A theatre can seat a total of $705$705 people. Tickets cost $\$2$$2 for children, $\$4$$4 for students, and $\$6$$6 for adults. At a recent sold out screening, the combined total of children and students was equal to twice the number of adults. The money made from ticket sales totalled $\$2842$$2842.

a

Let $x$`x` be the number of children, $y$`y` be the number of students, and $z$`z` be the number of adults.

Complete the system of equations:

$x$x |
$+$+ | $y$y |
$+$+ | $z$z |
$=$= | $\editable{}$ |

$\editable{}x$x |
$+$+ | $\editable{}y$y |
$+$+ | $\editable{}z$z |
$=$= | $\editable{}$ |

$x$x |
$+$+ | $y$y |
$=$= | $\editable{}z$z |

b

In the following subproblems we will complete one approach to solving the system of equations using the elimination method:

$x$x |
$+$+ | $y$y |
$+$+ | $z$z |
$=$= | $705$705 | ----- equation $1$1 |

$2x$2x |
$+$+ | $4y$4y |
$+$+ | $6z$6z |
$=$= | $2842$2842 | ----- equation $2$2 |

$x$x |
$+$+ | $y$y |
$=$= | $2z$2z |
----- equation $3$3 |

Notice that $x+y$`x`+`y` appears in equation $1$1 and in equation $3$3. In equation $3$3, $x+y=2z$`x`+`y`=2`z`, and so we can make a substitution in place of $x+y$`x`+`y` in equation $1$1.

Step 1: Substitute equation $3$3 into equation $1$1.

$2$2$\editable{}$ | $+$+ | $z$z |
$=$= | $705$705 |

$\editable{}$ | $=$= | $705$705 | ||

$z$z |
$=$= | $\editable{}$ |

c

Now that we have a value for $z$`z`, we can plug this value into equation $1$1.

Step 2: Substitute the value of $z$`z` into equation $1$1.

$x$x |
$+$+ | $y$y |
$+$+ | $\editable{}$ | $=$= | $705$705 | ----- equation $1$1 |

$x$x |
$+$+ | $y$y |
$=$= | $\editable{}$ | ----- equation $4$4 |

We can also plug this value of $z$`z` into equation $2$2.

Step 3: Substitute the value of $z$`z` into equation $2$2.

$2x$2x |
$+$+ | $4y$4y |
$+$+ | $6\times\editable{}$6× | $=$= | $2842$2842 | ---- equation $2$2 |

$2x$2x |
$+$+ | $4y$4y |
$+$+ | $\editable{}$ | $=$= | $2842$2842 | |

$2x$2x |
$+$+ | $4y$4y |
$=$= | $\editable{}$ | ----- equation $5$5 |

d

Now we have two linear equations in two unknowns ($x$`x` and $y$`y`).

We would like to use the elimination method here, so we need both of the equations to have the same coefficient for one of the unknowns. Let's make the $x$`x` coefficient the same by multiplying every term in equation $4$4 by $2$2.

Step 4: Multiply equation $4$4 by $2$2 so we can use the elimination method.

$x$x |
$+$+ | $y$y |
$=$= | $470$470 | ----- equation $4$4 |

$2x$2x |
$+$+ | $2y$2y |
$=$= | $\editable{}$ | ----- equation $6$6 |

e

Now that equation $4$4 and equation $6$6 have the same $x$`x` coefficient we can subtract one from the other to get an equation with only one unknown.

Step 5: Subtract equation $6$6 from equation $5$5.

$2x$2x |
$+$+ | $4y$4y |
$=$= | $1432$1432 | ----- equation $5$5 |

$2x$2x |
$+$+ | $2y$2y |
$=$= | $940$940 | ----- equation $6$6 |

$\editable{}y$y |
$=$= | $\editable{}$ | |||

$y$y |
$=$= | $\editable{}$ |

Now we have a value for $y$`y`, we can substitute this value back into equation $4$4.

Step 6: Substitute the value of $y$`y` into equation $4$4.

$x$x |
$+$+ | $\editable{}$ | $=$= | $470$470 | ----- equation $4$4 |

$x$x |
$=$= | $\editable{}$ |

f

Deduce the number of:

Children = $\editable{}$.

Students = $\editable{}$.

Adults = $\editable{}$.

Easy

Approx 9 minutes

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Form and use systems of simultaneous equations, including three linear equations and three variables, and interpret the solutions in context

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