2 1/2 1 1/16 = 3 1/16 3 1/2 3 9/16 3 3/16
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error is in constant term of quotient
Fries are $2.50 and burgers are $4.75.
Our system of equations for this situation would be
We want the coefficients of one of the variables to be the same in order to eliminate one of them; we will make the coefficients of f the same. To do this, multiply the top equation by 5 and the bottom equation by 3:
We subtract the bottom equation:
Divide both sides by 5:
5b/5 = 23.75/5
b = 4.75
Burgers are $4.75.
Substitute this into the first equation:
4(4.75) + 3f = 26.50
19 + 3f = 26.50
Subtract 19 from both sides:
19 + 3f - 19 = 26.50 - 29
3f = 7.50
Divide both sides by 3:
3f/3 = 7.50/3
f = 2.50
Our system of equations for this situation would be
We want the coefficients of one of the variables to be the same in order to eliminate one of them; we will make the coefficients of f the same. To do this, multiply the top equation by 5 and the bottom equation by 3:
We subtract the bottom equation:
Divide both sides by 5:
5b/5 = 23.75/5
b = 4.75
Burgers are $4.75.
Substitute this into the first equation:
4(4.75) + 3f = 26.50
19 + 3f = 26.50
Subtract 19 from both sides:
19 + 3f - 19 = 26.50 - 29
3f = 7.50
Divide both sides by 3:
3f/3 = 7.50/3
f = 2.50