The difference of the reciprocals of two consecutive integers is 1/72. What are the two integers?

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Answer 1 · 2017-04-26T12:42:46

$8, 9$ $8,9$

Let the consecutive integers be $x and x + 1$ $x and x+1$

The difference of their reciprocals is equal to $\frac{1}{72}$ $1/72$

$\to \frac{1}{x} - \frac{1}{x + 1} = \frac{1}{72}$ $rarr1/x-1/(x+1)=1/72$

Simplify the left side of the equation

$\to \frac{(x + 1) - (x)}{(x) (x + 1)} = \frac{1}{72}$ $rarr((x+1)-(x))/((x)(x+1))=1/72$

$\to \frac{x + 1 - x}{x^{2} + x} = \frac{1}{72}$ $rarr(x+1-x)/(x^2+x)=1/72$

$\to \frac{1}{x^{2} + x} = \frac{1}{72}$ $rarr1/(x^2+x)=1/72$

The numerators of the fractions are equal, so as the denominators

$\to x^{2} + x = 72$ $rarrx^2+x=72$

$\to x^{2} + x - 72 = 0$ $rarrx^2+x-72=0$

Factor it

$\to (x + 9) (x - 8) = 0$ $rarr(x+9)(x-8)=0$

Solve for the values of $x$ $x$

$\Rightarrow x = - 9, 8$ $color(green)(rArrx=-9,8$

Consider the positive value to get the correct answer

So, the integers are $8$ $8$ and $9$ $9$