Question 1099450
<br>
Let g, s, and b represent the numbers of gold, silver, and bronze tokens.<br>
{{{14g+20s+24b = 20}}}  (1)
{{{20g+15s+19b = 20}}}  (2)
{{{30g+5s+13b = 20}}}  (3)<br>
To solve the system by hand, I see that the coefficients of s in the three equations are multiples of 5, so I would use elimination to reduce the system to two equations in g and b.<br>
3 times equation (3) minus equation (2) gives {{{70g+20b = 40}}} or {{{35g+10b = 20}}}  (4)
4 times equation (3) minus equation (1) gives {{{106g+28b = 60}}} or {{{53g+14b = 30}}}  (5)<br>
5 times equation (5) minus 7 times equation (4) eliminates b, giving us {{{20g=10}}}, so g = 0.5<br>
Substituting back in earlier equations then gives us s = 0.35 and b = 0.25.