Math

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Arectangle has perimeter 64cm and area 23cmsquared. Solve the following system of equations to find the rectangle's dimensions.

  • Math -

    the equations are:
    l= (23/w)
    l+w=32

  • Math -

    so sub l= (23/w) into l+w = 32
    23/w + w = 32
    23 + w^2 = 32w
    w^2 - 32w + 23 = 0

    solve for w, then plug into l=23/w

    I got w = .7357
    l = 31.2643

    check: .7357x31.2643 = 23.001
    .7357 + 31.2643 = 32

  • Math -

    thankyou !

    the second part to the question is.:
    solve the system of equations:
    x^2 + y^2 = 1
    xy = 0.5

  • Math -

    do it the same way
    from xy = .5 = 1/2
    y = 1/(2x)

    sub into the first
    x^2 + 1/(4x^2) = 1
    4x^4 + 1 = 4x^2
    4x^4 - 4x^2 + 1 = 0
    let x^2 = p
    then your equation becomes
    4p^2 - 4p + 1 = 0
    (2p-1)(2p-1) = 0
    p = 1/2
    so x^2 = 1/2
    x = +/- 1/√2
    sub back into xy=1/2 to get the y

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