The length of a rectangle is 6in more than twice its width If the perimiter of the rectangle is 48 in what wid

Question

what is the width of the rectangle

Answer 1

basically the width = w
and the length= 6 + 2w

so the perimeter would equal= 2(6+2w)+2(w)
add each side, 2 sides equal to the length and two to the width

and since the perimeter = 48
then: 2(6+2w)+2(w)=48

and solve from there

(12+4w)+(2w)=48
12+4w+2w=48
12+6w=48 subtract 12 from each side
6w=36 divide each side by 6
w=6

so the width = 6

and then u can find the length if needed by plugging 6 in as w to the equation that defines length (see above)

hope that helps!

Answer 2

hey, first suppose width to be X (or any other variable of ur choice).
then,if width is X (or any other variable)
then length,which is 6in more than twice the width,wd be (2X+6)in.
n perimeter=2(length+width)
or 48=2(2X+6+X)
or 24=3X+6
or 18=3x
or X = 6 inches.
i.e width =6 inches.
n length= 2(6) +6= 18 inches.

p.s........u can cross check ur answer by putting the values u obtd in the formula u used for perimeter earlier.the answer wd be 48.

Answer 3

L=6+2w
2w+2L=48

Replace the L in the second statement with the equation that equals L (the first statement).

2w+2(6+2w)=48
2w+12+4w=48
2w+4w=36
6w=36
width=6