math question 4?

Question

Suppose that the width of a rectangle is 5 inches shorter than the length and that the perimeter of the rectangle is 50.
a) Set up an equation for the perimeter involving only L, the length of the rectangle.
Answer:
b) Solve this equation algebraically to find the length of the rectangle. Find the width as well.
Answer: Length ______, Width ______

Answer 1

Perimeter Formula

P = 2L + 2W

Let

L = length

L - 5 = Width

50 = 2L + 2(L- 5)

The distriputive prooperty on the right side of the equation

50 = 2L + 2L - 10

collecting Like Terms

50 = 4L - 10

Transpose 10

50 + 10 = 4L - 10 + 10

Collecting like terms

60 = 4L

Divide both sides of the equation by 4

60 / 4 = 4L / 4

60 / 4 = L

15 = L

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The length is 15 inches

The width is 10 inches

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Answer 2

Call the width of a rectangle "W", the lenght is "L"

Because the width is 5 inches shorter than the length: W = L - 5

Perimeter = 2 lengths + 2 widths
= 2 L + 2 W
= 2 (L + W)
= 2 [ L + (L - 5 ) ]
= 2 [ 2 L - 5 ]
= 4 L - 10

==> Perimeter = 4 L - 10

because perimeter given is 50
Thus : 50 = 4 L - 10
50 + 10 = 4 L
60 = 4 L
60/4 = L
15 = L

We found out that W = L - 5
So W = 15 - 5 = 10

so width is 10, length is 15

Answer 3

let w = width,and l = length w= l -5. perimeter = 2(l +w) and perimeter is also equal to 50.
2( l + l -5) = 20
2( 2l - 5) = 20
4l - 10 = 20
4l = 20- 10
4l = 10
l = 10/4
l= 2.4 hence length of rectangle is 2.4

Answer 4

Width (W) = Length (L) - 5
Perimeter (P) = 50

A) P= W+W+L+L = 2W + 2L = 2(L-5) + 2L

P= 2(L-5) + 2L

B. 50 = 2(L-5) + 2L first distribute
50 = 2L - 10 + 2L the simplify
50 = -10 + 4L add 10 to both sides
60 = 4L divide by 4 on each side
15 = L

W = L - 5
W = 15 - 5
W = 10

Answer 5

W = L - 5

2(L + W) = 50
2(2L - 5) = 50 .............A
4L = 60
L = 15

W = 15 - 5 = 10

Length = 15 inches , Width = 10 inches.

Answer 6

L=L
W=L-5
P=50
P=2L+2W
50=2L+2(L-5)......by substituting W interms of L
50=2L=2L-10
50+10=4L
4L=60
L=15
there fore W=10