# Thread: How to solve this two equations simultaneously?

1. ## How to solve this two equations simultaneously?

I have two equations with two unknowns x and y.
(r-yp)x^2 –{y^2 (b-a)-2yq+(a+c)}x+(y^2+1)( yp -r)=0 … … (1)
(q-xp)y^2 –{x^2 (a-b)-2xr+(b+c)}y+(x^2+1)( xp -q)=0 … … (2)
here a,b,c,p,q,r are some constants.
How to solve this ? Pls help me.

2. Why post it 3 times...expect 3 answers ?

Seems to be limitless solutions.
Here's one of them:
a=1, b=1, c=2, p=1, q=2, r=2, x=1, y=1

3. I didn't post it 3 times, it may happen due to disconnection of network while submitting. Sorry for the inconvenience.
However, I don't need typical solution, I need algebraic solution of x, y in terms of a,b,c,p,q,r. Can u suggest any such solution ?

4. Originally Posted by Robin
I didn't post it 3 times, it may happen due to disconnection of network while submitting. Sorry for the inconvenience.
However, I don't need typical solution, I need algebraic solution of x, y in terms of a,b,c,p,q,r. Can u suggest any such solution ?
All I can see is this:

(r-yp)x^2 – [y^2 (b-a) - 2yq + (a+c)]x + (y^2+1)(yp-r) = 0 … … (1)
(q-xp)y^2 – [x^2 (a-b) - 2xr + (b+c)]y + (x^2+1)(xp-q) = 0 … … (2)

I'm replacing p,q,r with d,e,f:
(f-dy)x^2 – [y^2 (b-a) - 2ey + a+c]x + (y^2+1)(dy-f) = 0 … … (1)
(e-dx)y^2 – [x^2 (a-b) - 2fx + b+c]y + (x^2+1)(dx-e) = 0 … … (2)

Let u = f-dy ; then -u = dy-f
Let v = e-dx ; then -v = dx-e
Let w = b-a ; then -w = a-b

ux^2 – (wy^2 - 2ey + a + c)x - u(y^2 + 1) = 0 … … (1)
vy^2 – (-wx^2 - 2fx + b + c)y - v(x^2 + 1) = 0 … … (2)

Rewriting (2):
(wy - v)x^2 + (2fy)x + v(y^2 - 1) - y(b + c) = 0 (3)

So we now have:
ux^2 – (wy^2 - 2ey + a + c)x - u(y^2 + 1) = 0 … … (1)
(wy - v)x^2 + (2fy)x + v(y^2 - 1) - y(b + c) = 0 ... ...(3)

Both can be solved for x using the quadratic formula.

I'll stop here for now.

5. Will this form give a correct root of x while you have taken v = e-dx ? without this x how would I get the root of x ?
Then how will I get the roots of y while you have taken u = f-dy ?

6. My bad; missed fact that 1st term of (3) which is (wy - v)x^2
will result in a cubic : dx^3 + (wy - e)x^2