Answer:
Seth would need 10 hours to paint the room.
Step-by-step explanation:
Let's define:
S = rate at which Seth works
T = rate at which Ted works
When they work together, the rate is S + T
And we know that when they work together they can pint one room in 5 hours, then we can write:
(S + T)*5 h = 1 room.
We also know that Ted alone would need one hour more than Seth alone.
Then if Seth can paint the room in a time t, we have:
S*t = 1room
and
T¨*(t + 1h) = 1room
Then we have 3 equations:
(S + T)*5 h = 1
S*t = 1
T¨*(t + 1h) = 1
(I removed the "room" part so it is easier to read)
We want to find the value of S.
First, let's isolate one variable (not S) in one of the equations.
We can isolate t in the second one, to get:
t = 1/S
Now we can replace it on the third equation:
T¨*(t + 1h) = 1
T¨*( 1/S + 1h) = 1
Now we need to isolate T in this equation, we will get:
T = 1/( 1/S + 1h)
Now we can replace this in the first equation:
(S + T)*5h = 1
(S + 1/( 1/S + 1h) )*5h = 1
Now we can solve this for S
(S + 1/( 1/S + 1h) )= 1/5h
S + 1/(1/S + 1h) = 1/5h
Now we can multiply both sides by (1/S + 1h)
(1/S + 1h)*S + 1 = (1/5h)*(1/S + 1h)
1 + S*1h + 1 = 1/(S*5h) + 1/5
S*1h + 2 = (1/5h*S) + (1/5)
Now we can multiply both sides by S, to get:
(1h)*S^2 + 2*S = (1/5h) + (1/5)*S
Now we have a quadratic equation:
(1h)*S^2 + 2*S - (1/5)*S - (1/5h) = 0
(1h)*S^2 + (9/5)*S - (1/5h) = 0
The solutions are given by the Bhaskara's formula:
[tex]S = \frac{-(9/5) \pm \sqrt{(9/5)^2 - 4*(1h)*(-1/5h)} }{2*1h} = \frac{-9/5 \pm 2}{2h}[/tex]
Then the solution (we only take te positive one) is:
S = (-9/5 + 2)/2h
S = (-9/5 + 10/5)/2h = (1/5)/2h = 1/10h
Then Seth needs a time t to paint one room:
(1/10h)*t = 1
t = 1/(1/10h) = 10h
So Seth would need 10 hours to paint the room.
