17. What are the solutions of the system? y=x^2+3x-4 y=2x+2

Set the two equal to each other. \(2x+2=x^2+3x-4\), then solve.

If y = x^2 + 3x - 4 and it ALSO equals 2x + 2, then: \(\Large \color{Black}{\Rightarrow 2x + 2 = x^2 +3x - 4 }\) Solve for x.

hold let me try to solve it.

what would be -x=x^2?

How did you get that?

Subtract the \(2x+2\) from both sides and you will get \(x^2+x-6=0\), then simply solve that quadratic.

k.. thanks.

still don't know. but thanks for trying to help.

I'm still here, don't give up lol. You've solved quadratics before, right?

no. i'm stupid in math literally ahha.

Math isn't about stupid or smart, it's about practicing until you've learned it. If you don't yet know how to solve a quadratic, it's a simple matter of practicing it until you do. Everyone can learn it, if they practice.

So, to solve \(x^2+x-6=0\) there are a few different techniques. Here, I would just look for two numbers that multiply to -6 and add to 1, which will tell you what the two binomials should be.

i still don't get it. this is useless. i'm worthless. and a failure. i'm just going to fail this final.

That's a decision that you can choose to make. You can decide to fail, or you can decide to succeed. Nobody else is going to decide that for you, it is your choice.

k. but clearly i have know idea what i'm doing so.lololol.

That is fine, you can learn it. Can you tell me two numbers that multiply together to give you -6 and add together to give you 1?

1+0=1? & 2x-3=-6

What I mean is the same two numbers. The same two numbers multiply to -6 and add to 1

i have know idea.

Well, think about it. There are only so many numbers that multiply together to -6. They are whole numbers. List them all out, and see which of them add to 1

There are a total of 8 possible pairs of numbers that it could be.

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