16 September, 03:13

# The admission fee at a small fair is \$1.50 for children and \$4.00 for adults. On a certain day, 2200 people enter the fair and \$5050 is collected. How many children and how many adults attended?

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1. 16 September, 05:11
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You'll need to write an equation for this. Let's name our adults y, and the children x. So x+y=2200, as 2200 people went to the fair and people are either x or y. We also know that kid admission is 1.50, and adult admission is 4, so use y and x to make an equation for this. 1.5x+4y=5050 as 5050 was collected. So, you have the system of equations

x+y=2200

1.5x+4y=5050

I'll use substitution to solve. Isolate x in the top equation by subtracting y. so x+y-y=2200-y, and x=2200-y. Substitute this value for x into the second equation, so 1.5 (2200-y) + 4y=5050. Use the distributive property for the parentheses, so 3300-1.5y+4y=5050. Add together the y values of - 1.5 and 4. You get 2.5, so 3300+2.5y=5050. Subtract 3300 from each side of the equation to begin isolating 2.5y, which gives you 1750. Divide 1750 by 2.5 to isolate y, which gives you 700. So you know y=700, so 700 adults attended. You know y=700 so put that into the first equation.

x+700=2200.

Isolate x by subtracting 700 from both sides, which gives you x=1500. So, 700 adults and 1500 children attended.