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?Choose 1 answer:(A) Yes; they should charge $3.00 for an apple and $2.30 for an orange.(B) Yes; they should charge $3.00 for an apple and $4.00 for an orange.(C) No; the system has many solutions.(D) No; the system has no solution.
Q. 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?Choose 1 answer:(A) Yes; they should charge $3.00 for an apple and $2.30 for an orange.(B) Yes; they should charge $3.00 for an apple and $4.00 for an orange.(C) No; the system has many solutions.(D) No; the system has no solution.
Equations Setup: Let's denote the price of an apple as A and the price of an orange as O. We can then write two equations based on the information given:1. For 1 apple and 1 orange, the cost is $5.30: A+O=5.302. For 2 apples and 2 oranges, the cost is $14.00: 2A+2O=14.00
Simplifying Equations: We can simplify the second equation by dividing each term by 2 to make it easier to compare with the first equation:2A+2O=14.00A+O=214.00A+O=7.00Now we have two equations:1. A+O=5.302. A+O=7.00
Comparison of Equations: By comparing the two equations, we can see that they cannot both be true at the same time because they give different sums for the same combination of one apple and one orange. This means that there is no unique solution to this system of equations.