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Mathematics 27 Online
OpenStudy (anonymous):

Harry can rake the leaves in the yard 8 hours faster than his little brother Jimmy can. If they work together, they can complete the job in 3 hours. Using complete sentences, explain each step in figuring out how to determine the time it would take Jimmy to complete this job on his own???

OpenStudy (anonymous):

please help!! @Muzzack @RadEn @iPwnBunnies

OpenStudy (raden):

(x + x- 8 )/x(x-8) = 3 solve for x

OpenStudy (anonymous):

?

OpenStudy (raden):

opsss, backward that should like x(x-8)/(x+x-8) = 3

OpenStudy (anonymous):

ok so cant we cancel some x's or no

OpenStudy (raden):

with x is the Jimmy's time if he complete the job alone x(x-8)/(x+x-8) = 3 (x^2 - 8x)/(2x-8) = 3 cross multiply then simplify and solve for x (notice that it must be positive)

OpenStudy (anonymous):

\[\frac{ x ^{2}-8x }{ 2x-8 }=\frac{ 3 }{ 1 }\]

OpenStudy (raden):

yes, just continue that ...

OpenStudy (anonymous):

so x^2-8x/6x-24?

OpenStudy (raden):

x^2-8x = 6x-24 simply by combine then solve for x

OpenStudy (anonymous):

x^2-14x+24=0? Then what

OpenStudy (raden):

looks it can be factored :) can you factorize the LHS ?

OpenStudy (anonymous):

(x-12)(x-2)=0

OpenStudy (raden):

yes, that's right. then set up each factor be equal zero then solve for x

OpenStudy (anonymous):

x=12 and x=2? is that what you meant?

OpenStudy (raden):

yep, but we have to recheck again which one of that numbers can be solution for this problem. see before i take by assumed x = jimmy's time and x-8 = harry's time if they do the job alone. now if you subtitute both value of x above, which one possible as the answer ? if x = 2, then x - 8 = 2 - 8 = -6 (it cant) if x = 12, then x-8 = 12-8 = 4 (nah, this is a positive number), so x = 12 is satisfies as the solution

OpenStudy (anonymous):

ok so x=12 is the solution? :)

OpenStudy (raden):

yep, because it would give us Jimmy = 12 Harry = 4 (see that harry 8 hours faster than Jimmy)

OpenStudy (anonymous):

ohhh thank you! :D

OpenStudy (raden):

you're welcome :)

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