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A fruit stand has to decide what to charge for their produce. They need $5 for 1 apple and 1 orange. They also need $15 for 3 apples and 3 oranges. 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 $2.00 for an apple and $3.00 for an orange. (B) Yes; they should charge $1.00 for an apple and $4.00 for an orange. (c) No; the system has many solutions. (D) No; the system has no solution.

A fruit stand has to decide what to charge for their produce. They need $5 \$ 5 for 11 apple and 11 orange. They also need $15 \$ 15 for 33 apples and 33 oranges. Put this information into a system of linear equations.\newlineCan we find a unique price for an apple and an orange?\newlineChoose 11 answer:\newlineA Yes; they should charge $2.00 \$ 2.00 for an apple and $3.00 \$ 3.00 for an orange.\newlineB Yes; they should charge $1.00 \$ 1.00 for an apple and $4.00 \$ 4.00 for an orange.\newline(C) No; the system has many solutions.\newlineD No \mathrm{No} ; the system has no solution.

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Q. A fruit stand has to decide what to charge for their produce. They need $5 \$ 5 for 11 apple and 11 orange. They also need $15 \$ 15 for 33 apples and 33 oranges. Put this information into a system of linear equations.\newlineCan we find a unique price for an apple and an orange?\newlineChoose 11 answer:\newlineA Yes; they should charge $2.00 \$ 2.00 for an apple and $3.00 \$ 3.00 for an orange.\newlineB Yes; they should charge $1.00 \$ 1.00 for an apple and $4.00 \$ 4.00 for an orange.\newline(C) No; the system has many solutions.\newlineD No \mathrm{No} ; the system has no solution.
  1. Equations based on fruit cost: Let's denote the cost of an apple as 'a' and the cost of an orange as 'o'. The information given can be translated into two equations based on the cost of the fruits.\newlineFirst equation: a+o=5a + o = 5 (since $5\$5 for 11 apple and 11 orange)\newlineSecond equation: 3a+3o=153a + 3o = 15 (since $15\$15 for 33 apples and 33 oranges)
  2. Simplifying the second equation: We can simplify the second equation by dividing all terms by 33 to make it easier to compare with the first equation.\newlineSimplified second equation: a+o=5a + o = 5
  3. Identical equations: Now we have two identical equations:\newlinea+o=5a + o = 5\newlinea+o=5a + o = 5\newlineThis means that we cannot find a unique solution for ′a′'a' and ′o′'o' because the two equations are the same and do not provide distinct information to solve for two variables.
  4. No unique solution: Since we have two identical equations, the system of equations does not have a unique solution. Instead, it has infinitely many solutions because any value of 'a' and 'o' that satisfies the equation a+o=5a + o = 5 will be a solution to the system.

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