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Posts Tagged ‘Simultaneous Equations’

Simultaneous Equations

March 10, 2019 3 comments

It’s been a while since i last wrote anything here. Which says more about how busy I’ve been than my desire to write, but I hope to start writing more regularly.

This week I was teaching simultaneous equations and a student asked a question that made me think about things so I thought i would share.

I was teaching elimination method and I had done some examples with the coefficients of y having different signs and I put one on the board with the same signs and asked the class to think how we may go about solving. One of the students in the class put uo his hand after a while and said he thought he had solved it.

5x + 4y = 13

2x + 2y = 6

I asked hime to talk us through his thinking and he said “first I multipled the bottom equation by -2”

5x + 4y = 13

-4x – 4y = -12

“then I added the equations as before”

x = 1

“Then I subbed in and solved.”

2 + 2y = 6

2y = 4

y = 2

“so the point of intersection is (1,2)”.

This wasn’t what I was expecting. I was expecting him to have spotted we could subtract instead, but this method was clearly just as correct. It wasn’t something I had considered as a method before this, but I actually really liked it as a method and it led to a good discussion with the class after another student interjected with her solution which was what I expected, to multiply by 2 and subtract.

It was a great start point to a discussion where the students were looking at the two methods, and understanding why they both worked, the link between addition of a negative and subtracting a positive and many more.

I was wondering, does anyone teach this as a method? Have you had similar discussions in your lessons? What do you think of it?

An interesting area puzzle 

July 5, 2017 12 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!

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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