Question

(y+1/3)+(y+1/9)+(y+1/27)+(y+1/81)=56/81

Answer 1

(y + \(\dfrac{1}{3}\)) + ( y + \(\dfrac{1}{9}\)) + ( y + \(\dfrac{1}{27}\)) + ( y + \(\dfrac{1}{81}\)) = \(\dfrac{56}{81}\)

( y + y + y + y ) + (\(\dfrac{1}{3}\)+ \(\dfrac{1}{9}\) + \(\dfrac{1}{27}\) + \(\dfrac{1}{81}\)) = \(\dfrac{56}{81}\)

4\(y\) + ( \(\dfrac{27}{81}\) + \(\dfrac{9}{81}\) + \(\dfrac{3}{27}\) + \(\dfrac{1}{81}\) ) = \(\dfrac{56}{81}\)

4y + \(\dfrac{40}{81}\) = \(\dfrac{56}{81}\)

4y = \(\dfrac{56}{81}\) - \(\dfrac{40}{81}\)

4y = \(\dfrac{16}{81}\)

y  = \(\dfrac{16}{81}\) : 4

y = \(\dfrac{4}{81}\)

Answer 2

\(\left(y+\dfrac{1}{3}\right)+\left(y+\dfrac{1}{9}\right)+\left(y+\dfrac{1}{27}\right)+\left(y+\dfrac{1}{81}\right)=\dfrac{56}{81}\)

\(\Rightarrow y+\dfrac{1}{3}+y+\dfrac{1}{9}+y+\dfrac{1}{27}+y+\dfrac{1}{81}=\dfrac{56}{81}\)

\(\Rightarrow4\times y+\dfrac{40}{81}=\dfrac{56}{81}\)

\(\Rightarrow4\times y=\dfrac{56}{81}-\dfrac{40}{81}\)

\(\Rightarrow4\times y=\dfrac{16}{81}\)

\(\Rightarrow y=\dfrac{16}{81}:4\)

\(\Rightarrow y=\dfrac{4}{81}\)

Answer 3

\((y+\dfrac{1}{3})+(y+\dfrac{1}{9})+(y+\dfrac{1}{27})+(y+\dfrac{1}{81})=\dfrac{56}{81}\)

\(=>4y+(\dfrac{1}{3}+\dfrac{1}{9]+\dfrac{1}{27}+\dfrac{1}{81})=\dfrac{56}{81}\)

\(=>4y+\dfrac{1}{3}(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27})=\dfrac{56}{81}\)

\(=>4y+\dfrac{1}{2}(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}-\dfrac{1}{3}-\dfrac{1}{9}-\dfrac{1}{27}-\dfrac{1}{81})=\dfrac{56}{81}\)

\(=>4y+\dfrac{1}{2}(1-\dfrac{1}{81})=\dfrac{56}{81}\)

\(=>4y+\dfrac{1}{2}\times\dfrac{80}{81}=\dfrac{56}{81}\)

\(=>4y+\dfrac{40}{81}=\dfrac{56}{81}\)

\(=>4y=\dfrac{56}{81}-\dfrac{40}{81}\)

\(=>4y=\dfrac{16}{81}\)

\(=>y=\dfrac{4}{81}\)

Answer 4

\(\left(y+\dfrac{1}{3}\right)+\left(y+\dfrac{1}{9}\right)+\left(y+\dfrac{1}{27}\right)+\left(y+\dfrac{1}{81}\right)=\dfrac{56}{81}\)

\(\Rightarrow4y+\left(\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}\right)=\dfrac{56}{81}\)

\(\Rightarrow4y+\left(\dfrac{27}{81}+\dfrac{9}{81}+\dfrac{3}{81}+\dfrac{1}{81}\right)=\dfrac{56}{81}\)

\(\Rightarrow4y+\dfrac{40}{81}=\dfrac{56}{81}\)

\(\Rightarrow4y=\dfrac{56}{81}-\dfrac{40}{81}=\dfrac{16}{81}\)

\(\Rightarrow y=\dfrac{16}{81}:4=\dfrac{16}{81}.\dfrac{1}{4}=\dfrac{4}{81}\)