I know the answer to this algebra problem is 2 hours and 24 minutes but how do you solve it? If Sally can paint a house in 4 hours, and John can paint the same house in 6 hour, how long will it take for both of them to paint the house together? 2 hours and 24 minutes 3 hours and 12 minutes 3 hours and 44 minutes 4 hours and 10 minutes 4 hours and 33 minutes We have two responses for you, one that uses algebra and another that doesn't use algebra. With algebra: If you multiply the speed of housepainting (houses/hour) by the number of hours, you get the number of houses, because the hours units cancel out. John and Sally, when painting together, are painting for the same amount of time, let's call that T. Then the total amount of houses that John paints in time T is T x 1/6 = T/6, because he paints 1 house in 6 hours alone. The total amount that Sally paints is T x 1/4 = T/4 for the same reason. So the total amount of houses they paint together is T/6 + T/4. But you know they together paint exactly one house in that time, so: 1 = T/6 + T/4. If you now solve for T, you will have your time (expressed in hours, so you may need to calculate the minutes from the fractional part of the answer). Without algebra: Sally can paint a house in 4 hours so she paints an the rate of 1/4 of a house per hour. John can paint a house in 6 hours so she paints an the rate of 1/6 of a house per hour. Thus together they paint at the rate of 1/4 + 1/6 of a house per hour. Express this sum as a single fraction. At that rate how long does it take to paint 1 house? Stephen and Penny