Question 159650
Call the 4 consecutive integers {{{n}}},{{{n+1}}},{{{n+2}}},{{{n+3}}}
given:
{{{2*((n+2)+(n+3)) = 3n + 91}}}
This is 1 equation and 1 unknown, so I should be able to solve
{{{2*(2n + 5) = 3n + 91}}}
{{{4n + 10 = 3n + 91}}}
{{{n + 10 = 91}}}
{{{n = 81}}}
The 4 consecutive integers are 81,82,83, and 84
check:
{{{2*((n+2)+(n+3)) = 3n + 91}}}
{{{2*((81+2)+(81+3)) = 3*81 + 91}}}
{{{2*(83 + 84) = 243 + 91}}}
{{{2*167 = 334}}}
{{{334 = 334}}}
OK