Algemawhatsamacallit

  • Embedded Gallery2 is not available.
  • Embedded Gallery2 is not available.

I was stumbling through fail blog today and came across a fail/win entry of an algebra problem.

The answer in the failblog post was given, along with a drawing, as 2 animals; 1 pheasant with 138 legs and a rabbit with 41 heads (no legs it seems), "unseen" due to being in a box. Double fail.

So the problem then is.

Rabbits have four legs
Pheasants have two legs
If I own a collection of rabbits and pheasants with 138 legs and 42 heads between them how many rabbits and how many pheasants do I own.

And the answer...

Count the animals by type and not the heads and legs of course.

I was completely stuck looking for the mathematical solution for a few minutes; not knowing at all how to approach it other than it's some kind of equality problem.

So I started with what I knew.

Heads 
42	= x + y
Legs 
138	= (x * 2) + (y * 4)
x 	= pheasants
y 	= rabbits.

Okay that's given me some kind of starting point. Now how do I use them to get to something like x = ... and y = ...

I knew I had to solve the equations and I knew that I could do things to both sides of an equation as long as I balance things correctly. I couldn't remember any of the rules for balancing them apart from the obvious if I add x to one side of the equator I need to add x to the other side and some niggling feeling that moving something from one side to the other side caused negation. I'm sure if I thought about it a little more I could come up with why that is; so I was fairly confident it was true and I could always work it out later if things didn't work out.

I started off down this route and figured I should check for easy simplifications.
Starting with.

138	= (x * 2) + (y * 4)

We can divide that by 2

138/2	= ((x * 2) / 2) + ((y * 4) / 2)

and give ourselves.

69	= x + (y * 2)

The other equation is about as simple as it gets.

42	= x + y

Then I was stuck again, until I remembered equations can have operators applied to them.

So I figured I'd add them

69 + 42	= (x + (y * 2)) + (x + y)
69 + 42	= (x * 2 + (y * 3))

This isn't really helping. I'm stuck again. It's true and I think what it gives me is the ratio between x and y.

   69
+  42
= 111
111/3	= (x * 2/3) + y
37	= (x * 2/3) + y
1	= (x * 2/3)/37 + y/37
1	= x/55.5 + y/37

we can check this later.

So try subtracting them.

69 - 42	= (x + (y *2)) - (x + y)

That's better.

69 - 42	= (x - x) + ((y + y) - y)
69 - 42	= y

and finally transpose.

y	= 69 - 42
    69
    42
y = 27

Heads

42	= x + 27
x	= 42 - 27
    42
  - 27
x = 15

42	= 15 + 27
  15
+ 27
= 42

Legs

138	= (15 * 2) + (27 * 4)
138	= 30 + 108
   30
+ 108
= 138

69	= 15 + (27 * 2)
69	= 15 + 54
  15
+ 54
= 69

And finally the addition equation we simplified out earlier, we can at least use that work as a check.

37	= (x * 2/3) + y
37	= (15 * 2/3) + 27
37	= 10 + 27
1	= x/55.5 + y/37
1	= 15/55.5 + 27/37
1	= 0.27027027027 + 0.72972972973

I am now confident to state there are 15 pheasants and 27 rabbits.

I'm pretty sure my head of year (Steve Dunn) would have been proud if I'd ever handed anything like this in at school. I must have been paying attention though.