Question 1195103
let blue envelopes be {{{b}}} and red envelopes {{{r}}}


John had a total of {{{319}}}: so, 


{{{b+r=319}}}.........eq..1


If {{{1/3}}} the number of blue envelopes was the same as {{{2/5}}} the number of red envelopes, we have


{{{(1/3)b=(2/5)r}}}

{{{b=((2/5)/(1/3))r}}}

{{{b=(6/5)r}}}


go to


{{{b+r=319}}}.........eq..1, substitute {{{b}}}

{{{(6/5)r+r=319}}}.........solve for {{{r}}}

{{{(11/5)r=319}}}

{{{r=319/(11/5)}}}

{{{r=319*(5/11)}}}

{{{r=145}}}


go to


{{{b=(6/5)r}}}.....substitute {{{r}}}

{{{b=(6/5)145}}}

{{{b=174}}}


John did have  {{{174}}} blue envelopes.