You can
put this solution on YOUR website!lets call the cost of singles s and the cost of doubles d
so:
5s+12d=390....eq 1
9s+10d=412....eq 2
lets eliminate one of the variables by multiplying eq 1 by -9 and eq 2 by 5
-45s-108d=-3510............now add these two together and the s terms are
45s+50d =2060.............eliminated.
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-58d=-1450
cost of a double
:
now plug in d's value into eq 1 or 2. I chose eq 1
5s+12(25)=390
5s+300=390
5s=90
cost of a single
