Simplify the sum. State any restrictions on the variables. (3 points)

x-2/x+3 + 10x/x²-9

The speed of the current in a river is 6 mph. A ferry operator who works that part of the river is looking to buy a new boat for his business. Every day, his route takes him 22.5 miles each way against the current and back to his dock, and he needs to make this trip in a total of 9 hours. He has a boat in mind, but he can only test it on a lake where there is no current. How fast must the boat go on the lake in order for it to serve the ferry operator's needs?

if you don't want to give answer can you use a similar problem and work it out. I wouldn't mind understanding what I'm doing instead of going to the internet but my teacher teaches with PowerPoints... I can't learn math from a power point I need step by step and pencil and paper. With this in mind, your help will be greatly appreciated

2 answers

You need brackets for this question to work out

(x-2)/(x+3) + 10x/(x²-9)
= (x-2)/(x+3) + 10x/((x-3)(x+3)) , so the LCD is (x+3)(x-3)
= ( (x-2)(x-3) + 10x) / ((x-3)(x+3))
= (x^2 - 5x + 6 + 10x) / ((x-3)(x+3))
= (x^2 + 5x + 6) / ((x-3)(x+3)) , x ≠ ±3

Your river question:
the unknown is the speed of the boat he wants to buy

let the speed of the boat be x mph
So the effective speed of the boat against the current will be x - 6 mph
the effective speed of the boat with the current will be x + 6 mph

time going with the current = 22.5/(x+6)
time going against the current = 22.5/(x-6)

we want:
22.5/(x+6) + 22.5/(x-6) = 9
multiply each term by (x=6)(x-6) , the LCD of our two fractions
22.5(x-6) + 22.5(x+6) = 9(x+6)(x-6)
22.5x - 135 + 22.5x + 135 = 6x^2 - 324
6x^2 - 45x - 324 = 0
2x^2 - 15x - 108 = 0
(x - 12)(2x + 9) = 0
so x = 12, or x = -9/2, which of course makes no sense

He should buy a boat that can do at least 12 mph
Ferry boat must go 9mph!!