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Posted by on Friday, November 9, 2012 at 1:50pm.

help me solve this equation by substitution.

y^2− 4x^2 = 7
9y^2+16x^2 = 323

  • Algebra - , Friday, November 9, 2012 at 3:33pm

    from the 1st: y^2 = 4x^2 + 7
    into the 2nd

    9(4x^2 + 7) + 16x^2 = 323
    36x^2 + 63 + 16y^2 = 323
    52y^2 = 260
    y^2 = 5
    y = ± √5

    sub y^2 = 5 into the 1st:
    5 - 4x^2 = 7
    -4x^2 = -2
    x^2 = 1/2
    x = ± 1/√2

    4 solutions:
    x=√5, y= 1/√2
    x = -√5, y = 1/√2
    x = √5, y = -1/√2
    x = -√5 , y = -1/√2

    your graph would show an ellipse intersecting with a hyperbola, with 4 symmetric points of intersection

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