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Math Help - System of second degree equations and inequalities?

  1. #1
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    System of second degree equations and inequalities?

    Can anyone help me answer these two questions?

    1. Find the coordinates of the point of intersection for the graphs
    x^2=25-9y^2
    xy=-4

    I am slightly confused on how I would find x and y

    The second question is graph each system of linear inequalities

    x^2+4y less than 16

    x^2 less than y^2+4

    What I do not understand in this question is how I could get it to y=mx+b

    Because whenever I move it I end up getting a negative?
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  2. #2
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    Quote Originally Posted by homeylova223 View Post
    Can anyone help me answer these two questions?

    1. Find the coordinates of the point of intersection for the graphs
    x^2=25-9y^2
    xy=-4

    I am slightly confused on how I would find x and y
    From xy= -4 you get y= -4/x. Put that into the first equation: x^2= 25- 144/x^2. Multiply through by x^2: x^4= 25x^2- 144 which is the same as x^4- 25x^2+ 144= 0. Let u= x^2 so the equation becomes u^2- 25u+ 144= 0. Solve that for two values of u. Solve x^2= u for each of those values of u to find four values of x. Finally, solve y= -4/x for each value of x to get four (x, y) pairs.

    The second question is graph each system of linear inequalities

    x^2+4y less than 16

    x^2 less than y^2+4

    What I do not understand in this question is how I could get it to y=mx+b
    You can't! Why would you think you could? y= mx+ b is linear and equations involving x^2 are not- their graphs are parabolas.

    Because whenever I move it I end up getting a negative?
    What's do you mean? What's wrong with negtives?

    The best way to solve an inequality is to start by solving the equation.
    x^2+ 4y= 16 is the same as 4y= 16- x^2 or y= 4- x^2/4. That's a parabola, opening downward, with vertex at (0, 4), and symmetric about the y-axis. Graph that parabola. The point is that "x^2+ 4y= 16" separates "x^2+ 4y> 16" from "x^2+ 4y< 16". That parabola separates the plane into two parts, one which satisfies x^2+ 4y> 16 and the other x^2+ 4y< 16. I see that (0, 0) (which is below the parabola) satisfies 0^2+ 4(0)= 0< 16 so every point below the parabola satisfies x^2+ 4y< 16 and every point above it satisfies x^2+ 4y> 16.
    Last edited by HallsofIvy; May 20th 2011 at 03:19 AM.
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  3. #3
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    I made an error when typing the problem I mean
    x^2+4y^2 less than 16

    I am sorry.
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  4. #4
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    Again, start by graphing the equality: what does the graph of x^2+ 4y^2= 16 look like? (It is the same as \frac{x^2}{16}+ \frac{y^2}{4}= 1.)

    As before points on that graph separate "> 16" from "< 16". Which points satisfy "< 16"?
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