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Posts Tagged ‘Simultaneous Equations’

An interesting area puzzle 

July 5, 2017 11 comments

Regular readers will know that I have a tendency to collect puzzles and I like to have a go at them. This evening I had a crack at this while my daughter was playing before bed.

It looked interesting when I saw it on Facebook a while ago and so I thought I’d have a crack. If you haven’t yet, do it now. I want to know if you took the same approach!

I wasn’t too sure where to start, so I drew it out, labelled some stuff and came up with some equations:

I thought if I multiples two of my equations together and rearranged I could get the yellow area as the subject:

Then I needed dw:

Then I needed cx:


I thought “now I need ay”, then realised I had it:

This meant I could.sub back in through the equations:

So the area:

I thought 27 was a nice answer, and I’m fairly sure it’s correct, however I have a feeling that I may have missed something blindingly obvious that would have gotten me there much quicker. 

If you did it, I’d love to hear your approach, especially if you spotted something I missed!

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Bicycle Puzzle

September 12, 2015 5 comments

This week’s puzzle from Chris Smith is a nice contextualised simultaneous equations puzzle I intend to use next week.

image

It boils down to:

2X + B + T = 135
2X + 2B + 3T = 269
X + B + T = 118

Via elimination using 1 and 3 we can see that X (number of tandems) is 17.

That gives:

34 + B + T = 135
34 + 2B + 3T = 269
17 + B + T = 118

1 and 3 rearrange to the same leaving:

B + T = 101
2B + 3T = 235

Elimination gives T (number of Tricycles) to be 33 which leave B (number of Bicycles) to be 68.

A nice little problem.

What a fantastic puzzle!

April 4, 2014 2 comments

When I logged on to twitter this evening I saw this tweet from Colin Beveridge (@icecolbeveridge):

image

Being the sort of person that seems a maths puzzle and finds it impossible not to have a crack at it I had a go.

xy=3 x+y=2 what is 1/x + 1/y?

My thoughts process was fairly straight forward:

xy=3 so it follows that  x=3/y and hence 1/x = y/3. Likewise xy=3 so y=3/x and hence 1/y = x/3. Thus, 1/x + 1/y = x/3 + y/3 = (x+y)/3 = 2/3 {as we know x+y=2}.

It seemed a straight forward puzzle, I noticed some tweets including complex numbers and thought they were odd, “Professor Yaffle” (@adamcreen) then tweeted a much simpler solution:

1/x + 1/y = (x+y)/xy =2/3

Which I thought was lovely. Then Colin asked “how would your students tackle it?” I thought “Grrr, it’s the holidays so I can’t try it on them for a fortnight!” Then I about it a bit and decided that on the whole they would probable try to solve the simultaneous equations using substitution.

x+y =2 so y=2-x
xy=3 so x(2-x)=3
2x – x^2 = 3
x^2 -2x + 3 = 0

Hang on, there are no real solutions to that quadratic! My non-further maths students would stop there stumped, my Further Students would work through using complex numbers. I thought I check another substitution:

xy = 3 so y=3/x
x+y=2
x+3/x=2
x^2+3=2x
x^2-2x+3=0

Yep, that’s the same quadratic so I haven’t made any silly errors. I figured that if you followed this through with complex numbers you must end up with the same answer, but wanted to check:

x=(2+(4-12)^1/2)/2
Or
x=(2-(4-12)^1/2)/2
So
x=(2+(-8)^1/2)/2
Or
x=(2-(-8)^1/2)/2

Root (-8) = i2root2

x=1+i(2)^1/2 or 1-i(2)^1/2

Meaning y is the complex conjugate of x in each case (by substitution back into original equations).

So 1/x + 1/y = 1/(1+i(2)^1/2) + 1/(1-i(2)^1/2) = (1+i(2)^1/2 + 1-i(2)^1/2)/(( 1+i(2)^1/2)(1-i(2)^1/2)

Which, of course, simplifies to 2/3.

What a delightful puzzle! There are no real values for x and y, but the answer is a lovely, real, rational number! I thoroughly enjoyed exploring it, and I hope my Further maths pupils will enjoy it too. I’m not sure whether to give it to my other A Level pupils or not, I will decide over the holidays. I am definitely going to give it to some of my year 11s though. We’ve just hammered algebraic fractions, and this is going to be an extension task!

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