A leaky valve on the water meter overcharges the residents for one gallon of water every 1 1 over 3 months. The overcharged amount w varies directly with time t. Find the equation that models this direct variation. How many months it will take for the residents to be overcharged for 5 gallons of water? w = 3 over 4t; 62 over 3 months w = 4 over 3t; 62 over 3 months w = 3 over 4t; 33 over 4 months w = 4 over 3t; 33 over 4 months
it overcharges \(1\) gallon for every \(\large 1 \frac{1}{3} \) months
oh ok...so then it'd be the last one?
ive second thoughts, scratch that
Ok..
\(\large 1 \frac{1}{3}= \frac{4}{3} \) if w is the gallons over charged, every 4/3 months u will be overcharged 1 gallone so, the required equation wud be \(\large \color{red}{w= \frac{3}{4}t} \) where, \(w\) is gallons over charged \(t\) is time in months
this will work, cus, when u put in t = 4/3 months, u wud get 1 gallon
Oh ok, I see now.
good :) next do the second part simply replace w by 5 and solve that equation for t
\(\large \color{red}{w= \frac{3}{4}t} \) u wanto knw how many months(t) it will take for residents to be overcharged 5 gallons so, put w = 5 \(\large \color{red}{5= \frac{3}{4}t} \) solve \(t\)
wouldn't I, multiply or something on those lines the 5 and the fraction?
multiply 4 both sides
\(\large \color{red}{5= \frac{3}{4}t} \) multiply 4 both sides \(\large \color{red}{5 \times 4= 4 \times \frac{3}{4}t} \) \(\large \color{red}{20 = 3t} \)
next divide 3 both sides
u wud get \(\large \frac{20}{3} = t \)
then woldnt I get 6
u can convert it to mixed fraction like below :- \(\large \frac{20}{3} = \frac{18+2}{3} = 6 + \frac{2}{3} = 6 \frac{2}{3}\)
how did you get the 18+2?
see if that makes more/less sense... to be clear, the answer is first option.
20 = 18+2
oh durr.. lol
i believe u asked me why i wrote 20 as 18+2, but not 17+3, or 16+4 right ?
yes
its just that i figured out 18 is the multiple of 3, so it cancels... once u knw how to convert to mixed fractions, u will see ...
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