Question

If the perimeter of a rectangle is 152m and the length is 7m more than twice the width how do I find the length and width?

Answer 1

P =(L+W)*2
L=7+2W
152= (7+2W)+W))*2
152/2= (7+2W)+W
76 = 7+ 3W
76-7 = 3W
69/3 = W
23= W
L = 7+(2*23)
L =53

Answer 2

Let L be the length and W be the width.
Well the perimeter is 2L + 2W, so
2L + 2W = 152

And the length is given to be 7 more than 2 times the width so:
2W +7 = L.
Therefore 2W = L - 7

Putting both those equations together you have:
\[2L + 2W = 152\]\[\implies 2L + (L - 7) = 152\]\[\implies 3L = 152 + 7\]\[\implies 3L = 159 \]\[\implies L = 53\]

And since
\[2W = L - 7\]\[\implies 2W = 53 - 7 = 46\]\[\implies W = 46/2 = 23\]

Answer 3

Thank you !

Answer 4

u r w come