# Kid's homework: 4 equations 5 unknowns? Going crazy!

I'm new here, and I'm hoping someone can help out. My 10 year old son has been set a maths problem, which I can't solve. I've got a PhD in neuroscience and do a fair amount of matlab stuff (data analysis, image processing) on a daily basis, but I can't work this out.

The problem is expressed in words, but I've read it through a dozen times and I'm sure it boils down to the following:

a + b = 55

b + c = 43

c + d = 42

d + e = 37

They are asked to find the value of e. But this is 4 equations with 5 unknowns. Is there really a unique solution for this system of equations? Where am I going wrong?

If you set one of the variable to 0 you can solve for the rest, of course, but I'm pretty sure this is not what they are meant to do. The hint says it's easiest to start by working out the value of c.

I'm lost, any help would be most appreciated!

The exact question is:

The following people take part in a school trip: 55 boys and girls; 43 girls and fathers; 42 fathers and mothers and 37 mothers and teachers. How many teachers took part in the school trips?

Assuming the classes are mutually exclusive (i.e. no teachers are also parents), I'm pretty sure that is the set of equations I posted. The other problems in the same homework are similar in form but all have 1 additional piece of information: the total number (e.g., a + b + c + d + e = 100). Those ones are solvable no problem.

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The best thing would be to include the original problem so that we have all the information. –  naslundx Apr 19 '14 at 9:21
There is no unique solution to that system of equations even if you limit yourself only to the natural numbers. I am posting just to confirm your observation, but concur with @naslundx. –  Jesko Hüttenhain Apr 19 '14 at 9:32
They haven't even been taught algebra yet, so if they are meant to solve it 'intuitively'. It's in 'Pisa-Training 5' (Mildenberger Press), p. 47, q. 21.3 :-) –  user144198 Apr 19 '14 at 9:47
You are right to be skeptical: this system of equations doesn't have a unique solution. In view of your comment to Sami's answer, it looks like they just forgot to specify $a+b+c+d+e=100$ as the fifth constraint. –  TonyK Apr 19 '14 at 9:48
Since when are fathers not also boys, and mothers not also girls? (And before anyone gets on about "boy implies young", look at the target for the question: a 10-year-old. At that age, "boy" meant "male" for me. It's also the type of word game that teachers seem to like in the early years of school) –  Izkata Apr 19 '14 at 20:36

We add these equalities in this manner $$a+b+43+c+d+37=55+b+c+42+d+e$$ now we cancel we find $$\require{cancel}a+\cancel{b}+43+\cancel{c}+\cancel{d}+37=55+\cancel{b}+\cancel{c}+42+\cancel{d}+e$$ hence $$a+80=e+97\iff e=a-17$$ so each time you take a value of $a$ we find a value of $e$. Can you now answer your son?

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Thanks everyone for the rapid responses! The exact question is: The following people take part in a school trip: 55 boys and girls; 43 girls and fathers; 42 fathers and mothers and 37 mothers and teachers. How many teachers took part in the school trips? Assuming the classes are mutually exclusive (i.e. no teachers are also parents), I'm pretty sure that is the set of equations I posted. The other problems in the same homework are similar in form but all have 1 additional piece of information: the total number (e.g., a + b + c + d + e = 100). Those ones are solvable no problem. –  user144198 Apr 19 '14 at 9:35
You're welcome and I am pleased to help you! –  Sami Ben Romdhane Apr 19 '14 at 9:37
Nice answer and feedback... –  amWhy Apr 19 '14 at 11:12
It doesn't help. Your answer suggests that e=a-17 makes five equations for five unknowns so that we can solve the system. However, this is false. Since you just manipulated the original equations, your new equation is linearly dependent with them. You won't be able to uniquely solve your five-equation system any more than OP could uniquely solve with four equations. –  krubo Apr 20 '14 at 7:20
"Your answer suggests that e=a-17 makes five equations for five unknowns so that we can solve the system." I don't see why you interpret it this way. To me, Sami Ben Romdhane's answer does not imply that. –  Joel Reyes Noche May 2 '14 at 9:25

We have:

$$b=55-a$$ $$c=43-b=43-(55-a)=a-12$$ $$d=42-c=42-(a-12)=54-a$$ $$e=37-d=37-(54-a)=a-17$$ For all these numbers to be non-negative we therefore need $$17\le a\le 54$$with the inequalities strict if we require all the numbers to be positive.

