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Answer:
loaf: $0.90; roll: $0.15
Explanation:
Let b and r represent the cost of a loaf of bread and a roll, respectively.
The two purchases can be written in equation form as ...
b +6r = 1.80
2b +4r = 2.40
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The usual ways to work problems like this are "substitution" or "elimination". The latter is sometimes called "addition."
Substitution
The way these equations are written, we can see that it is relatively easy to write an expression for b from the first equation:
b = 1.80 -6r
Substituting that into the second equation, we get ...
2(1.80 -6r) +4r = 2.40
3.60 -8r = 2.40 . . . . . . . collect terms
1.20 = 8r . . . . . . . . . . add 8r-2.40
0.15 = r . . . . . . . . . divide by 8
Then substituting this into the expression for b, we have ...
b = 1.80 -6r = 1.80 -6(0.15) = 0.90
One loaf costs $0.90; one roll costs $0.15.
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Elimination
The idea with "elimination" is to combine the equations in such a way that one of the variables is eliminated in the process. Here, there are several ways that can be done.
We recognize that the b-coefficient in the second equation is double that in the first equation. So, we can subtract the second equation from double the first equation to make the b-terms disappear.
2(b +6r) -(2b +4r) = 2(1.80) -(2.40)
8r = 1.20 . . . . . . . simplify
r = 0.15 . . . . . . . divide by 8
Using the first equation, we can find b:
b + 6(0.15) = 1.80
b = 0.90 . . . . . . . . . subtract 0.90
One loaf costs $0.90; one roll costs $0.15.
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Another way we can eliminate one of the variables is to rewrite the second equation without its excess factor of 2:
b +2r = 1.20
Then subtracting the first equation from 3 times this will eliminate the r variable:
3(b +2r) -(b +6r) = 3(1.20) -(1.80)
2b = 1.80 . . . . . . . . . simplify
b = 0.90
Then, using the above version of the second equation, we have ...
0.90 +2r = 1.20
2r = 0.30
r = 0.15
One loaf costs $0.90; one roll costs $0.15.