# Helping 7th grade with math question… I'm stumped.

1. First salesperson says 7 baubles together with 5 gewgaws is the same value as 6 trinkets
2. Second salesperson says 4 baubles with 9 trinkets has the same value as 5 gewgaws
3. Third salesperson says 6 trinkets with 3 gegwaws has the same values as 4 baubles.

One of the salespeople is wrong... which one?

We tried out a bunch of different substitution approaches but I can't figure this one out. If you can explain it to me, I can explain it to him.

• I get all my baubles and gewgaws from amazon :) – Neil Apr 10 '15 at 22:53
• Why downvote a question.... I mean, really. – brettwgreen Apr 10 '15 at 23:06
• I don't think there is a solution to any equation pair unless you allow the value of some of the items to be negative. – Joffan Apr 10 '15 at 23:24
• Questions sometimes get down voted when no work is shown in the question. Notice how even though rogerl solved the problem for you, he was unaware of where you were making mistakes, and so was unable to clarify the issue for you. – Jonny Apr 10 '15 at 23:27
• I was in 7th grade once and wished my parents could have helped me with my math homework :) +1 – Neil Apr 10 '15 at 23:52

Technically, none of the salespeople are wrong if trinkets, gewgaws, and baubles are worthless.

If this case is to be excluded, assign some arbitrary worth to one of the three items. For example, let trinkets be worth 1. Then we have three equations:

$$7b + 5g = 6$$ $$4b - 5g = -9$$ $$3g - 4b = -6$$

If all three salesmen valued baubles, gewgaws, and trinkets equally, then these three equations would represent lines that intersected at a single point: the agreed upon value for baubles and gewgaws when trinkets are assumed to be worth 1.

However, this is not the case. If salesman 1 and salesman 2 are correct, then baubles are worth -0.273 and gewgaws are worth 1.582. If salesman 1 and salesman 3 are correct, then baubles are worth 1.171 and gewgaws are worth -0.439. If salesman 2 and salesman 3 are correct, then baubles are worth 7.125 and gewgaws are worth 7.5.

If we further assume that no item is allowed to have negative value, then only salesman 2 and salesman 3 can be correct, but technically we still would have to prove this holds no matter what value we assign $t$.

However, value is subjective. I see no reason why three different salesmen can't have conflicting views about the values of objects. The fact that salesman one disagrees with two other salesmen about an exchange rate is insufficient proof that he is 'wrong' in my book.

• I think they want to know which salesperson is ripping you off for trinkets. – Neil Apr 10 '15 at 23:55
• I'm with you... thanks for your help. – brettwgreen Apr 11 '15 at 0:36

Statement 1 is the same as "6 trinkets minus 5 gewgaws has the same value as 7 baubles", which is clearly inconsistent with statement 3. So one of these two sales people is wrong.

Similarly, statement 2 is the same as "5 gewgaws minus 4 baubles has the same value as 9 trinkets", which is inconsistent with statement 1. So one of these two sales people is wrong.

Thus sales person 1 is wrong. Assuming that all three items have a positive value.

• I do the same reasoning between 2 and 3 and conclude they are inconsistent with each other (using the 4 baubles)... which leads me to conclude that 3 is wrong. – brettwgreen Apr 10 '15 at 22:54
• 2 and 3 are not inconsistent. Let 2 gewgaws equal 15 trinkets. – Jonny Apr 10 '15 at 23:19
• solve each for 4 baubles... #2 says that's 5 gewgaws minus 9 trinkets, while #3 says it's 3 gegaws plus 6 trinkets. These are not consistent. – brettwgreen Apr 10 '15 at 23:23
• Yes, they are consistent. Again, let 2 gewgaws equal 15 trinkets and you will see the two equations are interchangable. – Jonny Apr 10 '15 at 23:26