Is this solvable?

Question

x, y, and r are unknown. a, b, c, d, e, and f are constant.

1) ar + b = cy
2) dr + e = fx
3) r^2 = x^2 + y^2

Answer 1

Yes, there are 3 equations and 3 unknowns, but 1 of the equations isn't linear. You will not get unique solutions when you plug into the 3rd equation.

Lets go ahead and do it.

From 1) y = ar/c + b/c
From 2) x = dr/f + e/f

Plug into 3)

r^2 = (dr/f + e/f)^2 + (ar/c + b/c)^2

= d^2r^2/f^2 + 2dre/f^2 + e^2/f^2 + a^2r^2/c^2 + 2arb/c^2 + b^2/c^2

Lets clear fractions by multiplying through by f^2*c^2
r^2f^2c^2=
c^2d^2r^2 + 2c^2dre + c^2e^2 + f^2a^2r^2 + 2f^2arb +f^2b^2

Gathering like degree terms of r:

(f^2c^2 - c^2d^2 - f^2a^2)*r^2 - (2c^2de + 2f^2ab)*r - c^2e^2 - f^2b^2
=0.

So, it can be reduced to a 2nd degree polynomial in r. This will only have real solutions if the discriminant >= 0.

Lastly, imagine what each of these things is.

1) The equation of a plane in 3 D space.
2) The equation of another plane in 3 D space.
3) The equation of a sphere in 3 D space.

The intersection of 1 and 2 will be one of the following: empty (they are parrallel), a single line (nice solution), or they coincide.

The intersection of the sphere with the intersection of 1 and 2 will have solutions only if the line passes through the sphere. It might not.

So, if the disrimenent from that mess above is > 0, you get a nice line that passes through the sphere and there will be exactly 2 solutions.

Oh, and the answer to the question, 'When would you ever use this" is that it comes up in optimization and business problems when you want to find where various curves meet.

Answer 2

It should be soluable, though it'll take more than I can do offhand... Three equations in three unknowns are usually soluable...

What seems to me to be the approach is solve 1 and 2 for x and y, then put that into 3 to get a polynomial in r, solve that for r and plug that back into the other two...

Answer 3

Yes it is solvable.

To solve it, you first solve for x and y in the first two equations:
x = (dr+e)/f
y = (ar+b)/c

Substitute them into equation 3 to get a value for r, then substitute that value into 1 and 2 to get y and x.

Answer 4

yes, it is solvable, because you have three equations with three unknowns. Just use substitution until you have two equations with two unknowns and then one equation with one unknown and then substitute the known in the other equations.

Answer 5

Yes,
Mark the Red is right.
There are three equations with three unknowns.

Answer 6

yes it is solvable, and the answer about three above me is right on the money...just look at that one. i know my answer didnt help 2 much, but will you plz choose me as your best answer cuz i didnt make any wise cracks? thnx!

Answer 7

YES.
ar+b=cy..........(1)
dr+e=fx...........(2)
r^2=x^2+y^2.....(3)

From 1 & 2
(cy-b)/a=(fx-e)/d
=> y=((fx-e)*(a/d)*+b)/c
Further solving the answer with eqn 2. you wil arive at the solution.

Answer 8

The answer is window PANE!!! The first guy almost had it, but he forgot the pane.

Bye now

Answer 9

I showed this question to Bill Gates and asked him if he would hire you. He said you should keep your Mc Donald's job. SORRY.

Answer 10

well first solve for number three take the squareroot of them

so you have r=x+y

so now you have

a*(x+y)+b=cy

now you have

ax+ay+b=cy

ax+ay/y+b=cy/y

a+ax+b=c

d(x+y)+e=fx
dx+dy+e=fx
dx/x+dy+e=fx/x
d+dy+e=f

so now you have...
a+ax+b=c
d+dy+e=f
x+y=r

or now you have

r-x=y
r-y=x

so now ...
d+dy+e=f
d+d(r-y)+e=f
d+dr+-dy+e=f
d+dr+-dy+e=f
d/d+dr/d+-dy/d+e/d=f/d
r+-y+e/d=f/d
r+-y+e/d-e/d=f/d-e/d
r+-y=(e-f)/d
r+-y=(e-f)/d
r-y=(e-f)/d

a+ax+b=c

a+a(r-x)+b=c
a+ar+-ax+b=c
a/a+ar/a+-ax/a+b/a=c/a
r+-x+b/a=c/a
r+-x+b/a-b/a=c/a-b/a
r+-x=(b-c)/a
r+-x=(b-c)/a
r-x=(b-c)/a



so if
r-x=y
then
y=(b-c)/a

and if
r-y=x
then
x=(e-f)/d


jeeze i hope i'm right i'm gonna be a sophomore and jeeze i really hope this is right lol

check ya later ♥