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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+2x1. the second is y=3x+5> I don"t know how to begin.
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+2x1. the second is y=3x+5> I don"t know how to begin.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the >is supposed to be a period at the end of the second equation.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0set them equal and solve. x^2+2x1=3x+5 x^2x6=0 (x3)(x+2)=0 x=3, x= 2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, 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
 5 years ago
Best ResponseYou've already chosen the best response.0still looking at it...what is ^? and what happened to the "y"?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok, 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
 5 years ago
Best ResponseYou've already chosen the best response.0if the two equations are equal, then the 'y' values must be the same. That is why you can set them equal.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now from what you have there you can either solve by factoring, or use the quadratic formula to find your x values.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0OK, thank you! i got it!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no, I don"t have it yet. when I get to the part of the equation of: Xsquaredx6=0, how do I get rid of the squared X and get to the answer?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Quadratic equation.. \[ax^2 + bx + c = 0 \implies x = \frac{(b) \pm \sqrt{(b)^24(a)(c)}}{2(a)}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we aren"t doing that yet.....is there another way?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What two numbers add together to make 1 and multiply together to make 6

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(the coefficients of your middle and last terms respectively)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Correct. Therefore your equation factors into (x+2)(x3) = 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Which means x can be either 2 (making x+2 = 0) or 3 (making x3 = 0)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i 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
 5 years ago
Best ResponseYou've already chosen the best response.0I tried to apply the same formula on my next equation and got as far as the previous one....I have xsquaredx2=0. what am I missing?
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