Maths help please

[Lucyccfc]

I am aware that maths is taught quite differently than when I was in school (long time ago) and I sometimes struggle to help my DS with his homework, as he tells me 'that's not the way we do it'. So I thought I had better check out with any teachers or 'clued up' parents the correct method to work out the following:

You have 3 kids, let's call them Jane, Tom and Ben. They each have a number on their back. They cannot see their own number.

Jane can see two numbers that add up to 100
Tom says the numbers on Jane and Ben's add up to 120
Ben can see a total of 80

You need to work out the number on each child's back.

Thanks in advance.

T+B=100
J+B=120
J+T=80

T=100-120+80-T
2T=60
T=30

B=70

J=50

Ben + Tom = 100
Jane + Ben = 120
Jane + Tom = 80

Ben - Tom = 40

Ben = 40 + Tom

40 + Tom + Tom = 100

2x Tom = 60
Tom= 30

From that: Jane = 80-30=50
Ben= 100-30= 70

Similar but I did
T+B=100
J+B=120
so J=T+20

SO if J+T=80, then T+20+T=80
so 2T=60 and T=30
carry on from there!

Yes, the bottom line is being able to substitute one equation for a term in the other.

so, for instance, Ben= 40 + Tom allows you to put '40 +Tom' in place of 'Ben' in the equation 'Ben + Tom = 100' That way you've only got one 'unknown' so can work it out.

Hi - I did this slightly differently.

Using the initial equations Le Mous. came up with:

T+B=100
J+B=120
J+T=80 (because Jane & Tom are the two people Ben could see)

and then used lougle's substitution method.

So if we add the bottom two equations together

J + B + J + T = 200

or

2J + (B+T) = 200

We can replace B + T with 100 because we know T + B = 100

so we get

2J + 100 = 200

subtract 100 from both sides (remember you can carry out any function as long as you do the same to both sides of the equation).

2J + (100 - 100) = 200 - 100

2J + (0) = 100

2J = 100

now divide both sides by two (same principle - any function is o.k. as long as done to both sides of the equation).

J = 50.

Now knowing J = 50 - you can work out the rest.

Let's start with the second equation:

J + B = 120

replace B with 50

50 + B = 120

subtract (take away may be the term used in school - heaven forbid we teach proper mathematical terminology) 50 from both sides of the equation....

B = 70

Knowing that you can solve the first equation.

T + B = 100

replace B with 70

T + 70 = 100

subtract (take away) 70 from both sides

T + (70 - 70) = 100 - 70

T + (0) = 30
T = 30

Answers: Jane = 50
Ben = 70
Tom = 30

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although lougle and spaniel eyes approaches also work - I think this rather long winded method may make more sense for those just dealing with algebraic equations.

I know when I was 10/11 I'd look at those three equations and then Lougle's Ben - Tom = 40 and would be confused.

However, that's not to say these methods might make more sense to your DC - after all there are many ways of solving the same equation - it's just to say if they don't help initially - there is a more longwinded way of doing it which might make the substitutions a little clearer for a beginner.

HTH