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anonymous

  • one year ago

the sum of the reciprocals of two consecutive even integers is 9/40. this can be represented by the equation shown. 1/x + 1/x+2 = 9/40 use the rational equation to determine the integers

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  1. campbell_st
    • one year ago
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    well I would suggest to use a common denominator \[x \times (x + 2) \times 40\] that way the numerators become \[\frac{40(x+2)}{40x(x+2)} + \frac{40x}{40x(x + 2)} =\frac{9x(x +2)}{40x(x+2)}\] so you now need to just solve the numerator so its \[40x + 80 + 40x = 9x^2 + 18x\] so you have a quadratic... collect the like terms and solve for x. how it helps

  2. anonymous
    • one year ago
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    Lets arrange the items here: \[\frac{ 1 }{ x } + \frac{ 1 }{ x+2 } = \frac{9}{40}\] lets equalize the denominators: \[\frac{x+2}{x^2 + 2x} + \frac{x}{x^2 + 2x} = \frac{9}{40}\] \[\frac{2x + 2}{x^2 + 2x} = \frac{9}{40}\] Multiply both sides by the denominator to get rid of denominator: \[(x^2 + 2x) \times \frac{2x + 2}{x^2 + 2x} = \frac{9}{40} \times (x^2 + 2x)\] After simplification, we will get this: \[40(2x + 2) = 9(x^2 + 2x)\] and you will have a quadratic equation. Find 'x's.

  3. anonymous
    • one year ago
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    im still confused :/

  4. anonymous
    • one year ago
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    like i dont know how to solve for x

  5. campbell_st
    • one year ago
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    ok... so you could use the general quadratic formula the equation can be written as \[9x^2 - 62x - 80 = 0\] so a = 9, b = -62 and c = -80 the formula is \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] so when you substitute you get \[x = \frac{62 \pm \sqrt{(-62)^2 - 4 \times 9 \times (-80)}}{2 \times 9}\] calculate the values and only use the integer answer

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