Question

the length of a rectangular is one less than twice the width. Find the length + width, if the perimeter is 40

Answer 1

`\huge\boxed{\sf Length = 13, \ \ Width = 7}`

Explanation:

Let the length of rectangle be x and width be y.

Condition # 1:

x = 2y - 1 --------------(1)

Condition # 2:

2x + 2y = 40

• Take 2 as a common factor

2(x + y) = 40

• Divide both sides by 2

x + y = 40/2

x + y = 20 -----------(2)

• Put Eq. (1) in Eq. (2)

2y - 1 + y = 20

2y + y - 1 = 20

3y - 1 = 20

• Add 1 to both sides

3y = 20 + 1

3y = 21

• Divide both sides by 3

y = 21/3

y = 7

• Put y = 7 in Eq. (1)

x = 2(7) - 1

x = 14 - 1

x = 13

So,

Length = 13

Width = 7

`\rule[225]{225}{2}`