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A fruit stand has to decide what to charge for their produce. They need $5.30 for 1 apple and 1 orange. They also need $7.30 for 1 apple 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 $3.30 for an apple and $2.00 for an orange. (B) Yes; they should charge $2.00 for an apple and $3.30 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 $5.30 \$ 5.30 for 11 apple and 11 orange. They also need $7.30 \$ 7.30 for 11 apple and 22 oranges. We put this information into a system of linear equations.\newlineCan we find a unique price for an apple and an orange?\newlineChoose 11 answer:\newline(A) Yes; they should charge $3.30 \$ 3.30 for an apple and $2.00 \$ 2.00 for an orange.\newlineB Yes; they should charge $2.00 \$ 2.00 for an apple and $3.30 \$ 3.30 for an orange.\newline(C) No; the system has many solutions.\newline(D) 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 $5.30 \$ 5.30 for 11 apple and 11 orange. They also need $7.30 \$ 7.30 for 11 apple and 22 oranges. We put this information into a system of linear equations.\newlineCan we find a unique price for an apple and an orange?\newlineChoose 11 answer:\newline(A) Yes; they should charge $3.30 \$ 3.30 for an apple and $2.00 \$ 2.00 for an orange.\newlineB Yes; they should charge $2.00 \$ 2.00 for an apple and $3.30 \$ 3.30 for an orange.\newline(C) No; the system has many solutions.\newline(D) No \mathrm{No} ; the system has no solution.
  1. Define Prices: Let's define the price of an apple as AA and the price of an orange as OO. We can then translate the given information into two equations:\newline11 apple + 11 orange = $(5.30)\$(5.30)\newline11 apple + 22 oranges = $(7.30)\$(7.30)\newlineThis can be written as the system of linear equations:\newlineA+O=5.30A + O = 5.30\newlineA+2O=7.30A + 2O = 7.30
  2. Use Elimination Method: To solve this system, we can use the method of substitution or elimination. Let's use elimination. We'll subtract the first equation from the second to eliminate AA and find the price of an orange.(A+2O)−(A+O)=7.30−5.30(A + 2O) - (A + O) = 7.30 - 5.30This simplifies to:2O−O=2.002O - O = 2.00O=2.00O = 2.00
  3. Substitute to Find Apple Price: Now that we have the price of an orange, we can substitute it back into the first equation to find the price of an apple.\newlineA+O=5.30A + O = 5.30\newlineA+2.00=5.30A + 2.00 = 5.30\newlineSubtracting 2.002.00 from both sides gives us:\newlineA=5.30−2.00A = 5.30 - 2.00\newlineA=3.30A = 3.30
  4. Final Prices: We have found unique prices for both an apple and an orange:\newlineA (apple) = $3.30\$3.30\newlineO (orange) = $2.00\$2.00\newlineThis corresponds to answer choice (A)(A).

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