Math Is Fun Forum

What are the answers sarah? I don't.

I think I've been ignoring the - on the gradient and am visualising a reflection of the line.

I did indeed get to that but the gradient is supposed to be an integer (whole number).

I have to then work out the equation of the line passing through (2, -3), parralel to y = 3/2x + 3 I work this out as y = 3/2x - 6.  The book tells me that the answer is 3x + 2y = 0.

What is the first step for the following simultaneous equation:

2x - 3y = 2
3x + 2y = 16

>.<

I was just working it out as I went along, I've only been studying maths for 6 weeks or so

Ah thankyou, I did not know of this method of making a pair of variables match.

4x + 3y = 19
4x - 10y = 6

surely

0x + (3y - 10y) = 13
-7y = 13 and not 13y = 13 as you suggest.  Your answer is indeed correct but I'm not quite sure how you got this part.

is that 244,140,625 marbles or colour combinations?  I don't think I took repeats into account

you could have 12 marbles of all the same colour

12

or 2 colours, 11 of 1 colour and 1 of the other, 10 of 1 colour and 2 of the other..
11 : 1
10 : 2
9 : 3
8 : 4
7 : 5
6 : 6
5 : 7
4 : 8
3 : 9
2 : 10
1 : 11

Now we move on to 3 colours
10 : 1 : 1
9 : 1 : 2
9 : 2 : 1
8 : 1 : 3
8 : 3 : 1
7 : 1 : 4
7 : 4 : 1
6 : 1 : 5
6 : 5 : 1
5 : 1 : 6
5 : 6 : 1
4 : 7 : 1
4 : 1 : 7
3 : 1 : 8
3 : 8 : 1
2 : 9 : 1
2 : 1 : 9
1 : 1 : 10
1 : 10 : 1

4 colours...
9 : 1 : 1 : 1
8 : 1 : 1 : 2
8 : 1 : 2 : 1
8 : 2 : 1 : 1
7 : 1 : 1 : 3
7 : 1 : 3 : 1
7 : 3 : 1 : 1
6 : 1 : 1 : 4
6 : 1 : 4 : 1
6 : 4 : 1 : 1
5 : 1 : 1 : 5
5 : 1 : 5 : 1
5 : 5 : 1 : 1
4 : 1 : 1 : 6
4 : 1 : 6 : 1
4 : 6 : 1 : 1
3 : 1 : 1 : 7
3 : 1 : 7 : 1
3 : 7 : 1 : 1
2 : 1 : 1 : 8
2 : 1 : 8 : 1
2 : 8 : 1 : 1
1 : 1 : 1 : 9
1 : 1 : 9 : 1
1 : 9 : 1 : 1

We could continue in this way but if we look, we can see that with just one colour we have just the one combination, two colours has 12 combinations, three has 19 combinations and if we counted the rest we would see a pattern emerge.  From this we can say that;

n = (c - 1) * (12 - c) + 1

where c is the number of different colours and n is the number of combinations. Thus;

if we have just one colour
n = (1 - 1) * (12 - 1) + 1
n = 0 * 11 + 1
n = 1

two colours
n = (2 - 1) * (12 - 1) + 1
n = 1 * 11 + 1
n = 12

three colours
n = (3 - 1) * (12 - 3) + 1
n = 19

four colours
n = (4 - 1) * (12 - 4) + 1
n = 25

I know the answer was right, I looked it up
My question (which I realise is not that clear) was,  "How do I work it out?"

Okay, maybe I should not take the 2 from the x^2 and instead have;

-[2x^2 + 3x - 2].

I think this will work but I have to go out now, thanks anyway