$\displaystyle P=2l+ 2w$

We have the perimeter and the width, so we can plug those values into our equation and solve for our unknown. 

$\displaystyle 32=2l+2(9)$

$\displaystyle 32=2l+18$

Subtract $\displaystyle 18$ from both sides

$\displaystyle 32-18=2l+18-18$

$\displaystyle 14=2l$

Divide $\displaystyle 2$ by both sides

$\displaystyle \frac{14}{2}=\frac{2l}{2}$

$\displaystyle 7=l$

$\displaystyle P=2l+ 2w$

We have the perimeter and the width, so we can plug those values into our equation and solve for our unknown. 

$\displaystyle 30=2l+2(10)$

$\displaystyle 30=2l+20$

Subtract $\displaystyle 20$ from both sides

$\displaystyle 30-20=2l+20-20$

$\displaystyle 10=2l$

Divide $\displaystyle 2$ by both sides

$\displaystyle \frac{10}{2}=\frac{2l}{2}$

$\displaystyle 5=l$

$\displaystyle P=2l+ 2w$

We have the perimeter and the width, so we can plug those values into our equation and solve for our unknown. 

$\displaystyle 34=2l+2(8)$

$\displaystyle 34=2l+16$

Subtract $\displaystyle 16$ from both sides

$\displaystyle 34-16=2l+16-16$

$\displaystyle 18=2l$

Divide $\displaystyle 2$ by both sides

$\displaystyle \frac{18}{2}=\frac{2l}{2}$

$\displaystyle 9=l$

$\displaystyle P=2l+ 2w$

We have the perimeter and the width, so we can plug those values into our equation and solve for our unknown. 

$\displaystyle 18=2l+2(4)$

$\displaystyle 18=2l+8$

Subtract $\displaystyle 8$ from both sides

$\displaystyle 18-8=2l+8-8$

$\displaystyle 10=2l$

Divide $\displaystyle 2$ by both sides

$\displaystyle \frac{10}{2}=\frac{2l}{2}$

$\displaystyle 5=l$

$\displaystyle P=2l+ 2w$

We have the perimeter and the width, so we can plug those values into our equation and solve for our unknown. 

$\displaystyle 24=2l+2(5)$

$\displaystyle 24=2l+10$

Subtract $\displaystyle 10$ from both sides

$\displaystyle 24-10=2l+10-10$

$\displaystyle 14=2l$

Divide $\displaystyle 2$ by both sides

$\displaystyle \frac{14}{2}=\frac{2l}{2}$

$\displaystyle 7=l$

$\displaystyle P=2l+ 2w$

We have the perimeter and the width, so we can plug those values into our equation and solve for our unknown. 

$\displaystyle 20=2l+2(6)$

$\displaystyle 20=2l+12$

Subtract $\displaystyle 12$ from both sides

$\displaystyle 20-12=2l+12-12$

$\displaystyle 8=2l$

Divide $\displaystyle 2$ by both sides

$\displaystyle \frac{8}{2}=\frac{2l}{2}$

$\displaystyle 4=l$

$\displaystyle P=2l+ 2w$

We have the perimeter and the width, so we can plug those values into our equation and solve for our unknown. 

$\displaystyle 14=2l+2(5)$

$\displaystyle 14=2l+10$

Subtract $\displaystyle 10$ from both sides

$\displaystyle 14-10=2l+10-10$

$\displaystyle 4=2l$

Divide $\displaystyle 2$ by both sides

$\displaystyle \frac{4}{2}=\frac{2l}{2}$

$\displaystyle 2=l$

$\displaystyle P=2l+ 2w$

We have the perimeter and the width, so we can plug those values into our equation and solve for our unknown. 

$\displaystyle 22=2l+2(2)$

$\displaystyle 22=2l+4$

Subtract $\displaystyle 4$ from both sides

$\displaystyle 22-4=2l+4-4$

$\displaystyle 18=2l$

Divide $\displaystyle 2$ by both sides

$\displaystyle \frac{18}{2}=\frac{2l}{2}$

$\displaystyle 9=l$

The fence is going around the garden, so this is a perimeter problem. 

$\displaystyle P=2l+2w$

$\displaystyle P=2(6)+2(3)$

$\displaystyle P=12+6$

$\displaystyle P=18$

The fence is going around the garden, so this is a perimeter problem. 

$\displaystyle P=2l+2w$

$\displaystyle P=2(4)+2(10)$

$\displaystyle P=8+20$

$\displaystyle P=28$