1) Create a real-world problem involving a related set of two equations
George bought 5 apples and 2 peaches, he pays $22.00. Isabella bought 10 apples and 2 peaches, she pays $32.00. a) How much does an apple cost? b) How much does a peach cost?
2)Write the system of equations that would be used to solve them.
Let x be the cost of an apple
Let y be the cost of a peach
George bought 5 apples and 2 peaches, he pays $22.00:
[tex]5x+2y=22[/tex]Isabella bought 10 apples and 2 peaches, she pays $32.00:
[tex]10x+2y=32[/tex]System of equations:
[tex]\begin{gathered} 5x+2y=22 \\ 10x+2y=32 \end{gathered}[/tex]3) Show how to solve the same system using ALL 3 methods:
• Graphing
Find two points (x,y) for each equation:
-First equation:
[tex]\begin{gathered} When\text{ }x=0 \\ 5(0)+2y=22 \\ 2y=22 \\ y=\frac{22}{2} \\ y=11 \\ Point(0,11) \\ \\ When\text{ }x=6 \\ 5(6)+2y=22 \\ 30+2y=22 \\ 2y=22-30 \\ 2y=-8 \\ y=-\frac{8}{2} \\ y=-4 \\ Point(6,-4) \end{gathered}[/tex]-Second equation:
[tex]\begin{gathered} When\text{ }x=0 \\ 10(0)+2y=32 \\ 2y=32 \\ y=\frac{32}{2} \\ y=16 \\ Point(0,16) \\ \\ When\text{ }x=4 \\ 10(4)+2y=32 \\ 40+2y=32 \\ 2y=32-40 \\ 2y=-8 \\ y=-\frac{8}{2} \\ y=-4 \\ Point(4,-4) \end{gathered}[/tex]Use each pair of points to graph the corresponding line: Put the point in the plane and draw a line that passes through the corresponding pair of points:
The solution is the point of intersection (2,6)
• Substitution:
[tex]\begin{gathered} 5x+2y=22 \\ 10x+2y=32 \end{gathered}[/tex]1. Solve x in the first equation:
[tex]\begin{gathered} 5x+2y=22 \\ 5x=22-2y \\ x=\frac{22}{5}-\frac{2}{5}y \end{gathered}[/tex]2. Substitute x in the second equation with the value you get in the first step:
[tex]10(\frac{22}{5}-\frac{2}{5}y)+2y=32[/tex]3. Solve y:
[tex]\begin{gathered} \frac{220}{5}-\frac{20}{5}y+2y=32 \\ \\ 44-4y+2y=32 \\ \\ 44-2y=32 \\ \\ -2y=32-44 \\ \\ -2y=-12 \\ \\ y=\frac{-12}{-2} \\ \\ y=6 \end{gathered}[/tex]4. Use the value of y to solve x:
[tex]\begin{gathered} x=\frac{22}{5}-\frac{2y}{5} \\ \\ x=\frac{22}{5}-\frac{2(6)}{5} \\ \\ x=\frac{22}{5}-\frac{12}{5} \\ \\ x=\frac{22-12}{5} \\ \\ x=\frac{10}{5} \\ \\ x=2 \end{gathered}[/tex]The solution is (2,6)
• Elimination:
[tex]\begin{gathered} 5x+2y=22 \\ 10x+2y=32 \end{gathered}[/tex]1. Subtract the equations:
2. Solve x:
[tex]\begin{gathered} -5x=-10 \\ x=\frac{-10}{-5} \\ x=2 \end{gathered}[/tex]3. Use the value of x to solve y:
[tex]\begin{gathered} 5x+2y=22 \\ 5(2)+2y=22 \\ 10+2y=22 \\ 2y=22-10 \\ 2y=12 \\ y=\frac{12}{2} \\ y=6 \end{gathered}[/tex]The solution is (2,6)
4) Identify your solution & explain what it means in the context of your word problem!
