Question 978389


Let &nbsp;<B>x</B>&nbsp; and &nbsp;<B>y</B>&nbsp; be our numbers.


Then we have a system of two equations 


{{{system(xy = 405,
x = 2y-3)}}}.


Substitute the second equation into the first one, &nbsp;and you get

{{{(2y-3)*y}}} = {{{405}}}.


Simplify it:

{{{2y^2 - 3y - 405 = 0}}}.


Now apply the quadratic formula


{{{y[1]}}} = {{{(3 + sqrt(9+4*2*405))/4)}}} = {{{(3 + sqrt(3249))/4}}} = {{{(3 + 57)/4}}} = {{{15}}}, 


{{{y[2]}}} = {{{(3 - sqrt(9+4*2*405))/4)}}} = {{{(3 - sqrt(3249))/4}}} = {{{(3 - 57)/4}}} = {{{-13.5}}}. 


Since we are asked about the integer solutions, &nbsp;only &nbsp;{{{y[1]=3}}}&nbsp; satisfies. 

It gives &nbsp;{{{x}}} = {{{2y - 3}}} = {{{2*15-3}}} = {{{27}}}. 


<B>Answer</B>. &nbsp;The numbers &nbsp;15&nbsp; and &nbsp;27&nbsp; are the solution.