Question 358617
the sum of two numbers is the same as their product, and the difference of thier recoprocals is 3. find the numbers
<pre>
Let the numbers be x and y

The sum of the the numbers = x + y

The product of the numbers = xy

The reciprocal of x is {{{1/x}}}

The reciprocal of y is {{{1/y}}}

"the sum of two numbers x+y is the same as their product xy",

so  

{{{x + y}}}{{{""=""}}}{{{xy}}} 

"the difference of their recoprocals {{{1/x}}}{{{""-""}}}{{{1/y}}} is 3",

so  

{{{1/x}}}{{{""-""}}}{{{1/y}}}{{{""=""}}}{{{3}}}

So we have the system of equations 

{{{system(x+y=xy,1/x-1/y=3)}}}

Clear the second equation of fractions by multiplying through by LCD xy

{{{xy}}}{{{""*""}}}{{{1/x}}}{{{""-""}}}{{{xy}}}{{{""*""}}}{{{1/y}}}{{{""=""}}}{{{xy}}}{{{""*""}}}{{{3}}}

{{{cross(x)y}}}{{{""*""}}}{{{1/cross(x)}}}{{{""-""}}}{{{x*cross(y)}}}{{{""*""}}}{{{1/cross(y)}}}{{{""=""}}}{{{3xy}}}

{{{y-x}}}{{{""=""}}}{{{3xy}}}

So the system  of equations is now

{{{system(x+y=xy,y-x=3xy)}}}

Write the left side of the first equation in reverse order:

{{{system(y+x=xy,y-x=3xy)}}}

Adding the two equations term by term,

2y = 4xy

Get 0 on the right side:

2y - 4xy = 0

Divide through by 2

y - 2xy = 0

y(1 - 2x) = 0

Set each factor = 0

y = 0;   1 - 2x = 0

            -2x = -1

              x = {{{1/2}}}

y = 0 must be discarded because the reciprocal of 0 is not defined.

So x = {{{1/2}}}  

Substituting in 

{{{x+y}}}{{{""=""}}}{{{xy}}}
 
{{{1/2+y}}}{{{""=""}}}{{{expr(1/2)y}}}

Multiplying through by LCD of 2

1 + 2y = y

     y = -1

So the numbers are {{{1/2}}} and {{{-1}}}

Checking:

The sum of the the numbers = {{{1/2+(-1)=1/2-1=1/2-2/2=-1/2}}}

The product of the numbers = {{{expr(1/2)(-1) =-1/2}}}

That checks because they are both the same.

The reciprocal of {{{1/2}}} is 2

The reciprocal of -1 is {{{1/(-1)}}} or -1.

The difference of these reciprocals is {{{2-(-1)=2+1=3}}}

So that checks.

Edwin</pre>