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A fruit stand has to decide what to charge for their produce. They need $10 for 4 apples and 4 oranges. They also need $15 for 6 apples and 6 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 $1.00 for an apple and $1.50 for an orange. (B) Yes; they should charge $1.50 for an apple and $1.00 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 $10 \$ 10 for 44 apples and 44 oranges. They also need $15 \$ 15 for 66 apples and 66 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 $1.00 \$ 1.00 for an apple and $1.50 \$ 1.50 for an orange.\newlineB Yes; they should charge $1.50 \$ 1.50 for an apple and $1.00 \$ 1.00 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 $10 \$ 10 for 44 apples and 44 oranges. They also need $15 \$ 15 for 66 apples and 66 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 $1.00 \$ 1.00 for an apple and $1.50 \$ 1.50 for an orange.\newlineB Yes; they should charge $1.50 \$ 1.50 for an apple and $1.00 \$ 1.00 for an orange.\newline(C) No; the system has many solutions.\newline(D) No \mathrm{No} ; the system has no solution.
  1. Define Equations: Let's denote the price of an apple as AA and the price of an orange as OO. We can then write the given information as a system of linear equations:\newline11. 4A+4O=($)104A + 4O = (\$)10\newline22. 6A+6O=($)156A + 6O = (\$)15
  2. Simplify Equations: We can simplify both equations by dividing them by their common factors to make the coefficients smaller and the system easier to work with:\newline11. Divide the first equation by 44:\newlineA+O=$(2.50)A + O = \$(2.50)\newline22. Divide the second equation by 66:\newlineA+O=$(2.50)A + O = \$(2.50)
  3. Identify Identical Equations: We notice that after simplification, both equations are identical, which means they represent the same line. Therefore, there are infinitely many solutions to this system, as any point on the line A+O=$(2.50)A + O = \$(2.50) is a solution.
  4. Infinite Solutions: Since there are infinitely many solutions, we cannot find a unique price for an apple and an orange. The correct answer is (C)(C) No; the system has many solutions.

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