Write an algebraic equation to solve the problem

Question

The length of a rectangular room is 9 feet longer than twice the width. If the rooms perimeter is 174 feet what are the rooms dimensions?

 

A) Width =39ft; Length =48ft

B) Width=26ft; Length=61ft

C) Width= 52ft; Length= 122ft

D) Width=31ft; legth=71ft

Answer 1

First, read the problem and understand it.  The room is a rectangle.  A relationship between its length and width is given.  The perimeter is given.  What are the length and width?

 

Also, we know that:   P= L+W+L+W

 

Translate:

"The length of a rectangular room is 9 feet longer than twice the width"   means

           L                                     = 9               +                     2W

  "the rooms perimeter is 174 feet"   means      P=174=2L+2W

  "what are the rooms dimensions?"     means     report L,W

 

Now, the easy math:

  Solve for either L or W in one equation and substitute that into the other equation; or, use elimination (which I like):

      P = 174 = 2L + 2W

              -9 =  -L + 2W     (rearranged second equation)

          --------------------

            183  = 3L             (subtract equations)

              61 = L               (divide both sides by 3)

 

Now, substitute:  [ note: even though only one answer has a 61 ]

      L = 9 + 2W

      61 = 9 + 2W

      52 = 2W                 (subtract 9 from both sides)

      26 = W              (divide both sides by 2)

     

B) Width=26ft; Length=61ft

Answer 2

w represents the width

2w + 9 represents the length

 

P = 2(length + width)

 

174 = 2(2w + 9 + w)      divide both sides by 2

87 = 2w + 9 + w             combine like terms

87 = 3w + 9                   subtract 9 from both sides

78 = 3w                          divide both sides by 3

26 = w

 

The length is 61 ft and the width is 26 ft.

 

Check

2(61) + 2(26)

122 + 52

174