**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