How do you find two consecutive even integers, the sum of whose reciprocals is 3/4?

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How do you find two consecutive even integers, the sum of whose reciprocals is 3/4?

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Answer 1 · 2016-06-26T22:58:28

$\frac{1}{2} + \frac{1}{4} = \frac{3}{4}$ $1/2 + 1/4 =3/4$

Explanation:

Let the first number be $x$ $x$
The second is $x + 2$ $x+2$

$\frac{1}{x} + \frac{1}{x + 2} = \frac{3}{4}$ $1/x+ 1/(x+2) = 3/4$

$\frac{4 x (x + 2) \times 1}{x} + \frac{4 x (x + 2) \times 1}{(x + 2)}$ $color(red)(4x(x+2)xx1)/x + color(red)((4x(x+2)xx1)/((x+2))$
= $\frac{4 x (x + 2) \times 3}{4} multiply by 4 x (x + 2)$ $color(red)((4x(x+2)xx3)/4 " multiply by" color(red)(4x(x+2)$

$color(red)(4cancelx(x+2)xx1)/cancelx + color(red)((4xcancel(x+2)xx1)/(cancel(x+2))$
= $color(red)((cancel4x(x+2)xx3)/cancel4 " multiply by" color(red)(4x(x+2)$

$4(x+2) + 4x = 3x(x+2)$

$4x + 8 +4x =3x^2 + 6x$
$3x^2-2x-8=0$

$(3x+4)(x-2)=0$

$x=2 or x = -4/3 "(reject)"$

The numbers are 2 and 4.
Check:
$1/2 + 1/4 =3/4$

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How do you find two consecutive even integers, the sum of whose reciprocals is 3/4?

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