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Solving a Fruit Sale Equation: Matt and Ming's Fundraiser Challenge

In this algebraic problem, Matt and Ming are raising funds by selling boxes of oranges for their school. Matt sold 3 small and 14 large boxes, totaling $203, while Ming sold 11 small and 11 large boxes for $220. The challenge is to determine the cost of each type of box using elimination and substitution methods. This scenario illustrates how algebra can simplify complex problems, making them easier to solve. Explore how to find the prices of small and large boxes step-by-step!

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Solving a Fruit Sale Equation: Matt and Ming's Fundraiser Challenge

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  1. System of Equations Keyona Nettles

  2. LOGO

  3. Slogan • Algebra makes the simplest things complicated , and the complicated things simple .

  4. Problem • Matt and Ming are selling fruit for a school fundraiser. Costumers can buy small boxes of oranges and large boxes of oranges. Matt sold 3 small boxes and 14 large boxes of oranges for a total of $203. Ming sold 11 small boxes of oranges and 11 large boxes of oranges for a total of $220. Find the cost each of the one small box of oranges and one large box of oranges .

  5. Elimination • -11(3x+14=203 11y+11(13)=220 • -33y-154x=2233 11y+143=220 • 33y + 33x=660 -143 -143 • -121x=1573 11y=77 • x=13 y=7

  6. Substitution • 11x+14y=220 • 11

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