Solving Simultaneous Equations in INDIA | Grade 12

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SUMMARY

The discussion focuses on solving simultaneous equations involving square roots: √x + y = a and x + √y = b. Participants suggest transforming the variables by letting x = m² and y = n², leading to the equation (m-n)(1-m-n) = a-b. Further algebraic manipulation is required to isolate variables, ultimately resulting in a fourth-degree polynomial equation for m. The conversation emphasizes the necessity of substitution and algebraic techniques to simplify and solve the equations effectively.

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  • Understanding of algebraic manipulation techniques
  • Familiarity with polynomial equations
  • Knowledge of substitution methods in solving equations
  • Basic concepts of square roots and their properties
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  • Study techniques for solving fourth-degree polynomial equations
  • Learn advanced algebraic manipulation strategies
  • Explore substitution methods in greater depth
  • Practice solving simultaneous equations with square roots
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High school students, particularly those in Grade 12 mathematics, educators teaching algebra, and anyone interested in advanced algebraic problem-solving techniques.

AlbertEinstein
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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.
 
Last edited by a moderator:
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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