The length of a rectangular poster is 5 more inches than half its width. The area of the poster is 12 square inches. Solve for the dimensions (length and width) of the poster.

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MrRoyal MrRoyal · Answer 1 · 2022-08-05T18:34:45+02:00

The dimensions (length and width) of the poster are 6 inches and 2 inches

Represent the length with x and the width with y

From the question, we have the following parameters

x = 5 + 0.5y

Area, A = 12

The area of a rectangle is represented as:

A = xy

Substitute the known values in the above equation

(5 + 0.5y) * y = 12

Expand the bracket

5y + 0.5y^2 = 12

Multiply through by 2

10y + y^2 = 24

Rewrite as:

y^2 + 10y - 24 = 0

Expand

y^2 + 12y - 2y - 24 = 0

Factorize the expression

(y - 2)(y + 12) = 0

Solve for y

y = 2 or y = -12

The dimension cannot be negative.

So, we have

y = 2

Substitute y = 2 in x = 5 + 0.5y

x = 5 + 0.5 * 2

Evaluate

x = 6

Hence, the dimensions (length and width) of the poster are 6 inches and 2 inches