## anonymous 5 years ago i have to "solve the following system of equations algebraically". but there are two equations and two variables. the first is y=xsquared+2x-1. the second is y=3x+5> I don"t know how to begin.

1. anonymous

the >is supposed to be a period at the end of the second equation.

2. anonymous

set them equal and solve. x^2+2x-1=3x+5 x^2-x-6=0 (x-3)(x+2)=0 x=3, x= -2

3. anonymous

Well, when you are solving systems, you are looking for the points that all the equations have in common. If the points are common, then the y value in one equation will equal the y value in the other equation. Therefore you can set the two expressions for y equal to each other and solve for x.

4. anonymous

still looking at it...what is ^? and what happened to the "y"?

5. anonymous

Ok, you have this: $y = x^2+2x -1$$y=3x+5$ Set those two expressions for y in terms of x equal to each other.

6. anonymous

x^+2x-1=3x+5

7. anonymous

if the two equations are equal, then the 'y' values must be the same. That is why you can set them equal.

8. anonymous

Now from what you have there you can either solve by factoring, or use the quadratic formula to find your x values.

9. anonymous

OK, thank you! i got it!

10. anonymous

no, I don"t have it yet. when I get to the part of the equation of: Xsquared-x-6=0, how do I get rid of the squared X and get to the answer?

11. anonymous

Quadratic equation.. $ax^2 + bx + c = 0 \implies x = \frac{-(b) \pm \sqrt{(b)^2-4(a)(c)}}{2(a)}$

12. anonymous

we aren"t doing that yet.....is there another way?

13. anonymous

factor it

14. anonymous

What two numbers add together to make -1 and multiply together to make -6

15. anonymous

(the coefficients of your middle and last terms respectively)

16. anonymous

2 and -3

17. anonymous

Correct. Therefore your equation factors into (x+2)(x-3) = 0

18. anonymous

Which means x can be either -2 (making x+2 = 0) or 3 (making x-3 = 0)

19. anonymous

i don"t see where you got the -1 and 6....i have a -6....at the end of the second one, but a -x on the first

20. anonymous

I tried to apply the same formula on my next equation and got as far as the previous one....I have xsquared-x-2=0. what am I missing?