Answer:
[tex]\boxed{\frac{7}{5}y} \longrightarrow \boxed{y+\frac{2}{5}y}[/tex]
[tex]\boxed{0.68y} \longrightarrow \boxed{y-0.32y}[/tex]
[tex]\boxed{\frac{3}{5}y} \longrightarrow \boxed{y-\frac{2}{5}y}[/tex]
[tex]\boxed{1.32y} \longrightarrow \boxed{y+0.32y}[/tex]
Step-by-step explanation:
[tex]\begin{aligned}\implies y-0.32y & = 1y-0.32y\\& = (1-0.32)y\\& = 0.68y\end{aligned}[/tex]
[tex]\begin{aligned}\implies y+0.32y & = 1y+0.32y\\& = (1+0.32)y\\& = 1.32y\end{aligned}[/tex]
[tex]\begin{aligned}\implies y+\dfrac{2}{5}y & = 1y+\dfrac{2}{5}y\\& = \dfrac{5}{5}y+\dfrac{2}{5}y\\& = \left(\dfrac{5}{5}+\dfrac{2}{5}\right)y\\& = \left(\dfrac{5+2}{5}\right)y\\& = \dfrac{7}{5}y\\\end{aligned}[/tex]
[tex]\begin{aligned}\implies y-\dfrac{2}{5}y & = 1y-\dfrac{2}{5}y\\& = \dfrac{5}{5}y-\dfrac{2}{5}y\\& = \left(\dfrac{5}{5}-\dfrac{2}{5}\right)y\\& = \left(\dfrac{5-2}{5}\right)y\\& = \dfrac{3}{5}y\\\end{aligned}[/tex]
