Question

The length of a rectangular poster is 9 more inches than two times its width. The area of the poster is 161 square inches. Solve for the dimensions (length and width) of the poster.

Answer 1

Width = 7 inches

Length = 23 inches

Step-by-step explanation:

Let :

• length = 2x + 9
• width = x

Equating to area :

• (2x + 9)(x) = 161
• 2x² + 9x = 161
• 2x² + 9x - 161 = 0
• 2x² - 14x + 23x - 161 = 0
• 2x(x - 7) + 23(x - 7) = 0
• x - 7 = 0
• x = 7

Dimensions are :

• Width = 7 inches
• Length = 2(7) + 9
• Length = 23 inches

Answer 2

Length: 23 inches

Width: 7 inches

Step-by-step explanation:

• Width: w

• Length: 2w + 9

Area of rectangle: Length × Width

• (2w + 9) × (w) = 161

• 2w² + 9w - 161 = 0

• 2w² + 23w - 14w -161 = 0

• w(2w + 23) -7(2w +23) = 0

• w = 7, -11.5

As length measure cannot be negative. Width is +7 inches

Length: 2w + 9 = 2(7) + 9 = 14 + 9 = 23