Science

Math

Humanities

... and beyond

Socratic Meta
Featured Answers

Topics

The sum of the reciprocals of two consecutive even integers is 9/40, what are the integers?

1 Answer

May 30, 2015

If the smaller of the two consecutive even integers is $x$ $x$
then, we are told,
$\frac{1}{x} + \frac{1}{x + 2} = \frac{9}{40}$ $color(red)(1/x)+color(blue)(1/(x+2)) =9/40$

So
$XXXXX$ $color(white)("XXXXX")$ generating a common denominator on the left side:
$[\frac{1}{x} \cdot \frac{x + 2}{x + 2}] + [\frac{1}{x + 2} \cdot (\frac{x}{x})] = \frac{9}{40}$ $[color(red)(1/x*(x+2)/(x+2))] + [color(blue)(1/(x+2)*(x/x))]=9/40$

$[\frac{x + 2}{x^{2} + 2 x}] + [\frac{x}{x^{2} + 2 x}] = \frac{9}{40}$ $[color(red)((x+2)/(x^2+2x))] + [color(blue)((x)/(x^2+2x))]=9/40$

$\frac{(x + 2) + (x)}{x^{2} + 2 x} = \frac{9}{40}$ $(color(red)((x+2)) + color(blue)((x)))/(x^2+2x) = 9/40$

$\frac{2 x + 2}{x^{2} + 2 x} = \frac{9}{40}$ $(2x+2)/(x^2+2x) = 9/40$

$(40) (2) (x + 1) = 9 (x^{2} + 2 x)$ $(40)(2)(x+1) = 9(x^2+2x)$

$80 x + 80 = 9 x^{2} + 18 x$ $80x + 80 = 9x^2+18x$

$9 x^{2} - 62 x - 80 = 0$ $9x^2-62x-80 = 0$

$(9 x + 1) (x - 8) = 0$ $(9x+1)(x-8) = 0$

Since $x$ $x$ is an even integer
the two consecutive even integers are
$8$ $8$ and $10$ $10$

Related questions

See all questions in Applications Using Linear Models

Impact of this question

20780 views around the world

You can reuse this answer
Creative Commons License