HELP....math homework?

Question

Please help solve the following algebra problem:
Sue can paint a house in 4 hours. Joe can paint the same house in 6 hours. If Sue & Joe paint the house together how long will it take for them to finish painting the house?
PLEASE EXPLAIN YOUR ANSWER & HOW YOU CAME UP WITH IT!! THE ANSWER IS NOT 5 HOURS!!!!!

Answer 1

2.4 hours (2 hours, 24 minutes)

Sue paints the house in 4 hours so she paints ¼ of it in 1 hour. Similarly, Joe paints 1/6 of it in 1 hour. Working together means adding their rates together: ¼ + 1/6 = 3/12 + 2/12 = 5/12 of the house painted every hour. So, divide the whole job (1) by the rate (5/12) and you get 12/5 of an hour or 2 2/5 hours (= 2.4 hours).

Answer 2

The first answer is good... here's the equation for it:

x : represents how long it would take them to finish the job together

1/4 of x or x/4 : Sue's part of the work

1/6 of x or x/6 : Joe's part of the work

1 : represents the whole job

x/4 + x/6 = 1
multiply the whole equation by the LCM of 4 & 6 which would be 12
12(x/4 + x/6 = 1)
3x + 2x = 12
5x = 12
x = 2 and 2/5 hours

Answer 3

Suppose the total tobepainted area of the house is x.

Sue paints at the rate of x/4 units per hour. Joe paints at x/6 units per hour.

Now in two hours, Sue paints half the house, while Joe does 1/3rd.
So they did 5/6th of the total area in two hours, or 5x/6 units in 2 hours.

Their collective speed is 5x/12 units per hour.

So the total time taken would be total area/speed.
Which is 12x/5x, which is 12/5, which is 2.4 hours, which is 2 hours and 24 minutes.

Answer 4

Sue is faster than Joe. Failing that, write "Marmot".

Answer 5

i had that same question.... and there's a girl in my class named jodi too lol

Answer 6

im in algebra but i dont get it....sowwy...

Answer 7

i suck at math sorry :-)