# algebra 2

posted by Anonymous

three computers were on sale model E computer was priced at 1/3 the price model C and model P was priced at 1/2 the price of model E on business bought one of each type of computers on sale and paid a total \$1800

1. Steve

e = c/3
p = e/2
e+c+p = 1800

c/3 + c + c/6 = 1800
3c/2 = 1800

c=1200
e=400
p=200

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