# Thread: two equations

hi

2. ## Re: two equations

You have a circle and an ellipse there, and need to find their points of intersection. Fundamentally it is no different than other systems of two equations with two unknowns.

3. ## Re: two equations

$x^2 + y^2 = 12 - x - y \quad So, \quad 12 - x - y - xy = 9 \Rightarrow x + y + xy = 3 \Rightarrow x + y = 3 - xy \Rightarrow (x + y)^2 = (3 - xy)^2 \Rightarrow x^2 + 2xy + y^2 = 9 - 6xy + (xy)^2 \Rightarrow x^2 + y^2 + 8xy = 9 + (xy)^2$

You continue. At the end you will find: x = 0, y = 0, xy = 9. When x = 0 => y = +-3, when y = 0 => x = +-3, when xy = 9 => $x^2 + y^2 = 18 \quad and \quad x = \frac{9}{y} ....$

4. ## Re: two equations

here is my trial

5. ## Re: two equations

Originally Posted by abualabed
here is my trial
You can't really get anywhere from there. The usual case for that would be that the RHS is zero. Then we could say that since ab = 0, we have that a = 0 or b = 0. But you don't have that.

ManosG's method seems to be a good one.

-Dan

Hello

7. ## Re: two equations

Hello, abualabed!

$\text{If }x,y\text{ are integers, solve this system:}$

. . $x^2 + y^2 + x + y \:=\:12\;\;[1]$

. . . . $x^2 + y^2 - xy \:=\:9\;\;[2]$

Subtract [1] - [2]: . $x + y + xy \:=\:3$

Solve for $y\!:\;\;y \:=\:\frac{3-x}{1+x}$

Substitute into [2]: . $x^2 + \left(\frac{3-x}{1+x}\right)^2 - x\left(\frac{3-x}{1+x}\right) \:=\:9$

. . which simplifies to: . $x^4 + 3x^3 - 9x^2 - 27x \:=\:0$

. . and which factors: . $x(x-3)(x+3)^2 \:=\:0$

Hence: . $\begin{Bmatrix}x &=& 0,3,\text{-}3 \\ y &=& 3,0,\text{-}3 \end{Bmatrix}$

Solutions: . $(x,y) \;=\;(3,0),\;(0,3),\;(\text{-}3,\text{-}3)$

8. ## Re: two equations

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$\displaystyle b = \frac{\sum(x - \bar{x})(y - \bar{y})}{\sum(x - \bar{x})^2}$ and $a = \bar{y} - b\bar{x}$

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