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Selling Multiple Items - Problem 1

Teacher/Instructor Alissa Fong
Alissa Fong

MA, Stanford University
Teaching in the San Francisco Bay Area

Alissa is currently a teacher in the San Francisco Bay Area and Brightstorm users love her clear, concise explanations of tough concepts

I've been teaching for a lot of years and I’ve used lots of different Math textbooks and I’ll tell you guys pretty much all of the algebra books I’ve seen have problems about selling tickets. I'm not sure why but this loves to show up in your Math classes and in your Math tests.

So here we go, Fairview High School is selling tickets to the annual Shakespeare play. On the first day they sold 6 child tickets and 3 adult tickets for a total of $75. They are not doing so well but hopefully the second day gets better. On the second day they took in $102, okay for selling 2 child tickets and 8 adult tickets. Find the selling price of each ticket type.

Okay so here's what I’m going to do I want to use a system of equations and the process of elimination to solve for reach variable. Each one of these sentences is going to turn into one of my equations. Here's what I mean on the first day they sold 6 child tickets that's going to become important and 3 adult tickets for a total of $75 so I’m going to use a black marker to show just that sentence.

On the first day 6 child tickets plus 3 adult tickets and they brought in $75. I'm using letters to help me keep track, C is going to stand for child and A is going to stand for adult. You could use x and y if you want to just keep straight in your brain which represents child ad which represents adult tickets.

Okay now we go to the second day, on the second day they took in 102 for selling 2 child tickets and 8 adult tickets. Okay 2 child plus 8 adult equals 102. Find the selling p0rice of each ticket. Okay now it’s just a straight forward system of equations I have two equations with two unknowns or two variables and I’m ready to start solving. Since I don't have any coefficients that are 1 and -1 I’m thinking that substitution isn't going to be the easiest method to solve this system.

What I’m going to do instead is try to use elimination and when I’m doing elimination I want to look for coefficients that are additive inverses. So 8 and 3 are kind of unrelated with numbers but that doesn't help me too much. But 6 and 2 are in the same vein here's what I mean if I multiply the second equation everything by -3 then my coefficients for C would be 6 and -6. Those are additive inverses and that will help me do elimination.

So I’m just going to rewrite this first equation the exact same way as it was 3A equals 75 my second equation all three terms are going to be multiplied by -3. -6C, oops! It’s going to be negative now instead of +8 it’s going to be -24a equals -306. Okay now for elimination I’m going to add my equations vertically so that I’ll only have one of the letters cancelled out or be eliminated and I’ll only have one equation with one unknown.

So my Cs get eliminated I have -21A equals 306 take away, sorry let me see what this subtraction problem gives me. I get -231 divide both sides by -21 and I’ll get that a equals 11. What that tells me is that an adult ticket costs $11. Let's go back and find out how much child tickets cost because I’m not done yet.

Child tickets I can find by using either original equation and substituting in A equals 11. I'm going to choose to use the top one but I should get the answer even if I get the bottom one. 6 times my child number times 3 times 11, 3 times 11 is 33, is going to give me the answer 75 so let's go through and subtract 33 from both sides and I’ll get 6C equals 42 so a child ticket is equal to 42 divided by 6 which is $7.

Okay I’m all done before I move there I’m going to make sure that I’ve clearly labeled my answers even though I personally think C and A is pretty clear I’m going to write $7 for a child ticket and $11 for an adult ticket. If you wanted to you can go back and check your work and make sure you are going to get an A+ on this homework by substituting the pair of 7, 11 into both equations and making sure you get both true equality statements.

So you guys the last thing I want to leave you with is the idea that when you have problems of this type where you are selling tickets or selling some other two types of quantities, two types of items, each sentence represents one of the equations for your system. This first one had all the information that I needed for this first equation and then this next that I did in blue had all the information that I needed for the second. From there I used elimination to solve for both of the variables.

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