Question

Perimeter of a rectangle is 22 inches and its length is 4 inches less than twice its width. Set up a system of linear equations and solve to find the dimensions of the rectangle.

Answer 1

Systems of linear equations are 2(L + W)  = 22 and L = 2W - 4 and the dimensions of the rectangle are length 6 inches and width 5 inches.

Formula used:

• Perimeter of a rectangle, P = 2(length+ width)
• P = 2(L + W)

Given:

The perimeter of a rectangle = 22 inches

Length = 4 inches less than twice its width.

To find: The dimensions of the rectangle

Solution:

Step 1: Take some variables for the length and width of a rectangle:

Let W be the Width of the rectangle and L be the length of the rectangle

Step 2: Write an equation according to the given conditions:

Perimeter of a rectangle =  2(L + W)

2(L + W)  = 22 …. (1)

Length is 4 inches less than twice its width.

L = 2W - 4 …….(2)

Step 3: Substituting the value of L from eq. 2 in eq.1 and solve:

2(L + W) = 22

$L + W = \frac{22}{2}$

$L + W = 11\\\\(2W - 4 ) + W = 11\\\\2W + W - 4 = 11\\\\3W = 11 + 4\\\\3W = 15$

$W =$ $\frac{15}{3}$

W = 5 inches

Width of a rectangle = 5 inches

Step 4: Substitute the value of W in eq. 2 :

$L = 2W - 4\\\\L = 2(5) - 4\\\\L = 10 - 4\\\\L = 6\ inches$

Length of a rectangle = 6 inches

Hence, the system of linear equations are 2(L + W)  = 22 and L = 2W - 4 and the dimensions of the rectangle are length of 6 inches and a width of 5 inches.

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