Question 1043768
<pre><b>

Let y = the larger integer
Let x = the smaller integer

{{{system(y-x=5,(1/x)+2(1/y)=23/66)}}}

{{{y}}}{{{""=""}}}{{{5+x}}}

Substitute (5+x) for y in

{{{(1/x)+2(1/y)}}}{{{""=""}}}{{{23/66)}}}

{{{(1/x)+2(1/(5+x))}}}{{{""=""}}}{{{23/66)}}}

{{{(1/x)+2/(5+x)}}}{{{""=""}}}{{{23/66)}}}

LCD = 66x(5+x)

Multiply each numerator by every factor of
the denominator which is not under it:

{{{1(66(5+x))+2(66x)}}}{{{""=""}}}{{{23(x(5+x)))}}}

{{{66(5+x)+132x}}}{{{""=""}}}{{{23x(5+x)}}}

{{{330+66x+132x}}}{{{""=""}}}{{{115x+23x^2}}}

{{{330+198x}}}{{{""=""}}}{{{115x+23x^2}}}

Subtract 115x from both sides

{{{330+83x}}}{{{""=""}}}{{{23x^2}}}

Subtract 23x<sup>2</sup> from both sides

to get 0 on the right side

{{{-23x^2+330+83x}}}{{{""=""}}}{{{0}}}

Arrange left side in descending order:

{{{-23x^2+83x+330}}}{{{""=""}}}{{{0}}}

Multiply through by -1:

{{{23x^2-83x-330}}}{{{""=""}}}{{{0}}}

There may be a way to factor that but
when the numbers are that big, it's
easier to use the quadratic formula,
especially if you have a calculator:

{{{x }}}{{{""=""}}}{{{ (-b +- sqrt( b^2-4ac ))/(2a) }}}

{{{x }}}{{{""=""}}}{{{ (-(-83) +- sqrt( (-83)^2-4(23)(-330) ))/(2*(23)) }}}

{{{x }}}{{{""=""}}}{{{ (83 +- sqrt(6889+30360))/46 }}}

{{{x }}}{{{""=""}}}{{{ (83 +- sqrt(37249))/46 }}}

{{{x }}}{{{""=""}}}{{{ (83 +- 193)/46 }}}

Using the +

{{{x }}}{{{""=""}}}{{{ (83 + 193)/46 }}}

{{{x=276/46}}}

{{{x}}}{{{""=""}}}{{{6}}}

Using the -

{{{x }}}{{{""=""}}}{{{ (83 - 193)/46 }}}

{{{x}}}{{{""=""}}}{{{-110/46}}}

{{{x}}}{{{""=""}}}{{{-55/23}}}

The problem calls for an integer.  That's not
an integer, so we can discard that value of x.

So x = 6

Substitute into

{{{y}}}{{{""=""}}}{{{5+x}}}

{{{y}}}{{{""=""}}}{{{5+6}}}

{{{y}}}{{{""=""}}}{{{11}}}

Answers: Larger integer = 11, Smaller integer = 6

Edwin</pre></b>