Equationsystem problem


by nastyjoe
Tags: equations, equationsystem, polynomial, variables
nastyjoe
nastyjoe is offline
#1
Mar20-13, 04:19 PM
P: 2
My maths skills are so rusty that I can't figure out how I simplify these equations so that I get a formula for x and y... a,b,c,d,e,f are constants

y=[itex]\sqrt{b^{2} - (x-f)^{2}}[/itex] + e
x=[itex]\sqrt{a^{2} - (y-c)^{2}}[/itex] + d

Can anyone help me? And is this equationsystem even possible?
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slider142
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#2
Mar20-13, 04:26 PM
P: 878
If you subtract the constants from both sides and square both sides, you should be able to see that your equations can be graphed in the xy-plane as the upper hemisphere of a circle of radius b centered at (f, e) and the upper hemisphere of a circle of radius a centered at (d, c). Whether these two curve segments intersect or not is up to the values of the constants.
To start, you can just use substitution: substitute your expression for y as a function of x into the second equation.
nastyjoe
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#3
Mar20-13, 04:31 PM
P: 2
I tried substituting y as a function of x into the second equation but I got an awfully complicated equation which I was unable to solve as I'm not that good at maths... :( Are you able to get a solution?

berkeman
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#4
Mar20-13, 05:06 PM
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Equationsystem problem


Quote Quote by nastyjoe View Post
My maths skills are so rusty that I can't figure out how I simplify these equations so that I get a formula for x and y... a,b,c,d,e,f are constants

y=[itex]\sqrt{b^{2} - (x-f)^{2}}[/itex] + e
x=[itex]\sqrt{a^{2} - (y-c)^{2}}[/itex] + d

Can anyone help me? And is this equationsystem even possible?
Quote Quote by nastyjoe View Post
I tried substituting y as a function of x into the second equation but I got an awfully complicated equation which I was unable to solve as I'm not that good at maths... :( Are you able to get a solution?
Welcome to the PF.

What are these equations from?
AlephZero
AlephZero is offline
#5
Mar20-13, 06:40 PM
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If you square the first equation, you get
##(y-e)^2 + (x-f)^2 = b^2##

If you draw a graph of that equation, what shape of curve do you get? (If you can't see the answer to that, start with the simpler case when e = f = 0).

The easiest way to solve the two equations is using geometry, not algebra.


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