One sub costs $4.5 and one cookie costs $0.75 or 75 cents
Step-by-step explanation:
Let s be the cost of one sub
and
c be the cost of one cookie
Then according to given statements
"The cost of 3 subs and 7 cookies is $18.75"
[tex]3s+7c = 18.75\ \ \ Eqn\ 1[/tex]
"Five subs and 21 cookies cost $38.25"
[tex]5s+21c = 38.25\ \ \ \ Eqn\ 2[/tex]
Multiplying eqn 1 by 3
[tex]9s+21c = 56.25\ \ \ Eqn\ 3[/tex]
Subtracting Eqn 2 from Eqn 3
[tex]9s+21c- (5s+21c) = 56.25-38.25\\9s+21c-5s-21c = 18\\4s = 18[/tex]
Dividing both sides by 4
[tex]\frac{4s}{4} =\frac{18}{4}\\s = 4.5[/tex]
Putting s=4.5 in Eqn 1
[tex]3s+7c = 18.75\\3(4.5)+7c = 18.75\\13.5+7c = 18.75\\7c = 18.75-13.5\\7c =5.25[/tex]
Dividing both sides by 7
[tex]\frac{7c}{7}=\frac{5.25}{7}\\c = 0.75[/tex]
So,
One sub costs $4.5 and one cookie costs $0.75 or 75 cents
Keywords: Linear Equations, Variables
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