My son has come home with some homework tonight and I'm sure he's done this in class, but as you can expect from a typical 13 year old..... "Teacher hasn't taught this yet". I'm a bit on the rusty side with this having not done anything like this for 20 (ish) years.


Description

I've taken a flyer on the question - don't take the answers in pen to be correct - this my (very lame) effort.

Here's my thought pattern.

a)
4x2 + 12 = 20
5x2 + 12 = 22
6x2 + 12 = 24

I'm really looking for one number since the question says "how much do they pay to go to each club" ???


b) - working backwards using the same formula as part a)
27-9 /3 = 6
39-9 /3 = 10
21-9 /3 = 4


** This is where I get stuck **
c)
3X + Y = 19
5X + Y = 29
8X + Y = 44


But it's asking for functions and I thought functions were f(x) ???

d) **now i'm stuck **



Can someone from the collective please explain this to me so I can pass this knowledge on to my son.....(and prove to him that dad's do actually know EVERYTHING)

Cheers

[Edited on 13/11/13 by Barkalarr]

Makes it easy looking at it like this

3x = 19-y
5x = 29-y
8x = 44-y

The answers x = 5, y = 4

Sorry I just looked at it and knew the answer

x =5 and y=4

too slow again

old school maths says

a: 2n + 12
b: 3n + 9
c: 5n + 4

they are all: xn + y

quote:

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Originally posted by mark chandler
Makes it easy looking at it like this

3x = 19-y
5x = 29-y
8x = 44-y

The answers x = 5, y = 4

Sorry I just looked at it and knew the answer

---

Solve it as a simultaneous equation

3X + Y = 19
5X + Y = 29
8X + Y = 44

I just ignored the third line and subtracted the first line from the second, giving 2X + 0Y = 10, giving X=5.

Put the 5 back into the first line and you've got 3X + Y = 19, giving 15 + Y = 19, giving Y = 19 - 15 = 4.

As above, there are lots of ways of solving it, including if you have a somewhat odd mind, just looking at it until you brain goes 'well thats easy...'

As said, there methods for solving simultaneous equations, and you actually only needs one equation per variable so as there are only tow variables you can start of if you want be ignoring one of the three.

The other, and incorrect way of doing it when its quite simple and you expecting whole number, is just to do it by trial and error. But I cannot recommend this as it misses the point and doesn't really do the work!


Daniel

what I would do as a first pass would be rearrange 2 of the equations and make say Y the subject of the formula

3X + Y = 19 would become Y=19-3X
5X + Y = 29 would become Y=29-5X

then make them equal to one another

19-3X=29-5X

and solve for x

29-19=5x-3x

10=2x

x=5

Then plug x=5 into any of the other 3 equations for y=4

Thanks all... Top of the class !!

Here's a simple way to visualise it:

3X + Y = 19
5X + Y = 29
8X + Y = 44

The difference between the first and second expression left hand side is just 2x (5x - 3x) and this makes a difference of (29-19) = 10 on the right hand side. So, if 2x is worth 10, then x must be worth 5 (10/2). Once you know x then you can easily work out y.