Help with another math problem?

Question

If the perimeter of a Rectangular courtyard is 34yd and the length is 2yd longer than twice the width, then what ar the length and width?

P=2L+2W is the formula

Answer 1

Its very simple. P=34 so

2L + 2W = 34 =>  L + W = 17 .....eq(1)

now

L = 2W +2    replace this value of L in eq(1) and find out W. and then L.

Ans: W=5, L=12.

Answer 2

let the width be x

hence length is 2x+2

P= 2L + 2W

also P= 34 yd.

hence 34 yd= 2(2x+2) +2x

34 yd = 4x+4+2x

hence 34 -4=4x+2x

hence 30 yd= 6x

hence x=5 yd

therefore

the length=2x+2= 2*5 +2= 12yd

hence width is 5yd

hope this helps.

Answer 3

P = 2(2W+2) + 2W

P= 6W+4

34=6W+4

30=6W

W=5yds

L=12yds

Answer 4

L = 2W + 2yd

P = 2L + 2W

34 yd = 2(2W + 2yd.) + 2W

34 yd = 4w + 4yd. + 2W

30 yd = 6W

W = 5

L = 12

2(5) + 2(12) = 34