Melody

All the points on the green lines satisfy the first condition

 and

All the points on the orange lines satisfy they second condition.

All the points that are on both the green and the orange lines satisfy both conditions.

I have labelled 4 of the poiints but I missed one.  Which one did I miss ?    

What you have here is a locus (set) of points such that  

x=2t+3      and      y=3-3t   for t is any real value.

so what you need to do it make t the subject of both these equations

So now you know that 

\(t=\frac{x-3}{2}=\frac{3-y}{3}\\ \frac{x-3}{2}=\frac{3-y}{3}\\ 6 \cdot \frac{x-3}{2}=6 \cdot \frac{3-y}{3}\\ 3(x-3)=2(3-y)\\ 3x-9=6-2y\\ 3x+2y-15=0 \qquad \text{which is a line}\\ or\\ 2y=-3x+15\\ y=\frac{-3x}{2}+\frac{15}{2}\)

which is still the same line.      gradient = -1.5    and    y intercept = -7.5

6 are 18 years old so if you add all thier ages together you get 18+18+18+18+18+18=6*18= 108

this about this and also that

the mean is the average and the average is the sum of everybodies ages divided by how many people that there are.

Now can you give it a go ?

Heuristics. That is a new word for me. Thanks.

Thanks to Rom and EP too  :)

Thank you   

It would be good if another mathematician checks this.  I think it is ok. 

Heureka taught me just a little while ago.

Find the coefficient of   \(a^4b^4\)   in the expansion of   \((\frac{a^2}{3}-2b)^6\)

what you need first is the coefficient of      \((\frac{a^2}{3})^2\;(-2b)^4\)

it will be      \(\frac{6!}{2!4!}=15\) 

so we have       \(15\cdot (\frac{a^2}{3})^2\;(-2b)^4\)

\(15\cdot (\frac{a^2}{3})^2\;(-2b)^4\\ =15\cdot (\frac{a^4}{9})\;\cdot 16b^4\\ =5\cdot (\frac{a^4}{3})\;\cdot 16b^4\\ =\frac{80}{3} a^4b^4\\ \)

so the coefficient is      \(\frac{80}{3}\)

This explanation, if that is what it is is absolute nonsense.

Becasue I did the algebraic division first (which I did not display) and x-2 was the remiander.

\((x^2+x) \text{ divided into } (x^3-2)  \;\;\;goes \;\;\;  (x-1) \text{ times  with a remainder of   } (x-2)\)

\(so\\ \frac{x^3-2}{x^2+x}=x-1+\frac{x-2}{x^2+x} \)

Maybe this will help

\(7 \div 2=3\;\;remainder 1\\ so\\ \frac{7}{2}=3+\frac{1}{2}\)

I hope I have not made any typing errors this time.   

Do you know how to do the algebraic division?

Put your answer where your mouth is!  

Or just break your fingers.

LOL this must not be my question! it is jinxed   

Oh well doesn't matter, the technique is the same.

Mathbum can copy the technique and get the original answer correct :)