# The play at the theater charges $7 for adults,$5 for school-age children, and $3 for babies? ## The play at the theater charges$7 for adults, $5 for school-age children, and$3 for babies. A group of people went to see the play. There were the same number of school-age children as babies and the number of adults was the same as school-age children and babies combined. If the group paid $1562, how many babies were there? ##### 2 Answers Dec 19, 2017 There were 71 babies. #### Explanation: First define some variables: Let $x$be the number of adults. Let $y$be the number of children. Let $z$be the number of babies. Now write some equations. (1) Money: $7 x + 5 y + 3 z = 1562$(2) Children vs. babies: $y = z$(3) Adults vs. Children&Babies: $x = y + z$Subbing (2) into (3) we get that $x = y + y \setminus \rightarrow x = 2 y$, which we'll call (4) for reference. Subbing (2) and (4) into (1) we get: $7 \left(2 y\right) + 5 y + 3 \left(y\right) = 1562$$14 y + 5 y + 3 y = 1562$$22 y = 1562$$y = 71$If $y = 71$, then by (2), $z = 71$also. By (4), $x = 142$. Subbing these values back into (1) confirms that this solution works. Since $z$is the number of babies, there were 71 babies. Dec 19, 2017 There were 71 babies. #### Explanation: Problems like this can be confusing for two reasons: 1. It's hard to find a good way to express all those amounts 2. There are two kinds of data to consider: . . . ~ The NUMBER of tickets of each type that were sold . . . ~ The monetary VALUE of each type of ticket ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ First find a way to express the NUMBER sold of each type of ticket Let $x$represent the number of babies' tickets that were sold Babies' tickets . . . . . . . . . . . .$x$$\leftarrow$number of babies' tickets Same number . . . . . . . . . . . ..$x$$\leftarrow$number of children's tickets The combined number . . . $2 x$$\leftarrow$number of adults' tickets ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ Now find a way to express the VALUE of the tickets sold x tickets @ $3   ea . . . . . . $3 x$ $\leftarrow$ value of babies' tickets

x  tickets  @ $5 ea . . . . . . $5 x$$\leftarrow$value of children's tickets 2x tickets @ $7  ea . . . .$7 \left(2 x\right)$ $\leftarrow$ value of adults' tickets
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Now you can set up the equation

[babies' value] plus [children's value] plus [adults' value] = $1562 [ . . . . . $3 x$. . . .] .$+$.[ . . . . . $5 x$. . . . . ] .$+$. [ . . . $7 \left(2 x\right)$. .] = 1562 Solve for $x$, already defined as "the number of babies' tickets" 1) Clear the parentheses by distributing the 7 $3 x + 5 x + 14 x = 1562$2) Combine like terms $22 x = 1562$3) Divide both sides by 22 to isolate $x$, already defined as "the number of tickets for babies" $x = 71$$\leftarrow$the number of tickets for babies Answer: There were 71 babies ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ Check: 71 babies @ $3  ea . . . . . . . $213 $71$$c h$$i l \mathrm{dr} e n$@ $5  ea . . . . . . . $355 142 ad$u$$l$$t s$@ $7  ea . . . . . . . $994 $\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .}$.............. $\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots .}$$T o t a l$. . . . . . $1562

$C h e c k$ !