What are the possible lengths and widths of a rectangle if its length is 8 centimeters longer than the width, and the area is less than or equal to 33 square centimeters?

The perimeter is twice the length plus twice the width, or 2x+2y. 

Therefore, 2x+2y = 38 as given.

Also, the length is 5 ft longer than the width so x = y + 5

Knowing these equations are true, you can solve.

Substitute for x: 2(y+5) + 2y = 38

Distribute: 2y + 10 + 2y = 38

Simplify: 4y + 10 = 38

Subtract 10: 4y=28

Divide by 4: y=7

x=y+5 = 12

The length is 12, which is 5 more than the width of 7. The perimeter is 2(12+7) = 2(19) = 38

really really sorry if i'm

-12 -5(x + 4) = 3x

-12 - 5x - 20 = 3x

-12 - 20 - 5x = 3x

-32 - 5x = 3x

    + 5x   + 5x

-32 = 8x

8         8

- 4= x