Question

What would be the equation (using x) to solve "What are the dimensions of a rectangle whose length is 4 more than twice the width and whose perimeter is 3 less than 7 times the width?"

Answer 1

Answer:Let w = the width

Let 2w + 4 = the length

Let 7w-3 = the perimeter

P = 2l + 2w

7w-3 = 2w + 2(2w+4)

7w-3 = 2w + 4w + 8

w = 11

length = 2w + 4 = 2(11) + 4

length = 26

P = 7(11) -3 = 77 - 3

Perimeter = 74

Explanation:

Answer 2

Let w = the width

Let 2w + 4 = the length

Let 7w-3 = the perimeter

P = 2l + 2w

7w-3 = 2w + 2(2w+4)

7w-3 = 2w + 4w + 8

w = 11

length = 2w + 4 = 2(11) + 4

length = 26

P = 7(11) -3 = 77 - 3

Perimeter = 74