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A fruit stand has to decide what to charge for their produce.They decide to charge $5.30 for 1 apple and 1 orange. They also plan to charge $14 for 2 apples and 2 oranges. We put this information into a system of linear equations. Can we find a unique price for an apple and an orange?

User Jpnavarini
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Final answer:

We can solve for the unique prices of an apple and an orange using a system of linear equations, but in this case, the equations given are inconsistent, indicating an error in the fruit stand's pricing strategy.

Step-by-step explanation:

Yes, we can find a unique price for an apple and an orange using the given system of linear equations. The first equation is formed from the statement 'They decide to charge $5.30 for 1 apple and 1 orange,' which can be written as 'A + O = 5.30,' where A represents the price of an apple and O represents the price of an orange. The second equation comes from 'They also plan to charge $14 for 2 apples and 2 oranges,' which gives us '2A + 2O = 14.' This can be simplified to 'A + O = 7' by dividing each term by 2.

When we look at these two equations:
1. A + O = 5.30
2. A + O = 7
We see immediately that they are inconsistent and actually describe two parallel lines, which means there's no unique solution to this system; thus, something might be wrong with the fruit stand's pricing strategy.

However, if the statement meant to communicate different prices for different quantities—which is a common scenario in pricing strategies—we would have distinct equations and could proceed with solving the system for unique prices of apples and oranges.

User Artem Makarov
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Answer:

No, we cannot.

Step-by-step explanation:

Let a represent the number of apples and r represent the number of oranges.

The first equation would then be

1a + 1r = 5.30

The second equation would be

2a + 2r = 14

We can see that the coefficients in the second equation are double those in the first equation. However, if we multiply the first equation by 2, we get

2(1a + 1r = 5.30) → 2a + 2r = 10.60

The constant at the end of the equation is not 14, but the coefficients of the variables are the same. This means this describes a line parallel to this line; this means they will not intersect and thus there is no solution to this system.

User Qts
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