Awesomeguy

We can multiply the first equation by \(pq\),

\(1/p + 1/q = 1\)

\(q+p = pq\)

We know that \(pq\) is 9, therefore q + p, or p + q, is 9.

ab^5 is basically ab^4 * b. Plugging in known values and simplifying,

\(ab^5 = ab^4 \cdot b\)

\(4608 = 384\cdot b\)

\(b = 12\)

Plug b into one of the known equations,

\(a \cdot 12^4 = 384\)

\(a \cdot 20736 = 384\)

\(a = 1/54\)

So the value of a is 1/54.

15 kids and need to choose 3, 15c3 = 455

Please look at this:

https://youtu.be/eF6zYNzlZKQ

Oh haha that's smart I didn't need to use the quadratic formula lol

Using distance formula,

\(\sqrt{(x-4)^2 + 196} = 10\sqrt{3}\)

Square both sides and subtract 196 on both sides,

\((x-4)^2 + 196 = 300\)

\((x-4)^2 = 104\)

Expand left side and subtract 104 on both sides,

\(x^2 - 8x +16 = 104\)

\(x^2 - 8x -88 = 0\)

Apply quadratic formula to get x values of,

\(x=4±2\sqrt{26}\)

Add both possible values of x together,

\((4+2\sqrt{26}) + (4-2\sqrt{26}) = 8\)

So the final answer is 8.

Multiply first equation by 3,

\(3x + 9y = -12\)

\(3x -7y = 36\)

Subtract,

\(16y = -48\)

\(y = -3\)

Plug y back into one of the equations,

\(x -9 = -4 \)

\(x = 5\)

So (x, y) = (5, -3)

http://www.stutzfamily.com/mrstutzblog/wp-content/uploads/2012/02/2009_Chapter_Solution_Set_2009_-_v2.pdf 

Page 7

In one method, you can use x and y, and figure out based on the expansion of (x+y)^3

But you can also notice that an x and y that satisfy \(x+y=9\) and \(xy=18\) is (6, 3)

Plugging these values in,

\(x^3+y^3=216+27=243\)

\(\)

Oh I completely forgot about arranging the two boys, thank you!