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I know the answers i just need help with how to get there
@Shalante any ideas??
\[x^2\] is x^2 not x2
yes, sorry it copy and pasted wrong
if y=-x^2+4 and y=2x+1 then -x^2+4=2x+1 Know how to solve for x?
not really lol
if you do know how to solve for x, plug the x value to any of the equation to get y
we don't know the x value
so solve and that is the solution?
\[x^2+2x-3\] move it over
yes you know how to solve that?
there should be two solutions with both having (x,y) form
what is the first step to solve that? lost again
where do i go after the x^2+2x-3
factor it ( )( )
(x-1) (x+3) ? sorry I'm confused
Yea you are right.
dunno why it took me back to my old reply
forgot to tell you its \[x^2+2x-3=0\]
so quadratic ?
that is what happens when you move all the numbers to the other side of an equation. After everything is gone, you a zero to it.
yes i know
so you get -3 after solving
so those are the x's what about the y's??
Then plug both this x back into the first 2 equation using both x=-3 and x=1
they should both equal right?
yes they both get -5
ANSWER #1 (x,y) plug in first x ANSWER #2 (x,y) plug in second x
oops my orignial was wrong hold on
so answer #1 is (-3,-5). you are correct on first one.
I got everything under control! thanks so much!
Yea, no problem.