Question 796197
To solve {{{5&5/7 : 3 = 2&3/8 : X}}} I would first convert those mixed numbers into improper fractions.
{{{575/7=5+5/7=35/7+5/7=40/7}}} and {{{2&3/8=2+3/8=16/8+3/8=19/8}}}
so the proportion can be written as
{{{(40/7) : 3 = (19/8) :X}}}
 
Solving:
{{{(40/7) : 3 = (19/8) :X}}}
Sometimes teachers tell you that in a proportion the product of the extreme numbers (the ones on the ends) is equal to the product of the middle numbers.
That would bring you to
{{{(40/7)X  = (19/8)3 }}} --> {{{(40/7)X  = 57/8 }}}
Then, to find {{{X}}}, you would divide {{{57/8}}} by {{{40/7}}}
{{{(40/7)*X  = 57/8 }}}
{{{X  = 57/8}}}÷{{{40/7}}}
{{{X  = 57/8}}} × {{{7/40}}}
{{{X  = 57*40/(8*7)}}} but {{{40=8*5}}}, so we can simplify that to
{{{X  = 57*5/7}}}
{{{X  = 285/7=280/7+5/7=40+5/7=40&5/7}}}