Question 704957

The sum of two digits is {{{66}}}. If the smaller number is subtracted from two thirds of the larger number, the result is one-third the positive difference of the original number. 

if {{{x}}} is 1st number, and {{{y}}} 2nd number,you are given that       

{{{x+y=66}}}.........eq.1


If the smaller number (let {{{x}}} be smaller number)  is subtracted from two thirds of the larger number (let {{{y}}} be larger number), the result is one-third the positive difference (which is {{{y-x}}}) of the original number.

{{{(2/3)y-x=(1/3)(y-x).......eq.2 


solve the system:


{{{x+y=66}}}.........eq.1
{{{(2/3)y-x=(1/3)(y-x).......eq.2 
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{{{x+y=66}}}.........eq.1..solve for {{{x}}}

{{{x=66-y}}}.......substitute in eq.2

{{{(2/3)y-(66-y)=(1/3)(y-(66-y)) }}}

{{{(2/3)y-66+y=(1/3)(y-66+y))}}}

{{{2y/3-66+y=y/3-66/3+y/3}}}

{{{2y/3-66+y=2y/3-22}}}

{{{cross(2y/3)-66+y=cross(2y/3)-22}}}

{{{-66+y=-22}}}

{{{y=-22+66}}}

{{{highlight(y=44)}}}


{{{x=66-y}}}

{{{x=66-44}}}

{{{highlight(x=22)}}}