I don't know how to solve this question. Help me out please.

Question

A rectangle has perimeter 80 cm and its length is 1 cm more than twice its width. Find the dimensions of a rectangle given that its perimeter is 

 80 cm and its length is 1  cm more than twice its width.

 

Set up your solution using the variables L for the length, W for the width, and P for the perimeter. 

Part a: Using the definition of perimeter, write an equation for P 

 in terms of L and W .

 

Part b: Using the relationship given in the problem statement, write an equation for L in terms of W .

 

Solve the equations from parts a and b.

Answer 1

Hi Abriana F.

a.

P = 2L + 2W

80 = 2L + 2W

b.

L = 2W + 1

c.

since L = 2W + 1 we can substitute

P = 2(2W + 1) + 2W

80 = 2(2W + 1) + 2W

Divide the equation by 2

40 = 2W + 1 + W

40 = 3W + 1

40 - 1 = 3W

39 = 3W

Divide both sides by 3

13 = w

2W + 1 = L = 2(13) + 1 = 26 + 1 = 27

Checking

80 = 2(27) + 2(13)

80 = 54 + 26

80 = 80

Hope this helps

Answer 2

L = 1 + 2W

P = 80 = 2L + 2W

40 = L + W

W = 40-L

L = 1 + 2(40-L)

L = 1 + 80-2L

3L = 81

L = 81/3 = 27 cm Long

W = 40-L = 40-27 = 13 cm Wide

P = 2(L+W) = 2(27+13) = 2(40) = 80

L = 1+2(13) = 1+26 = 27 cm