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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+2x-1. the second is y=3x+5> I don"t know how to begin.

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  1. anonymous
    • 5 years ago
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    the >is supposed to be a period at the end of the second equation.

  2. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    still looking at it...what is ^? and what happened to the "y"?

  5. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    x^+2x-1=3x+5

  7. anonymous
    • 5 years ago
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    if the two equations are equal, then the 'y' values must be the same. That is why you can set them equal.

  8. anonymous
    • 5 years ago
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    Now from what you have there you can either solve by factoring, or use the quadratic formula to find your x values.

  9. anonymous
    • 5 years ago
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    OK, thank you! i got it!

  10. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    Quadratic equation.. \[ax^2 + bx + c = 0 \implies x = \frac{-(b) \pm \sqrt{(b)^2-4(a)(c)}}{2(a)}\]

  12. anonymous
    • 5 years ago
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    we aren"t doing that yet.....is there another way?

  13. anonymous
    • 5 years ago
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    factor it

  14. anonymous
    • 5 years ago
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    What two numbers add together to make -1 and multiply together to make -6

  15. anonymous
    • 5 years ago
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    (the coefficients of your middle and last terms respectively)

  16. anonymous
    • 5 years ago
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    2 and -3

  17. anonymous
    • 5 years ago
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    Correct. Therefore your equation factors into (x+2)(x-3) = 0

  18. anonymous
    • 5 years ago
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    Which means x can be either -2 (making x+2 = 0) or 3 (making x-3 = 0)

  19. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    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?

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