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Marc sold 500 tickets for the school play. Student tickets cost $2 and adult tickets cost $5. Marc's sales totaled $1,789.a.) Write a system of equations that can be used to determine how many of each type of ticket Marc sold.b.) Solve the system to determine how many adult tickets and how many student tickets Marc sold.

Sagot :

Given that:

- Marc sold 500 tickets for the school play.

- Student tickets cost $2.

- Adult tickets cost $5.

- The total amount of money from the sale was $1,789.

a.) Let be "x" the number of student tickets Marc sold for the school play, and "y" the number of adult tickets Marc sold for the school play.

You can represent the total number of tickets he sold with this equation:

[tex]x+y=500[/tex]

You can represent the total amount of money from the sale with this equation:

[tex]2x+5y=1789[/tex]

Now you can set up this System of Equations that can be used to determine how many of each type of ticket Marc sold:

[tex]\begin{cases}x+y={500} \\ 2x+5y={1789}\end{cases}[/tex]

b.) You can solve the System of Equations using the Elimination Method:

1. Multiply the first equation by -2:

[tex]\begin{cases}(-2)(x+y)=(-2)({500)} \\ 2x+5y={1789}\end{cases}[/tex]

[tex]\begin{cases}-2x-2y={-1000} \\ 2x+5y={1789}\end{cases}[/tex]

2. Add the equations:

[tex]\begin{gathered} \begin{cases}-2x-2y={-1000} \\ 2x+5y={1789}\end{cases} \\ ---------- \\ 0+3y=789 \\ 3y=789 \end{gathered}[/tex]

3. Solve for "y":

[tex]\begin{gathered} y=\frac{789}{3} \\ \\ y=263 \end{gathered}[/tex]

4. Substitute the value of "y" into the first original equation and solve for "x":

[tex]\begin{gathered} x+263=500 \\ x=500-263 \\ x=237 \end{gathered}[/tex]

Hence, the answers are:

a.)

[tex]\begin{cases}x+y={500} \\ 2x+5y={1789}\end{cases}[/tex]

b.)

[tex]x=237\text{ \lparen Number of student tickets Marc sold\rparen}[/tex][tex]y=263\text{ \lparen Number of adult tickets Marc sold\rparen}[/tex]

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