obtain a set of solution , x=4, y = 9.
Check: x + y^(1/2) = 4+ 9^(1/2) = 4+3 =7,,,(OK)
y + x^(1/2) = 9 + 4^(1/2) = 9 + 2 =11...(OK)
Note: there is another positive real solution x of(**) between 3 & 4,
but it isin very complicated from.

Kenny

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