Question

2+y/3=4y/9-3

Answer 1

\(\bf 2+\cfrac{y}{3}=\cfrac{4y}{9}-3\quad ?\)

Answer 2

solve for variable then check your answer

Answer 3

\(\bf 2+\cfrac{y}{3}=\cfrac{4y}{9}-3\qquad {\color{red}{ \times 9}}\\ \quad \\

{\color{red}{ 9\times}} 2+{\color{red}{ 9\times}} \cfrac{y}{3}={\color{red}{ 9\times}} \cfrac{4y}{9}-{\color{red}{ 9\times}} 3\)

that'd get rid of the denominator
then just solve for "y"

Answer 4

yes go on

Answer 5

\(\bf {\color{red}{ 9\times}} 2+\cancel{{\color{red}{ 9\times}}} \cfrac{y}{\cancel{3}}=\cancel{{\color{red}{ 9\times}}} \cfrac{4y}{\cancel{9}}-{\color{red}{ 9\times}} 3\implies 18+3y=4y-27\)

Answer 6

surely you can take it from there

Answer 7

no

Answer 8

ok then \(\bf 18+3y=4y-27\qquad {\color{red}{ -3y}}\\ \quad \\

18\cancel{+3y-3y}=4y-3y-27\implies 18=y-27\qquad {\color{red}{ +27}}\\ \quad \\
18+27=y\cancel{-27+27}\)