SOLUTION: A painter can paint a room three hours less than an apprentice. Together they can pain the room in two hours. How long will it take the apprentice working alone to paint the room?

Question 172994: A painter can paint a room three hours less than an apprentice. Together they can pain the room in two hours. How long will it take the apprentice working alone to paint the room?
Found 2 solutions by Mathtut, gonzo:
Answer by Mathtut(3670) (Show Source): You can put this solution on YOUR website!
lets call the appretices time per hour 1/a and painter 1/a-3
:
is what they can do together per hour.
:
if it takes 2 hours together we have
:

:
multiplying each term by a(a-3)
:
distributing
:
combining like terms
a=1 and 6
:
throw out 1 because that would make the painters time negative so
:
hours alone for the apprentice to paint the room

Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=25 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 6, 1. Here's your graph:

Answer by gonzo(654) (Show Source): You can put this solution on YOUR website!
formula to work with is:
rate * time = number of units
the number of units is 1 room.
it takes the painter x - 3 hours to paint a room.
it take the apprentice x hours to paint the same room.
-----
let Rp = rate of the painter.
let Ra = rate of the apprentice.
-----
since rate * time = number of units,
for the painter:
Rp * (x-3) = 1
for the apprentice:
Ra * (x) = 1
-----
if they are both working together, it takes 2 hours to paint the room.
working together their rates are additive.
since rate * time = number of units,
for both:
2 * (Ra + Rp) = 1
-----
for the apprentice:
Ra * x = 1
which means:
Ra = 1/x
-----
for the painter:
Rp * (x-3) = 1
which means:
Rp = 1/(x-3)
-----
substituting for Ra and Rp, the formula:
2 * (Ra + Rp) = 1
becomes:

which becomes:

-----
if we multiply both sides of the equation by x*(x-3), we get:

-----
this becomes:

which becomes:

-----
if we subtract 4*x from both sides of the equation, and if we add 6 to both sides of the equation, we get:

which is the same as:

-----
this factors out to be:

which give us:
x = 1
or:
x = 6
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x cannot be 1, since x-3 would be negative which is not possible.
x must be 6.
-----
if x is 6, then
x-3 is 3.
-----
since x * Ra = 1
then Ra must be 1/6
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since (x-3)*Rp = 1
then Rp must be 1/3
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to prove this is true, we substitute in the original equations to see if they are true.
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first equation:
x * Ra = 1
6 * 1/6 = 1 is true.
second equation:
(x-3)*Rp = 1
(6-3)*1/3 = 1
3*1/3 = 1 is true.
third equation:
2 * (Ra + Rp) = 1
2 * (1/6 + 1/3) = 1
2 * (1/6 + 2/6) = 1
2 * (3/6) = 1
6/6 = 1 which is true.
-----
answer of x = 6 looks good.
apprentice takes 6 hours to paint a room.
painter takes 6-2 = 3 hours to paint the same room.
together they can paint the room in 2 hours.
-----

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