# Pool

If water flows into the pool by two inlets, fill the whole for 8 hours. The first inlet filled pool 6 hour longer than second. How long pool take to fill with two inlets separately?

**Result****Leave us a comment of this math problem and its solution (i.e. if it is still somewhat unclear...):**

**Showing 3 comments:**

**Math student**

1/t1+1/(t1-10)=1/18

multiply each term by18(t1)(t1-10)

that results in

18(t1-10)+18t1=t1(t1)(t1)-10t1

using the quadratic formula results in t1=-49.6 and 3.63

ubless i made a mistake, your calculations need reexamination!!! Correct me, please.

multiply each term by18(t1)(t1-10)

that results in

18(t1-10)+18t1=t1(t1)(t1)-10t1

using the quadratic formula results in t1=-49.6 and 3.63

ubless i made a mistake, your calculations need reexamination!!! Correct me, please.

10 months ago 2 Likes

**Dr Math**

right side of equation is wrong - should be t1*(t1-10) = t1

^{2}- 10*t1 now t1^{3}-10t1**Math student**

the problems seems to have changed - - - t2 is now equal t1-6

therefore 1/t1+1/(t1-6)=1/18

multiplying each term by18(t1)(t1-6) ==== 18(t1-6)+18t1=t1(t1-6), simplifying further 18t1-108+18t1=t1

or 0=t1

graphing y=18(t1-6)+18t1-t1(t1-6) results in t1=39.25 hours and t2=39.25-6=33.25 hours (same as your NEW answer!!!!

therefore 1/t1+1/(t1-6)=1/18

multiplying each term by18(t1)(t1-6) ==== 18(t1-6)+18t1=t1(t1-6), simplifying further 18t1-108+18t1=t1

^{2}-6t1or 0=t1

^{2}-6t1-18t1+108graphing y=18(t1-6)+18t1-t1(t1-6) results in t1=39.25 hours and t2=39.25-6=33.25 hours (same as your NEW answer!!!!

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