The length of a rectangular room is 6 feet longer than twice the width. if the room perimeter is 168 feet, what are the room's dimensions,?

The dimensions are: "length = 58 ft. and width = 26 ft." ;

or; write as: " 58 ft. × 26 ft. " .

Explanation:

To solve:

Note:

The formula for the perimeter of a rectangle:

P = 2L + 2w ;

in which:

"P = perimeter;

"L = length ;

"w = width ;

The dimensions are the "Length" {"L"] by the width {"w"} ; or; " L × w " ;

Given:

L = 6 + 2w ;

P = 2L + 2w ;

Plug in: "168" (given) for "P".

and: "(6 + 2w)" for "L" ;

to solve for: "w" {the "width"} ;

→ P = 2L + 2w ;

→ 168 = 2(6 + 2w) + 2w ;

Factor out a "2" in the "right hand side" of the equation:

→ 168 = 2 [(6 + 2w) + (w)]

Divide each side o the equation by "2" ;

→ 168 / 2 = { 2 [(6 + 2w) + (w)] } / 2 ;

to get:

→ 84 = [(6 + 2w) + (w)] ;

Note: Simplify the "right-hand side" of the equation:

→ [(6 + 2w) + (w)] = 6 + 2w + 1w = 6 + 3w ;

and rewrite the equation:

→ 84 = 6 + 3w ;

↔ 3w + 6 = 84 ;

Subtract "6" from each side of the equation:

→ 3w + 6 − 6 = 84 − 6 ;

to get:

→ 3w = 78 ;

Divide each side of the equation by "3" ;

to isolate "w" on one side of the equation; & to solve for "w" [the "width"] ;

→ 3w / 3 = 78 / 2 ;

to get:

→ w = 26.

Now:

Given: " L = 6 + 2w " ;

Solve for: "L" .

Plug in "26" for "w" in the equation; to solve for "L" :

→ " L = 6 + 2w ;

→ " L = 6 + 2(26) ;

→ " L = 6 + 52 ;

→ " L = 58 " .

So, we have: " L = 58 ft. " ; and "w = 26 ft."

So the dimensions are: "length = 58 ft. and width = 26 ft." ;

or; write as: " 58 ft. × 26 ft. " .

→ Do not forget the units of "feet" (abbreviated "ft.") !!!

Let us check our

Use the formula:

P = 2L + 2w ;

→ When "P = 168" ; (as given in the problem); does the equation hold true when "L = 58" and when "w = 26" ??

Let us check! ;

→ 168 =? 2(58) + 2(26) ??

→ 168 =? 116 + 52 ??

→ 168 =? 168 ??

→ 168 = 168 !! Yes!

Hope this answer—and explanation—is of help to you.

Best wishes in your academic endeavors—and within the community!

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