I will give 30 points to whoever answers this correctly and within 5-10 mins.

The perimeter of a rectangular room Is 24 feet. Twice the length decreased by three times the width is 4 feet. What are the dimensions of the room?

Question 1
Reread the question. What dimensions are you solving for?
Group of answer choices

perimeter

length

width

diagonals

Area

Question 2

Choose variables and write a system of equations.
Use elimination method to add or subtract equations.
Substitute the solution into one of the original equations and solve for the remaining variable.
Check answer by substituting solution into original equations.

Question 3
Interpret the answer in terms of the original question.

The length of the room is___feet and the width of the room isfeet.

1)The dimensions you  are solving for

length of the rectangular room

width of the rectangular room

2)Now let the length be L and the width be w then

the perimeter is 2L+2W

2L+ 2W  = 24(1)

Also, "Twice the length decreased by three times the width is 4 feet" can be written as

2L - 3W = 4(2)

Solving (1) and (2) by elimination method, by subtraction

2L+ 2W  = 24

2L - 3W = 4

(-)    (-)       (-)

0L  +5W = 20

5W = 20

W = 4(3)

Now by substitution method,substituting (3) in(1)

2L+ 2(4)  = 24

2L + 8 = 24

2L = 24 -8

2L = 16

L = 8

Substituting in the original equation and rechecking the perimeter

=2(8) + 2(4)

=16 + 8

= 24

Thus the found dimensions are correct

3)The length of the room is 8 feet and the width of the room is 4 feet.

just do 5.2 + 2 ( 7.2) and 1/3 x 3 (.9) and youv'e got your equation. 0.9( 9/10) + 7.2 ^2 = 16.4 hope i could !

step-by-step explanation:

step-by-step explanation:

the true statements are:

r = set of real numbers

n = set of natural numbers, 1,2,

w = n u{0} = {0,1,}