1.The length of a rectangular room is 9 feet longer than twice the width. If the room's perimeter is 210 feet, what are the room's dimensions?

rooms-dimensions

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1.The length of a rectangular room is 9 feet longer than twice the width. If the room’s perimeter is 210 feet, what are the room’s dimensions?
A) Label the two unknowns you are asked to find in the problem with variables and give a sentence stating what each variable represents. (worth 2 points)
B) Give the two equations you will use to solve for the unknown variables. (worth 2 points)
C) Solve the system of two equations with two unknowns by either the substitution or the elimination method and show your work/steps. (worth 2 points)
D) State the answer to the bolded question (or the answer to what you are asked to find in bold if there is no direct question) in the form of a sentence with appropriate units of measure on numeric values (examples of units: miles, ounces, dollars, etc.). (worth 2 points)
E) Check or verify your answers in the scenario; explain why they make sense for the situation in the problem. (worth 2 points)

Answer:

The length of a rectangular room is 9 feet longer than twice the width. If the room’s perimeter is 210 feet, what are the room’s dimensions?
A) Let the length of the rectangular room be x
Let the width of the rectangular room be y
B) We shall convert the word problem into equations form
Now x = y + 9
Also
Perimeter = 210
2(x+y) = 210
Dividing by 2 we get
x + y = 105
The equations are
x + y = 105
x = y+9
C)
Substituting
x= y+9 in
x + y = 105
y + 9 + y = 105
re-arranging and adding the like terms we get
2y = 105 -9
2y = 96
y = 48
Hence x = 48 + 9 = 57
D)
Length = x = 57 ft
Width = y= 48 ft
E) Here
Length - width = x-y= 57 - 48 = 9ft
So length is 9 more than width.
Also
Perimeter
= 2(57 + 48)
=2(105)
=210 ft
hence the lengths are verified