It was just possible that the constraints that all the numbers are non-negative (or positive) integers would have fixed a value for $a$, but they don't, and any $a$ which satisfies the constraints gives a consistent solution.

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The following people take part in a school trip: 55 boys and girls; 43 girls and fathers; 42 fathers and mothers and 37 mothers and teachers. How many teachers took part in the school trips? Assuming the classes are mutually exclusive (i.e. no teachers are also parents)

If you assume there were zero teachers, you get 37 mothers, 5 fathers, 38 girls and 17 boys.

If, on the other hand, you assume there were 37 teachers, you get 0 mothers, 42 fathers, 1 girl and 54 boys.

Everything between 0 and 37 teachers should admit a solution too, so the solution is rather non-unique, even if everything is constrained to be a positive integer.

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Edit. By back substitution, one can easily express $a,b,c,d$ in terms of $e$: \begin{cases} d=37-e,\\ c=e+5,\\ b=38-e,\\ a=e+17. \end{cases} Therefore $a+b+c+d+e=97+e$. It is very likely that they have simply forgotten the constraint that there are $100$ participants.

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I think it probably assumes:

Fathers are the same as mothers:

So, - Mothers = fathers = 21 - Girls are 22 - Boys are 33 - Teachers are 16

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Then you have 58 adults to supervise 55 children, which is an insane ratio for a school trip. –  tobyink Apr 19 '14 at 18:44
I don't think # mothers = # fathers would ever be a reasonable assumption for a school trip. –  Dustan Levenstein Apr 19 '14 at 20:16
We call that a group vacation. Time share, anyone? –  Patrick M Apr 20 '14 at 4:06

This may be a little much but you could set up the matrix below:

$$A = \begin{pmatrix}1{} & 1{} & 0{} & 0{} & 0{} & 55 \\0 &1 & 1 & 0 & 0 & 43 \\ 0 & 0 & 1 & 1 & 0 & 42 \\0 & 0 & 0 & 1 & 1 & 37\end{pmatrix}$$

Solve the matrix to get the following solution set:

{($a$, $b$, $c$, $d$, $e$)= (($17$ + $e$), ($38$ - $e$), ($5$ + $e$), ($37$ - $e$), $e$) | $e$ $\in$ $\mathbb{R}$}

By plugging in $0$ through $37$ for $e$ as pointed out by user fgp above you will get the various combinations of $a$, $b$, $c$, $d$, and $e$ that meet the constraints without giving you any negative participants.

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Not really an answer, but I couldn't post an image as a comment. Looks like you already have some nice answers, but a visual always helps me :D

Edit: Again, this wasn't intended to solve the question, just offer another point of view to play with. Here is a rough Legend for my comically grotesque visual :)

Teachers   Mothers    Fathers
boys       girls


Each number is the quantity of persons in the group(s) it overlaps:

37 Teachers and Mothers
42 Mothers and Fathers
54 Mothers and boys
97 Mothers, Fathers, boys, and girls (all members except Teachers.)
43 Fathers and girls
55 boys and girls

100? = OP suggests this may be the 'total' accidentally omitted from the problem.
3? = rough guess at an answer (based on 100 total.)

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This is very confusing. –  user13107 Apr 21 '14 at 3:13
I added some description, hopefully it's more clear. –  CodeShane Apr 23 '14 at 9:15

$\begin{eqnarray}{\bf Hint}\ &&a+b &-& (b+c) &+& (c+d) &-& (d+e) &\,=\,& a-e\\ = && \ \ \ 55 &-&\ \ \ 43 &+&\ \ \ 42 &-&\ \ \ 37 &\,=\,&\ \ 17\end{eqnarray}$

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The OP's problem is clearly not manipulating the equations - it's that this is a underdetermined system of equations which doesn't have a unique solution. –  CodesInChaos Apr 20 '14 at 17:52
@Codes That point has already been addressed in other answers. The point I emphasize has not, viz, that the innate symmetry in the system enables efficient elimination via telescopic cancellation. You can find many other applications of telescopy in posts by my students and I. –  Bill Dubuque Apr 20 '14 at 18:18