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.

Question

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.

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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.

anonymous · Answer

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

anonymous · Answer

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.

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

OK, thank you!  i got it!

anonymous · Answer

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?

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

factor it

anonymous · Answer

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

anonymous · Answer

(the coefficients of your middle and last terms respectively)

anonymous · Answer

2 and -3

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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?