• Apr 23rd 2008, 07:39 PM
cyy
Hi below is the problem:

3x2 – 4x + 3y = 0
3y2- 4y + 3x = 0

thx u!
• Apr 23rd 2008, 07:46 PM
Jhevon
Quote:

Originally Posted by cyy
Hi below is the problem:

3x2 – 4x + 3y = 0
3y2- 4y + 3x = 0

thx u!

solve the first equation for y and then plug in the expression you get into the second. you will then be able to solve for x, and then go back and solve for y
• Apr 23rd 2008, 08:00 PM
cyy
if i make 3x^2 - 4x + 3y = 0 ---- 1
3y^2 - 4y + 3x = 0 ---- 2

(1) - (2)
hence i gt: 3x^2 - 3y^2 + 3y - 3 x + 4y - 4 x = 0
3( x - y ) ( x + y ) - 7( x - y ) = 0

how shd i continue from here thx...
• Apr 23rd 2008, 08:24 PM
Soroban
Hello, cyy!

Quote:

$\displaystyle \begin{array}{cccc}3x^2 - 4x + 3y &=& 0 & {\color{blue}[1]} \\ 3y^2 - 4y + 3x &=& 0 &{\color{blue}[2]} \end{array}$

Subtract [2] from [1]: .$\displaystyle 3x^2-3y^2 - 4x + 4y + 3y - 3x\;=\;0$

Factor: . $\displaystyle 3(x^2-y^2) - 7x + 7y \;=\;0$

Factor: . $\displaystyle 3(x-y)(x+y) - 7(x-y) \;=\;0$

Factor: . $\displaystyle (x-y)\,[3(x+y) - 7] \;=\;0$ . .
Ah! I see that you already did this!

And we have two equations to solve . . .

$\displaystyle (A)\;\;x - y \:=\:0\quad\Rightarrow\quad y \,=\,x$

. . Substitute into [1]: .$\displaystyle 3x^2 - 4x + 3x \:=\:0\quad\Rightarrow\quad x(3x-1)\:=\:0$

. . And we have: .$\displaystyle \boxed{x \;=\;0,\:\frac{1}{3}} \quad\Rightarrow\quad\boxed{ y \;=\;0,\:\frac{1}{3}}$

$\displaystyle (B)\;\;3(x + y) - 7 \:=\: 0 \quad\Rightarrow\quad y \:=\:\frac{7-3x}{3}$

. . Substitute into [1]: .$\displaystyle 3x^2 - 4x + 3\left(\frac{7-3x}{3}\right) \;=\;0$

. . which simplifies to: .$\displaystyle 3x^2 - 7x + 7 \:=\:0$ .which has no real roots.

Therefore: . $\displaystyle (x,y) \;=\;(0,0),\;\left(\frac{1}{3},\;\frac{1}{3}\right )$