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FRUIT The price of 2 pears and 6 apples is $14. The price of 3 pears and 9 apples is $21. Can you determine the unit prices for pears and apples? Explain.

55. FRUIT The price of 22 pears and 66 apples is $14 \$ 14 . The price of 33 pears and 99 apples is $21 \$ 21 . Can you determine the unit prices for pears and apples? Explain.

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Q. 55. FRUIT The price of 22 pears and 66 apples is $14 \$ 14 . The price of 33 pears and 99 apples is $21 \$ 21 . Can you determine the unit prices for pears and apples? Explain.
  1. Set up equations: Let's call the price of one pear pp and the price of one apple aa. We can set up two equations based on the information given:\newlineEquation 11: 2p+6a=$(14)2p + 6a = \$(14)\newlineEquation 22: 3p+9a=$(21)3p + 9a = \$(21)
  2. Simplify Equation 22: First, let's simplify Equation 22 by dividing everything by 33 to make it easier to compare with Equation 11:\newline3p+9a=$(21)3p + 9a = \$(21) becomes p+3a=$(7)p + 3a = \$(7)
  3. Subtract equations: Now we have two simplified equations:\newlineEquation 11: 2p+6a=$(14)2p + 6a = \$(14)\newlineEquation 22: p+3a=$(7)p + 3a = \$(7)
  4. Correction: Subtract Equation 22 from Equation 11 to eliminate aa:(2p+6a)−(p+3a)=($)14−($)7(2p + 6a) - (p + 3a) = (\$)14 - (\$)7This simplifies to p+3a=($)7p + 3a = (\$)7Oops, we made a mistake here. We should have gotten p+3a=($)7p + 3a = (\$)7 after subtracting, but we already have that as Equation 22. Let's correct this.

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