Respuesta :
Look, Jazly....
Let the width is w
and
the length is l
Then the perimeter is P = 2(l+w)
BUT
The width of the rectangle is 10 more than the length
That means: w=l+10
so we can write:
P = 2(l + w)
= 2( l + l + 10)
= 2(2l+10)
= 4l + 20
and for P = 120
you want to solve for l:
120 = 4l + 20 -20
120-20 = 4l + 20 -20
100 = 4l / 4
l=25
and remember that::: w = l+10
so: w = 25+10=35
w=35
Let the width is w
and
the length is l
Then the perimeter is P = 2(l+w)
BUT
The width of the rectangle is 10 more than the length
That means: w=l+10
so we can write:
P = 2(l + w)
= 2( l + l + 10)
= 2(2l+10)
= 4l + 20
and for P = 120
you want to solve for l:
120 = 4l + 20 -20
120-20 = 4l + 20 -20
100 = 4l / 4
l=25
and remember that::: w = l+10
so: w = 25+10=35
w=35
I hope you got the idea
if you have 120 =2L +2(L+10)
distribute the 2nd part to get:
120 = 2L +2L +20
120 = 4L+20
100=4L
L = 100/4 = 25
length = 25
width = 25+10 = 35