Algebra help, I suck at this!?

Question

The length of a rectangle is fixed at 29 cm. What widths will make the perimeter greater than 80c?

What I have gotten is this:

a rectangle has 4 sides, so I would have to do this 4L + 4W = 80

L=4W+29

4(4W+29)+4W=80

16W+116+4W=80

20W= -36

W = -1.8, so L=4(-1.8)+29

L = 21.8

Am I doing this all wrong, cuz it does not look right to me?

Answer 1

A rectangle perimeter is 2L+2W

Since L=29cm, 2L=58cm

so 58cm+2W>80cm

or 2W > 22cm

W > 11cm

Answer 2

The formula for perimeter is 2L + 2W.  Try that.  So you are looking at 2(29) + 2L > 80

Answer 3

Alright first off I'll address your misunderstanding of the perimeter formula. The perimeter

For example a normal piece of paper is 8.5x11 inches so if you were to picture an ant (bear with me here) walking along the edge of the paper he would travel: 8.5 inches then 11 inches then 8.5 inches then 11 inches and then he'd be back where he started. That means on the trip around the paper he travelled 11 inches twice and 8.5 inches twice. That's how you know that the perimeter is 2l + 2w: it's the sum of the sides.

For an equilateral triangle: side 1 is x inches long, side 2 is x inches long and side 3 is x inches long, so the perimeter is 3x.

For a square: side 1 is x inches long, side 2 is x inches long, side 3 is x inches long and side 4 is x inches long, so P=4x

For a rectangle: side 1 is x inches long, side 2 is y inches long, side 3 is x inches long and side  4 is y inches long so P = 2x + 2y   (we usually use the variables 'l' and 'w' though)

As for the actual answer.. well the width is fixed so it can't be changed. That means you want to find the value for the length that makes the perimeter equal 80cm and therefore any number greater than that will be OK.

so here's how to find what length makes P=80:

we know P=80, P=2l + 2w and w=29

so 80 = 2l + 2w

80 = 2l + 2(29)

80 = 2l + 58

80 - 58 = 2l

22 = 2l

l = 11

therefore any length of 11cm or more will work

{xeIR | l >=11}

Answer 4

A rectangle has 4 sides - opposing sides are of equal length.

2L + 2W  = 80

2(29) + 2W = 80

The perimeter is the sum of all sides.

So we know that one side of the rectangle is 29cm - so it's opposite side is also 29 cm = 29 x 2 = 58 cm.

The perimter wanted is 80 cm.  So   80 - 58 = 22 cm

22 / 2 = 11.  So the other two sides have to be greater than 11 cm

Answer 5

a rectangle has 4 sides, 2 lengths and 2 widths

so,

2L + 2W > 80 where L = 29cm

2*29 + 2W > 80

2W > 22

W > 11

Answer 6

yeah you're kinda off. a rectangle has 4 sides; 2 lengths and 2 widths.

so p(erimeter) = 2W +2L

we know L = 29

so p = 2W +2(29)

p = 2W+58

and p > 80

so 2W +58>80

subtract 58 from both sides:

2W >22

divide by 2:

W >11