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## #1 2005-09-19 09:19:16

squall
Member
Registered: 2005-09-19
Posts: 1

### equation system problem

hi,
does someome can explain me how to solve the following equaction system:

``````x/(a+b)+y/(a-b)=2
(x+y)/(x-y)=a/b``````

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## #2 2005-09-19 11:35:52

kylekatarn
Member
Registered: 2005-07-24
Posts: 445

### Re: equation system problem

``````Substitutions: α = a+b; β = a-b; λ = a/b
Restrictions: α ≠ 0; β ≠ 0; x ≠ y

x   y       x+y
― + ― = 2 ٨ ――― = λ
α   β       x-y

Lets work each equation sep. in terms of x
-----------------------------
x   y
― + ― = 2
α   β

x       y
― = 2 - ―
α       β

α
Eq.1) x = 2α - y.―
β

-----------------------------
x+y
――― = λ
x-y

x+y
――― - λ = 0
x-y

x+y     x-y
――― - λ.――― = 0
x-y     x-y

x+y-(x-y).λ
――――――――――― = 0
x-y

x+y-(x-y).λ = 0
x+y-x.λ+y.λ = 0
x(1-λ)+y(1+λ) = 0
x(1-λ) = -y(1+λ)
y(1+λ)
Eq.2) x = - ――――――
1-λ
-----------------------------
Combine Eq.1 & Eq.2

α     y(1+λ)
2α - y.― = - ――――――
β      1-λ

α
2α.(1-λ) - y.―.(1-λ) = - y(1+λ)
β

α
2α.(1-λ)  = - y(1+λ) - y.―.(1-λ)
β

┌               ┐
│        α      │
2α.(1-λ)  = - y.│(1+λ) - ―.(1-λ)│
│        β      │
└               ┘

2α.(1-λ)
y = - ―――――――――――――――――
┌               ┐
│        α      │
│(1+λ) - ―.(1-λ)│
│        β      │
└               ┘

2αβ.(λ-1)
y = ―――――――――――――
α(λ-1)+β(λ+1)

-----------------------------
Now after doing the same for x...

2αβ.(λ+1)
x = ―――――――――――――
α(λ-1)+β(λ+1)

-----------------------------

So every pair...
┌                               ┐
│   2αβ.(λ+1)       2αβ.(λ+1)   │
[ x , y ] = │ ――――――――――――― , ――――――――――――― │
│ α(λ-1)+β(λ+1)   α(λ-1)+β(λ+1) │
└                               ┘
...with α ≠ 0; β ≠ 0; x ≠ y

is a solution of the system in question.``````

Last edited by kylekatarn (2005-09-19 12:09:32)

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