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Math Help - Solve Proportions using cross products

  1. #1
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    Solve Proportions using cross products

    My sister thinks these problems are unsolvable for x. I can't find any examples in my book. I know how to solve basic proportion problems, but these two are hard.

    1. x : x-3 = x+4 : x

    2. x+1 : 6 = x-1 : x

    Thanks!
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  2. #2
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    Re: Solve Proportions using cross products

    Hey BrokenRitual.

    Try getting rid of the denominators and collect the terms.

    For the first one we have x^2 + 4x =x^2 - 3x or 7x = 0 => x = 0 but this is a contradiction since x is on the denominator and can't be zero. So no solution exists.

    You try the same kind of technique for the second one.
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  3. #3
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    Re: Solve Proportions using cross products

    Hello, BrokenRitual!

    Why do you say these are "hard"?


    [1]\;\;x : x-3 \:=\: x+4 : x

    We have: . \frac{x}{x-3} \:=\:\frac{x+4}{x}

    Cross product: . (x)(x) \:=\:(x-3)(x+4)

    n . . . . . . . . . . . . . x^2 \:=\:x^2 + x - 12

    . . . . . . . . . . . . . . . 0 \:=\:x - 12

    . . . . . . . . . . . . . . . x \:=\:12




    [2]\;\; x+1 : 6 \:=\: x-1 : x

    We have: . \frac{x+1}{6} \:=\:\frac{x-1}{x}

    Cross product: . (x+1)(x) \:=\:6(x-1)

    n . . . . . . . . . . . . . x^2 + x \:=\:6x-6

    . . . . . . . . . . . x^2 - 5x + 6 \:=\:0

    . . . . . . . . . (x-2)(x-3) \:=\:0

    . . . . . . . . . . . . . . . . . . x \:=\:2,\:3
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