The length of a rectangle is six units longer than the width.?

Question:Find the length and width if the perimeter of the rectangle is 60 units? Translate to an algebraic equation.

Answers:
2w + 2L = 60
L = w + 6

so substitute...

2w + 2(w+6) = 60 --> 2w + 2w + 12 = 60
4w = 48
w = 12, L = 18
lenght is 25 and width is 35 so...
so, width = x, length = x + 6

perimeter = 2x + (2(x+6))
60 =4x+12
48 = 4x
x =12

Length = 18, width 12
x = one side
x + 6 = the longer side

So

2(x + x + 6) = 60

Reduced:
2(2x+6)=60


...I think. It's been a few years since I needed that info.
length =l=width + 6 units
width = w
perimeter is 2L+2W=60 units

2(w+6)+2w=60
2w+12+2w=60
4w+12=60
4w=48
w=12
l = 12+6 =18

12+12+18+18=60
24+36=60
60=60
Let L=Length, and W=Width. P=Perimeter.

Equation
P=2(L+W)

Known values
P=60, L=W+6

Substitution
60=2(W+6+W)

Combine like terms
60=2(2W+6)

Multiply all terms within brackets by 2
60=4W+12

Subtract 12 from both sides
48=4W

Divide by 4 on both sides
12=W

Thus, the width is 12 units, and the length is six more, 18 units.
You need the length and width (L & W).

If L is six longer than W, than L=W+6.
Perimeter (P) is 2L + 2W. .

P=60, so 60=2(L) + 2(W). L=W+6, so plug that in for L on the perimeter equation.

60=2(W+6) + 2W... multiply it out, so 2W+12+2W =60

Combine like terms and you get 4W+12=60

Subtract 12 from both sides and you get
4W=48

Divide by 4
W=12.

If L=W+6, and W=12, then 12+6=18

L=18 and W=12

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