Question 1148560
Let the numbers be {{{ n }}}, {{{ n+2 }}}, {{{ n + 4 }}}
{{{ (3/5)*n + (1/2)*( n + 2 ) + (3/8)*( n + 4 ) = 32 }}}
Multiply both sides by {{{ 40 }}}
{{{ 24n + 20*( n + 2 ) + 15*( n + 4 ) = 1280 }}}
{{{ 24n + 20n + 40 + 15n + 60 = 1280 }}}
{{{ 59n = 1180 }}}
{{{ n = 20 }}}
{{{ n + 2 = 22 }}}
{{{ n + 4 = 24 }}}
The numbers are 20, 22, and 24
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check:
{{{ (3/5)*n + (1/2)*( n + 2 ) + (3/8)*( n + 4 ) = 32 }}}
{{{ (3/5)*20 + (1/2)*22 + (3/8)*24 = 32 }}}
{{{ 12 + 11 + 9 = 32 }}}
{{{ 32 = 32 }}}
OK