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A swimming pool chargesv$6 for adults and $4 for children. On a particular day they made $1326 in admissions. If a tital of 289 people paid to swim, there must have been

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To solve this we can use the following system of equations:
6a + 4c = 1326
a + c = 289
When a = adults and c = children.
First we can take the second equation and subtract a from both sides to get
c = 289 - a. Then substitute that into the first equation to get
6a + 1156 - 4a = 1326. Next combine like terms to get 2a + 1156 = 1326.
Now subtract 1156 from both sides to get 2a = 170. To get the final answer for a divide both sides by 2 to get a = 85.
To get the value for c, now substitute it into the second equation to get
85 + c = 289. Subtract 85 from both sides to get c = 201.
85 adults and 201 children went swimming that day. Hope this helps!
User Erikbozic
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