Suppose that there are two types of tickets to a show: advance and same-day. The combined cost of one advance ticket and one same-day ticket is $35 . For one performance, 40 advance tickets and 30 same-day tickets were sold. The total amount paid for the tickets was $1250 . What was the price of each kind of ticket?

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[latex]Advance=x[/latex] [latex][/latex] [latex]x+y=35[/latex] so [latex]x=35-y[/latex] [latex]40x+30y=1250[/latex] [latex]40(35-y)+30y=1250[/latex] [latex]-10y=-150[/latex] [latex]y=15[/latex] [latex]x+15=35[/latex] [latex]x=20[/latex] Cost for Advance ticket: $20 Cost for Same-day ticket: $15


Let's set up variables and equations for things that we know from the question, with a standing for advance tickets, and s standing for same day tickets.  [latex]40a + 30s = 1250[/latex]  [latex]a + s = 35[/latex] Let's solve it through substitution. Isolate either variable, a or s. I'll choose to isolate a. Subtract s from both sides. [latex]a = -s + 35[/latex] Plug the new knowledge into the first equation, since we now know the value of a [latex]40(-s +35) + 30s = 1250[/latex]. Distribute the 40 to the -s and 35 [latex]-40s + 1400 + 30s = 1250[/latex] Combine the negative and positive s [latex]-10s + 1400 = 1250[/latex] Subtract 1400 from both sides. [latex]-10s = -150[/latex] Divide both sides by -10 [latex]s = 15 [/latex] The total price of an advanced ticket and same-day ticket is 35, and we know the value of a same-day ticket, so the price of an advance ticket must be 20.  The final answers are: advanced ticket = 20 same day ticket = 15

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