If John paints a house in 3 hours and Jane paints a house in 2 hours, how long does it take them to paint 1 house together?

We rephrase the problem:

John can paint one third of the house per hour.

Jane can paint one half of the house per hour.

To find out how much they can do together per hour, we make the necessary assumption that their labors are additive (in other words, that they never get in each other's way in any manner, they don’t stop and talk, etc.), and we add together what they can do individually per hour. So, per hour, their labors are:

¹/₃ + ¹/₂ = ²/₆ + ³/₆ = ⁵/₆ of the job per hour.

But the exercise didn't ask us how much they can do per hour; it asked us how long they'll take to finish one whole job, working together. So now let’s pick the variable "t" to stand for how long they take (that is, the time they take) to do the job together. Then they can do ¹/_t of the whole job per hour.

This gives us an expression for their combined hourly rate. We already had a numerical expression for their combined hourly rate. So, setting these two expressions equal, I get:

¹/_t = ⁵/₆

We can solve for t by flipping the equation; we get:

^t/₁ = ⁶/₅=1¹/₅

An hour has sixty minutes, so ¹/₅ of an hour has twelve minutes. Then:

They can complete the job of painting one house together in 1 hour and 12 minutes.

How long does it take for John and Jane to paint 5 houses together?

We simply multiply ⁶/₅ by 5. It takes them 6 hours. (This is probably unrealistic, because they most likely couldn’t keep up a constant rate for five hours!)

(See also https://www.purplemath.com/modules/workprob.htm for many similar problems.)

The important thing to understand about the above example is that the key was in converting how long each person took to complete the task into a rate.

Each person took a certain number of hours to complete the task:

hours to complete a whole job:

first painter: 3

second painter: 2

together: t

Since the unit for completion was "hours", we converted each time to an hourly rate; that is, we restated everything in terms of how much of the entire task could be completed per hour. To do this, we simply inverted each value for "hours to complete job":

completed per hour:

first painter: 1/3

second painter: 1/2

together: 1/t

Then, assuming that their per-hour rates were additive, we added the portion that each could do per hour, summed them, and set this equal to the "together" rate:

adding their labor:

¹/₃ + ¹/₂ = ⁵/₆ = ¹/_t

^t/₁ = ⁶/₅

As you can see in the above example, "work" problems commonly create rational equations. But the equations themselves are usually pretty simple to solve.

How about this one?

Jose can mow a 100 ft by 100 ft plot of grass in one and a half hours, and Maria can do it in 2 and a half hours. How long would it take for them to mow the lawn working together?

(Using algebra, the answer is ¹⁵/₁₆ hour, or about 56 minutes.)

And another:

Roger can wax a car in 2 hours. Working with Sue, the two can wax a car in 75 minutes. How long would it take Sue to do it alone?

(Using algebra, the answer is that Sue takes 3 hours and 20 minutes alone.)