Melody

Nice try Supernova :)

But you messed up a bit on the numerator. 

It is multiply not plus. 

Becasue.. for EVERY pair of democrats, there are 5 choices of republican - hence multiply.

also

Choosing R1 then R2 is the same as choosing R2 then R1 so you would have to halve that part of your answer.

Alternatively, you could say there are  6C2 ways to chose the democrats and 5 ways to chose the republican so the numerator would be 

6C2*5 = 15*5 = 75

This sequence is a GP

does that help?

What is a?  What is r?  What is the formula for the sum of a GP (n terms)?

Simplify the number expression but keep it as a fraction.

Put the 2 equal to each other and solve.

If you get stuck show us what you have done.

Wolfram alpha solution but not in the form you want.

I graphed it and I see 4 real solutions.

Oh, sorry,   x and y must both be positive  

like this

x^2-2mx+(m^2+4m+5)=0

\(x^2-2mx+(m^2+4m+5)=0\\ x=\frac{2m\pm \sqrt{4m^2-4(m^2+4m+5)}}{2}\\ x=m\pm \sqrt{m^2-m^2+4m+5}\\ x=m\pm \sqrt{4m+5}\\ \mbox{real roots when }\\ 4m+5\ge0\\ m\ge-1.25 \)

Not finished, someone else is welcome to continue if they want to.

Coding:

x^2-2mx+(m^2+4m+5)=0\\
x=\frac{2m\pm \sqrt{4m^2-4(m^2+4m+5)}}{2}\\
x=m\pm \sqrt{m^2-m^2+4m+5}\\
x=m\pm \sqrt{4m+5}\\
\mbox{real roots when }\\
4m+5\ge0\\
m\ge-1.25

b^2-4ac=0

b=c+7

I have not finished it but it looks easy enough.  Solve simultaneously.

half

Nice try, but...

What you have found is the prob that he chooses a number divsiable by 3 both times.  

Can you fix it?

beginning of 1st year  =  2800 vehicles / hour

beginning of second year  =  2800  *  1.08 vehicles / hour

beginning of third year  =  2800  *  1.08^2 vehicles / hour

beginning of nth year = 2800  * 1.08^(n-1)

You should be able to do it now.