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At Imelda's fruit stand, you bought 5 apples and 4 oranges for $10, and your friend bought 5 apples and 5 oranges for $11. Using this information, is it possible to determine the cost of one apple and one orange from the fruit stand? If so, what do they cost? If not, why not? Choose 1 answer: (A) Yes; they should charge $1.00 for an apple and $1.25 for an orange. (B) Yes; they should charge $1.20 for an apple and $1.00 for an orange. (c) No; the system has many solutions. (D) No; the system has no solution.

At Imelda's fruit stand, you bought 55 apples and 44 oranges for $10 \$ 10 , and your friend bought 55 apples and 55 oranges for $11 \$ 11 .\newlineUsing this information, is it possible to determine the cost of one apple and one orange from the fruit stand? If so, what do they cost? If not, why not?\newlineChoose 11 answer:\newline(A) Yes; they should charge $1.00 \$ 1.00 for an apple and $1.25 \$ 1.25 for an orange.\newlineB Yes; they should charge $1.20 \$ 1.20 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. At Imelda's fruit stand, you bought 55 apples and 44 oranges for $10 \$ 10 , and your friend bought 55 apples and 55 oranges for $11 \$ 11 .\newlineUsing this information, is it possible to determine the cost of one apple and one orange from the fruit stand? If so, what do they cost? If not, why not?\newlineChoose 11 answer:\newline(A) Yes; they should charge $1.00 \$ 1.00 for an apple and $1.25 \$ 1.25 for an orange.\newlineB Yes; they should charge $1.20 \$ 1.20 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. Write equations: Write the system of equations based on the information given.\newlineYou bought 55 apples and 44 oranges for $10\$10.\newlineYour friend bought 55 apples and 55 oranges for $11\$11.\newlineLet xx be the cost of one apple and yy be the cost of one orange.\newlineFirst equation: 5x+4y=105x + 4y = 10\newlineSecond equation: 5x+5y=115x + 5y = 11
  2. Subtract equations: Subtract the first equation from the second equation to solve for yy.
    (5x+5y)−(5x+4y)=11−10(5x + 5y) - (5x + 4y) = 11 - 10
    5x+5y−5x−4y=15x + 5y - 5x - 4y = 1
    y=1y = 1
  3. Solve for y: Substitute the value of yy into the first equation to solve for xx.5x+4(1)=105x + 4(1) = 105x+4=105x + 4 = 105x=10−45x = 10 - 45x=65x = 6x=65x = \frac{6}{5}x=1.20x = 1.20
  4. Substitute for x: Verify the solution by substituting the values of x and y into the second equation.\newline55(11.2020) + 55(11) = 1111\newline66 + 55 = 1111\newline1111 = 1111\newlineThe solution checks out.

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