# SOLUTION: My parents have to paint their house. If my son painted it alone it would take him 4 hours longer than I would take. However, if he worked with me, he would work twice as fast.

Algebra ->  Algebra  -> Rate-of-work-word-problems -> SOLUTION: My parents have to paint their house. If my son painted it alone it would take him 4 hours longer than I would take. However, if he worked with me, he would work twice as fast.       Log On

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 Click here to see ALL problems on Rate-of-work-word-problems Question 239722: My parents have to paint their house. If my son painted it alone it would take him 4 hours longer than I would take. However, if he worked with me, he would work twice as fast. Develop a formula for how long it would take both of us working together. Simplify/reduce to lowest terms.Answer by ankor@dixie-net.com(15645)   (Show Source): You can put this solution on YOUR website! If my son painted it alone it would take him 4 hours longer than I would take. However, if he worked with me, he would work twice as fast. Develop a formula for how long it would take both of us working together. Simplify/reduce to lowest terms. : Let x = time required by you working alone then (x+4) = time required by son working alone .5(x+4) = time if he worked twice as fast : Let y = time working together : Let the completed job = 1 : + = 1 : multiply both sides .5x(x+4) .5(x+4)y + xy = .5x(x+4) Factor out y y(.5(x+4) + x) = .5x(x+4) : y(.5x + 2 + x) = .5x^2 + 2x : y(1.5x + 2) = .5x^2 + 2x : y = Where y = hrs when working together using Dad's time working alone (x) ; : Prove that, assume dad can do it in 12 hrs, x = 12, find y y = y = y = y = 96/20 y = 4.8 hrs working together : : Check this in the shared work equation (son's time .5(12+4) = 8 hrs + .4 + .6 = 1; the completed job! : Interesting that the son does more work than the dad except when dad does the job in 4 hrs, then they share the work equally, and do it in two hours : : An interesting problem, thanks for submitting it.