# Thread: Algebraically find the coordinates.

1. ## Algebraically find the coordinates.

Find algebraically the coordinates of the points of intersection of the graph with this equation:

$\displaystyle y=3x-1 y=3+3/(x-1)$

this is what i did (but turned out wrong?

$\displaystyle 3x-1=3+3/(x-1)$

multiplied everything by x-1
$\displaystyle (x-1)(3x-1)= 3x-3 +2$

$\displaystyle (X-1)(3X-1)=3x-1$

$\displaystyle x-1=0$

$\displaystyle x=1$

but that doesn't work, and either way the answer is meant to be:

$\displaystyle (2,5) (1/3,0)$

No clue what to do.
Thanks.

2. Originally Posted by juliak
F
$\displaystyle y=3x-1 y=3+3/(x-1)$

multiplied everything by x-1
$\displaystyle (x-1)(3x-1)= 3x-3 +2$

$\displaystyle (X-1)(3X-1)=2x-1$
3x - 3 + 2 = 3x - 1 not 2x -1

Bobak

3. oooops.
typo.

edited.

4. Originally Posted by juliak
$\displaystyle (x-1)(3x-1)= 3x-3 +2$

$\displaystyle (X-1)(3X-1)=3x-1$

$\displaystyle x-1=0$ this is wrong

$\displaystyle x=1$

Two things here, firstly if you do cancel a factor of 3x-1 from both side the right hand side becomes 1 not zero, also by doing that your losing a solution to the equation. your working should look something like this...

$\displaystyle (x-1)(3x-1)=3x-1$
$\displaystyle (x-1)(3x-1) - (3x-1)=0$
$\displaystyle (3x-1)((x-1) -1)=0$
$\displaystyle (3x-1)(x-2)=0$

Bobak

5. oooo
thanks!