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Mush · #6 January 7th, 2009, 12:01 AM

Quote:

Originally Posted by Chizum

I had to solve two equations for y and then equate those values and solve and plug the result(s) back into the one of the original equations (to find the points of intersection of these two equations if graphed). The equations are $x = 3 - y^2$ and $y = x - 1$ . I'm on the step where I need to equate them and solve. I just don't remember at all going over how to solve equalities when +/- is involved.

$x= 3-y^2$ . Just plug this into the other equation $y = x-1$ , to get:

$y = 3-y^2-1$

Rearrange:

$y^2+y-2 = 0$

However for future reference, if you have the equation $y = \pm \sqrt{3-x}$ and the equation $y = x-1$ , then you would take the first equation and split it into two. And then do the usual process twice for each equation.

For example:

For $y = + \sqrt{3-x}$

$\sqrt{3-x}=x-1$

$3-x = (x-1)^2$

$3-x = x^2-2x+1$

$x^2-x-2 = 0$

For $y = - \sqrt{3-x}$

$-\sqrt{3-x}=x-1$

$\sqrt{3-x}=1-x$

$3-x = (1-x)^2$

$3-x = 1-2x+x^2$

$x^2-x-2 = 0$

It just so happens that in this case, both equations give identical results

The following users thank Mush for this useful post:
Chizum
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