Solve the following system of equations and show all work.
y = −x2 + 4
y = 2x + 1

- anonymous

Solve the following system of equations and show all work.
y = −x2 + 4
y = 2x + 1

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- anonymous

@Preetha

- anonymous

@ganeshie8

- anonymous

I know the answers i just need help with how to get there

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## More answers

- anonymous

@Hero

- anonymous

@Shalante any ideas??

- anonymous

\[x^2\] is x^2 not x2

- anonymous

yes, sorry it copy and pasted wrong

- anonymous

if y=-x^2+4
and y=2x+1
then -x^2+4=2x+1
Know how to solve for x?

- anonymous

not really lol

- anonymous

if you do know how to solve for x, plug the x value to any of the equation to get y

- anonymous

we don't know the x value

- anonymous

\[-x^2+4=2x+1\]

- anonymous

so solve and that is the solution?

- anonymous

\[x^2+2x-3\]
move it over

- anonymous

yes you know how to solve that?

- anonymous

there should be two solutions with both having (x,y) form

- anonymous

what is the first step to solve that? lost again

- anonymous

where do i go after the x^2+2x-3

- anonymous

factor it ( )( )

- anonymous

(x-1) (x+3) ?
sorry I'm confused

- anonymous

?

- anonymous

Yea you are right.

- anonymous

then what?

- anonymous

oops

- anonymous

dunno why it took me back to my old reply

- anonymous

forgot to tell you its
\[x^2+2x-3=0\]

- anonymous

so quadratic ?

- anonymous

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.

- anonymous

yes i know

- anonymous

so you get -3 after solving

- anonymous

and 1

- anonymous

so those are the x's what about the y's??

- anonymous

Then plug both this x back into the first 2 equation using both x=-3 and x=1

- anonymous

they should both equal right?

- anonymous

yes they both get -5

- anonymous

ANSWER #1 (x,y) plug in first x
ANSWER #2 (x,y) plug in second x

- anonymous

oops my orignial was wrong hold on

- anonymous

so answer #1 is (-3,-5). you are correct on first one.

- anonymous

I got everything under control! thanks so much!

- anonymous

Yea, no problem.

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