Factoring and all that (finding room's dimensions)

[Guest]

I really don't understand this exercise. I am in an "Independant" math group so I need some help. The problem is:

A room's length is 3 feet less than twice its width. The area of the room is 135 square feet. What are the room's dimensions?

So i figured that:

length=2w-3

...but that's about all I can get. Any help would be super! Thanks!

 

[stapel]

So you've defined "w" to stand for "width" and "2w - 3" to stand for "length". You've actually done most of the hard stuff! :wink:

Now plug the "wight" and the "length" into the "area" formula, plug "135" in for the "area" A, and solve the resulting equation for the value of "w". Then back-solve for the value of the length.

Eliz.

 

[cole92]

I really don't understand this exercise. I am in an "Independant" math group so I need some help. The problem is:

A room's length is 3 feet less than twice its width. The area of the room is 135 square feet. What are the room's dimensions?

So i figured that:

length=2w-3

...but that's about all I can get. Any help would be super! Thanks!

ok, first...yes, the length would be:

L=2w-3

for the widht, it would just be w.....

the area is 135, so the equation would be:

(2w-3)(w)=135

now, i think you can solve from there