Solving Simultaneous Equations in INDIA | Grade 12

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:smile: Hi Everbody,
I am a new member to this forum.I am from INDIA and studying in grade 12.
I was struck in a question and need help.Hope someone could solve it.
The question is to solve these simultaneous equations:-
√x + y = a -------- (i)
x +√y = b ---------- (ii)

I have a few hints such as making a change in variable by introducing
x = m^2 and y= n^2 and then doing some algebraic manipulations to get
(m-n)(1-m-n)=a-b
But I don't know know what to do next. Plz help.
 

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AlbertEinstein said:
:smile: Hi Everbody,
I am a new member to this forum.I am from INDIA and studying in grade 12.
I was struck in a question and need help.Hope someone could solve it.
The question is to solve these simultaneous equations:-
√x + y = a -------- (i)
x +√y = b ---------- (ii)

I have a few hints such as making a change in variable by introducing
x = m^2 and y= n^2 and then doing some algebraic manipulations to get
(m-n)(1-m-n)=a-b
But I don't know know what to do next. Plz help.
Those "algebraic manipulations" don't help because you are left with one equation in two variables. After you have m2+ n= b and m+ n2= a, you can solve the first for n: n= b- m2 and then substitute in the second: m+ (b-m2)2= a. That gives a single, fourth degree, equation for m.
 
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sqrt(x)+y=a
Sqrt(y)+x=b

I'd simply...use a substitution after working around w/ them

sqrt(x)=a-y
sqrt(y)=b-x

x=a^2 - 2ay + y^2
y=b^2 -2bx + x^2

And now...stick the x into the y=

y=B^2 -2b(a^2 - 2ay + y^2) + (a^2 - 2ay + y^2)^2
And algebrate. it's ugly but you can't dodge the forth power I don't think. SUbstitute y from both sides and you'll have the =0
 
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