rectangle: perim 36, length 3 less than 2X width; find width

[fresh83]

the perimiter of a rectangle is 36 . the length is 3 less then twice the width .

what is the width

 

[Deleted member 4993]

Re: rectangle problem

the perimiter of a rectangle is 36 . the length is 3 less then twice the width .

what is the width

If you had 2*(x+5) - how would you distribute 2?

If you had a*(x+b) - how would you distribute a?

You'll distribute 4/5 the same way.

However, this question (on your image) has nothing to do with problem you have "typed".

for that problem, start defining your variables.

Let

the width of the rectangle be = W

then

the Length of the rectangle be = L = 2*W - 3

Now continue.....

 

[fresh83]

Re: rectangle problem

srry my mistake on the picture , i replaced it with the correct one .

 

[Deleted member 4993]

Re: rectangle problem

srry my mistake on the picture , i replaced it with the correct one .

If a rectangle has length = L and width = W, then

Perimeter = 2L + 2W