# If the side of a square is doubled and then increased by 7, the new perimeter is 8 more than 3 times the old perimeter. ?

What is the side length of the original square??

Relevance
• 1 month ago

First stage, the side is s₁. The perimeter is:

p₁ = 4.s₁

Second stage, the side is doubled:

s₂ = 2.s₁ and then increased by 7

s₂ = 2.s₁ + 7

The perimeter is:

p₂ = 4.s₂

p₂ = 4.(2.s₁ + 7)

p₂ = 8.s₁ + 28

The new perimeter is 8 more than 3 times the old perimeter.

p₂ = 3.p₁ + 8 → recall: p₂ = 8.s₁ + 28

8.s₁ + 28 = 3.p₁ + 8

8.s₁ - 3.p₁ = - 20 → recall: p₁ = 4.s₁

8.s₁ - 3.(4.s₁) = - 20

8.s₁ - 12.s₁ = - 20

- 4.s₁ = - 20

s₁ = 5 ← this is the side length of the original square

• 1 month ago

2S + 7 so that 4(2S + 7) = 8*4S find S = ?  We have 8S + 28 = 32S; so that 28 = 24S and S = 7/6 ANS.

• 1 month ago

If the side of a square is doubled and then increased by 7,

the new perimeter is 8 more than 3 times the old perimeter.

The old perimeter is 4s and the new perimeter is 12s + 8

What is the side length of the original square?

• david
Lv 7
1 month ago

old perimeter  =  4x

new per. = 4(x + 7) = 8  + 3(4x)

, , ,   4x + 28  =  12x + 8

. . .   8x = 20

. . .   x = 2.5  <<<  answer