SOLUTION: the length of a rectangle is 4 inches greater than the width the perimeter is 16 inches what is the width of the rectangle 2(w+4)+2(w)=16 i know that you multiply and come out with

Algebra ->  Rectangles -> SOLUTION: the length of a rectangle is 4 inches greater than the width the perimeter is 16 inches what is the width of the rectangle 2(w+4)+2(w)=16 i know that you multiply and come out with      Log On


   

Question 1007659: the length of a rectangle is 4 inches greater than the width the perimeter is 16 inches what is the width of the rectangle 2(w+4)+2(w)=16 i know that you multiply and come out with 2w+8+2w=16 combine the w's and you get 4w+8=16 subtract 8 from both sides and 16-8=8 so would it be 4/8 i'm not sure
Found 2 solutions by josgarithmetic, addingup:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
w for width and L for length;
system%28L=w%2B4%2C+2w%2B2L=16%29

system%28L=w%2B4%2Cw%2BL=8%29-----do you know how?

w%2B%28w%2B4%29=8-------simple substitution.

Answer by addingup(3677) About Me  (Show Source):
You can put this solution on YOUR website!
2L+2W= Perimeter, in this case 16
And:
L= W+4
So in the first formula let's substitute L with this value:
2(W+4)+2W= 16 Multiply on left to eliminate parenthesis
2W+8+2W= 16 Subtract 8 on both sides and ad W on left:
4W= 8 Divide both sides by 4
W= 2 This is your width, and your length, the problem says:
L= W+4
L= 2+4
L= 6
---------------------------
Check:
2L+2W= 16
2(6)+2(2)= 16
12+4= 16
16= 16 We have the correct answer