Question

Find the length of the rectangle where the length is 12 units more than the width and the perimeter is 8 times the width.

Answer 1

perimeter = 8w
2(l+w) = 8w
2(w+12+w) 8w
2w+24+2w=8w
solve for w....
then subtitute back w to l=w+12

Answer 2

so when i substitute w to l=w+12, i don't understand what to do after that...

Answer 3

Solve the following for w and L, the width and length respectively:\[\{L=12+w,2L+2w=8w\}\]\[\{w=6,L=18\} \]

Answer 4

@RadEn

Answer 5

2w+24+2w=8w
4w+24=8w
24 = 8w-4w
24=4w
w=24/4=6
l=w+12=6+12=18

Answer 6

oh okay, thank you(: @RadEn