Seth and Ted can paint a room in 5 hours if they work together. If Ted were to work by himself, it would take him 1 hours longer than it would take Seth working by himself. How long would it take Seth to paint the room by himself if Ted calls in sick

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:

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.

f

step-by-step explanation:

f

a ≈ 172.05

step-by-step explanation:

the pictures.