The rectangle shown has a perimeter of 80 cm and the given area. Its length is 4 more than twice its width. Write and solve a system of equations to find the dimensions of the rectangle.

Question

I am not sure how to solve this, nor set up the equation. Please help.

Answer 1

Let x=width

length = 2x+4 (This is 4 more than twice the width)

Perimeter = all 4 sides added together

80 = x + x + (2x+4)+ (2x+4) Simplify by combining like terms

80 = 6x+8 Subtract 8 from both sides

80-8 = 6x+8-8 Simplify

72 = 6x Divide both sides by 6

72/6 = 6x/6 Simplify

x=12 cm is the width

Solve for length

length = 2x+4 = 2(12)+4 = 28 cm is the length

Answer 2

L = length. and W = width

Convert sentences into equations:

"rectangle shown has a perimeter of 80 cm "

2L + 2W = 80 <<eqn 1

"length is 4 more than twice its width."

L = 2W + 4

L - 2W = 4 <<eqn2

Add equations 1 and 2 to get:

3L = 84

L = 28

From equation 2, 2W = L - 4 = 24, so W = 12