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anonymous

  • one year ago

Help Please!!! Solve the following system of equations and show all work. y = x2 + 3 y = x + 5

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  1. anonymous
    • one year ago
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    @mathmale

  2. anonymous
    • one year ago
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    Or anyone that can help me

  3. anonymous
    • one year ago
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    what do you mean by "solve?"

  4. anonymous
    • one year ago
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    I'm not sure thats all they told us.

  5. anonymous
    • one year ago
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    you would write it: x+5=x2+3 and then solve from there. the reason that is corrret is because youre just repacing y with what y is equal to- y

  6. anonymous
    • one year ago
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    Okay thank you

  7. anonymous
    • one year ago
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    Do I solve for x?

  8. anonymous
    • one year ago
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    do the substitution method where you multiple one of the equations by -1

  9. NotTim
    • one year ago
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    Do not do what ew.sorry said. it may work, but @arek 's method is much, much safer.

  10. anonymous
    • one year ago
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    Okay I will. For the substitution method do I make y -1? Sorry I'm not that great at math.

  11. whpalmer4
    • one year ago
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    \[y=x^2+3\]\[y=x+5\] Those are two equations to solve. A couple of routes you can take: substitution setting the two right hand sides equal to each other substitution means solving one equation for one variable in terms of the other. Then you substitute for that variable in the other equation. In this problem, this would a fine way to go because the first step has already been done for you: you have \(y=x+5\) as one of the equations, and that gives you \(y\) in terms of \(x\). Take the other equation \((y = x^2+3)\) and replace \(y\) with \(x+5\), then solve for the values of \(x\). There will be two solutions thanks to the \(x^2\) term, although in some cases they may turn out to be the same value. another approach (which is really substitution as well) is to notice that we have two equations with the same thing on one side. Here, both equations have \(y\) by itself on one side of the equals sign. We can simply set the other two sides equal to each other, because they are equal to the same thing. After that, you solve just as before. In both approaches, after you find the value(s) of \(x\), you plug them into one of the original equations to find the corresponding values of \(y\).

  12. whpalmer4
    • one year ago
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    Another approach would be to graph the two equations and observe where they intersect. \[y=x^2+3\]is a parabola, and \[y=x+5\]is a line. The line crosses the parabola in two locations, which are the solutions to this system.

  13. whpalmer4
    • one year ago
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    And yet another approach, if you had the luxury of a multiple-choice problem where you could know that the answers are provided and you merely need to identify the correct answers, would be to plug each set of answers into all of the equations and select the answer choice where all of the equations work. Note well that you can have "answers" that work for some equations but all...the technical term for them is "wrong" :-)

  14. anonymous
    • one year ago
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    Wow! Thank you so much!

  15. anonymous
    • one year ago
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    |dw:1433532657818:dw|

  16. anonymous
    • one year ago
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    I'm stuck at this point. Did I do it wrong?

  17. whpalmer4
    • one year ago
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    \[x^2+3=x+5\] collect everything on one side, what do you get?

  18. anonymous
    • one year ago
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    0 = -x^2 +x + 2

  19. whpalmer4
    • one year ago
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    Yes, or \[x^2-x-2 = 0 \] Do you know how to factor that equation, or solve it with the quadratic formula, or complete the square?

  20. whpalmer4
    • one year ago
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    @torvia you still here?

  21. anonymous
    • one year ago
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    Yes sorry my dad had to use the computer

  22. anonymous
    • one year ago
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    I know how to solve using quadratic formula

  23. anonymous
    • one year ago
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    |dw:1433535440928:dw|

  24. anonymous
    • one year ago
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    did i do that right?

  25. whpalmer4
    • one year ago
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    close, but not close enough. I'm intrigued by how you went from \[\sqrt{1+8}\] to \[\sqrt{4}\]in particular :-)

  26. whpalmer4
    • one year ago
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    I think you probably just didn't read your handwriting...

  27. anonymous
    • one year ago
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    Yeah sorry it was really messy.

  28. whpalmer4
    • one year ago
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    correct answer is...

  29. anonymous
    • one year ago
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    My solutions then would be 2 and -2?

  30. whpalmer4
    • one year ago
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    let's try them out! \[x^2-x-2=0\]\[(2)^2-(2)-2=0\]\[4-2-2=0\checkmark\] you do the other one (x=-2)

  31. whpalmer4
    • one year ago
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    we are just verifying that we solved for \(x\) correctly here...

  32. anonymous
    • one year ago
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    \[x^2-x-2 = 0\] \[-2^2-(-2)-2 =0\] 4+2-2 = 0 4 = 0

  33. whpalmer4
    • one year ago
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    Hmm. maybe one of those solutions is incorrect :-)

  34. anonymous
    • one year ago
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    -2 won't be a solution

  35. whpalmer4
    • one year ago
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    by the way, you really need to write \[(-2)^2\]instead of \[-2^2\] The first one is \((-2)*(-2) = 4\) and the second one is \(-(2)(2) = -4\)

  36. anonymous
    • one year ago
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    Oh, sorry.

  37. whpalmer4
    • one year ago
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    I'll speed things along a little bit: \[\frac{1\pm\sqrt{9}}{2} = \frac{1+3}{2},\frac{1-3}{2} = \frac{4}{2},\frac{-2}{2} = 2,-1\]

  38. anonymous
    • one year ago
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    Thanks a million!

  39. whpalmer4
    • one year ago
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    and if we try \(-1\) in the equation: \[(-1)^2 -(-1)-2=0\]\[1+1-2=0\checkmark\] Now, you can plug those two values of \(x\) into one of the equations and find the corresponding values of \(y\).

  40. anonymous
    • one year ago
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    okay so then y will equal 7 and -4

  41. anonymous
    • one year ago
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    (2,7) (-1, -4)

  42. whpalmer4
    • one year ago
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    Yep! And if you plug them all in to all the equations, they should all work.

  43. anonymous
    • one year ago
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    Great! Thank you :)

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