A few days ago

# ii need help on a math prob using formulas to solve a problem.?

the length of a rectangle exceeds its width by 4 ft. if the width is doubled and the length is decreased by 2 ft, a new rectangle is formed whose perimeter is 8 ft more than the perimeter of the original rectangle.Find the dimensions of the original rectangle.

If someone culd help that would be great and can u show the work so i can understand.

A few days ago
violingirl7777

L=length, W=width

L=w+4

2w=new width, L-2=new length

Since L=w+4, you can substitute that so that the new length is (w+4)-2 or w-2.

So the perimeter is (2w)+(2w)+(w+2)+(w+2) which equals 4w+2w+4 or 6w+4

The old perimeter was 2w+2(w+4) which equals 4w+8

So 6w+4=4w+8+8

Combine like terms: 6w+4=4w+16

Move the w’s to the left side: 2w+4=16

Subtract 4 from both sides: 2w=12

Divide both sides by 2: w=6

The original dimensions were W=6, L=10

Hope this helps! 😀

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A few days ago
whats my name again
Length = l

Width = w

Perimeter = p

l = w + 4

2w + l – 2 = p + 8

Each rectangle will have 2 sides of length and 2 of width

So for the original the perimeter is:

2w + 2(w+4) = p

2w + 2w + 8 = p

p = 4w +8

In the second rectangle

2(2w) + 2 (l-2) = p + 8

p + 8 = 4w + 2l – 4

p = 4w + 2l – 12

Now compare both p equations

p = 4w +8

p = 4w + 2l – 12

4w + 8 = 4w +2l -12

2l = 20

l = 10

Now put l into original equation

l = w+ 4

10 = w + 4

w = 6

Perimeter is 2l + 2w

20 + 12

32

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A few days ago
Marley K
let w = width of original rectangle

w + 4 = length

2w + 2w + 8 = 4w+8 = original perimeter

2w = new width

w+2 = new length

4w + 2w + 4 = 6w + 4 = new perimeter

new = 8 + original

6w + 4 = 8 + 4w + 8

2w = 12

w = 6

original width is 6 ft.

original length is 10 ft.

That’s it!

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A few days ago
chasrmck
w = width of 1st rectangle

w+4 = length

2w = width of new rectangle

w+4-2

2(w+w+4)+8=2(2w+w+4-2)

2w+4+4=3w+2

6=w

10= length

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