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

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Sharon and Susie are selling holiday wreaths. Sharon sold 10 ivy wreathes and 9 holly wreaths for a total of $282 hopefully that's going to give me one equation just from that sentence. Moving on Susie sold 5 of each type for $150 total. How much would it cost to buy one of each wreath?

What I’m going to do is make sure that at the end of my problem I go through and make sure I answer the question they are asking; how much would it cost to but 1 of each wreath? Also before we do this problem there is a secret like shortcut way that some of you guys might be thinking of already for how to solve this problem this particular one.

But I want to show you how you could solve any one problem of this type. So here we go Sharon I’m going to get her equation from this whole first sentence she sold 10 ivy wreathes and she sold 9 holly wreathes for a total of $282. That represents Sharon.

Now I’m going to have an equation for Susie coming out of this whole sentence about Susie. Susie sold 5 wreaths of each type, 5 ivies +5 holly reeds and she had a $150 total. Now I’m ready to boogie, I have two equations with two unknowns and I can go through and solve using either elimination substitution, graphing or some of you guys even know matrices.

So while I’m trying to decide what method to use I’m going to look at the coefficients in front of my variables since I don't have anything with a coefficient of 1, I’m thinking I’m not going to be doing either graphing or substitution. What I do notice is that I almost have additive inverses from my ivy rates I would if this number right here instead of being 5 were -10, then I would have 10 and -10. In order to make that into -10 I’m going to multiply this entire second red equation by -2 then I’ll have additive inverses and I can use elimination.

The first guy is the same the second guy I’m going to be distributing that -2, -10I, -10H -300. This is the exact same system just rewritten so that now I have additive inverses. For elimination you add your answers vertically so that the Is cancel out and I’m left with -1H is equal -18. That means a +1H is equal to 18 or I can say a holly wreath costs $18. I'm half way done I already know how much holly wreaths cost.

Then I can go back and use either original problem to solve for how much an Ivy wreath costs. I'm going to choose to use this bottom equation, no real reason why either one should give me the same answer. 5 times whatever my Ivy cost is plus 5 times instead of holly I’m going to use 18 gives me the answer of 150 go through and give me the calculation.

5 times 18 is 90, subtract 90 from both sides so that 5I is equal to 60 or I is equal to 12 what that means is that an Ivy wreath costs $12. The problem asked me in the very beginning how much would it cost to buy one of each if I want to buy one of each I will have to add those together. 1 of each costs 18 plus 12 is $30.

That's the final answer to this problem. A lot of times the multiple-choice test teachers will be cruel and they'll have like answer a 15, answer b 12, answer c 30, answer d something bonehead answer like the trick problem, the trick answer. And somebody would have to have read this sentence really carefully in order to get that problem correct. They mighty have done all these Mathy stuff right but then they put the wrong bubble answer because they forgot to answer the last sentence how much would it cost to but one of each, that's why it’s 30.

Okay so that's how you do the problem the long way, had any of you guys figured out the short way? The short way is just looking at Susie’s sentence; Susie sold 5 of each type of wreath for 150. If we need to figure out how much it will cost to buy one of each, that's like a one fifth of what Susie sold right, so my answer here should be one fifth of Susie's total and it is one fifth of 150 equals $30.

I could have done that one in my head jut like that probably some of you guys did when you solved this problem on the board but I wanted to show you the long way so that you'll be able to solve any system of equations of this type.