Dimensions - Mat

Algebra

Dimensions - Mat

Postby Guest » Thu Jul 16, 2015 9:04 pm

A door mat's area is 882 sq. in. If the length had been 6 in. less and width 5.5 in. more, the mat would have been square. Determine dimensions.

Area = Length x Width

L x W = 882

L - 6

W + 5.5

Unsure how to proceed. Thanks.
Guest
 

Re: Dimensions - Mat

Postby Guest » Fri Jul 17, 2015 11:06 am

Please reply. Thank you.
Guest
 

Re: Dimensions - Mat

Postby Guest » Fri Jul 17, 2015 8:09 pm

Let X = length
Let Y = width
Question say X times Y = 882
if length is 6 inches less = X - 6
If width is 5.5 inches more = Y + 5.5
If now square then square sides are equal
that gives you 2 unknowns X and Y and 2 equations to use to solve for X and Y......etc.
Guest
 

Re: Dimensions - Mat

Postby Guest » Fri Jul 17, 2015 9:00 pm

X * Y = 882

X - 6
Y + 5.5

Not sure how to set up to solve.
Guest
 

Re: Dimensions - Mat

Postby Guest » Sat Jul 18, 2015 4:59 am

X - 6
Y + 5.5
These are both equal if they are the sides of a square. put them equal and that makes an equation

as in previous post...
"If now square then square sides are equal
that gives you 2 unknowns X and Y and 2 equations to use to solve for X and Y......etc."
Guest
 

Re: Dimensions - Mat

Postby Guest » Sat Jul 18, 2015 11:47 am

X * Y = 822

X - 6 = Y + 5.5

??
Guest
 

Re: Dimensions - Mat

Postby Guest » Sat Jul 18, 2015 12:36 pm

I think I figured it out.

Check my calculations:

X * Y = 822

X - 6 = Y + 5.5

X = 822/Y

822/Y - 6 = Y + 5.5

822 - 6Y = Y + 5.5

-6Y - Y = 5.5 - 822

-7Y = -816.5

-7Y/-7 = -816.5/-7

Y = 116.64 = 117

822/117 = 7.02 = 7

117 * 7 = 819

7 - 6 = 1 (X)

117 + 5.5 = 122.5 (Y)

1 * 122.5 = 122.5
Guest
 

Re: Dimensions - Mat

Postby Guest » Sat Jul 18, 2015 2:00 pm

Should be 882

X * Y = 882

X - 6 = Y + 5.5

X = 882/Y

882/Y - 6 = Y + 5.5

882 - 6Y = Y + 5.5

-6Y - Y = 5.5 - 882

-7Y = -876.5

-7Y/-7 = -876.5/-7

Y = 125.21 = 125

882/125 = 7.05 = 7

125 * 7 = 875

7 - 6 = 1 (X)

125 + 5.5 = 130.5 (Y)

1 * 130.5 = 130.5

Re: Dimensions - Mat
Guest
 

Re: Dimensions - Mat

Postby Guest » Sat Jul 18, 2015 3:58 pm

882/Y - 6 = Y + 5.5

882 - 6Y = Y + 5.5....you must multiply the Y right across both sides of the equation to keep the equation balanced
as below.....
882 - 6Y = Y^2 + 5.5Y
Guest
 

Re: Dimensions - Mat

Postby Guest » Sat Jul 18, 2015 4:55 pm

"882 - 6Y = Y^2 + 5.5Y"

I don't know how to solve.
Guest
 

Re: Dimensions - Mat

Postby Guest » Sat Jul 18, 2015 5:19 pm

882 - 6Y = Y^2 + 5.5Y

This is a quadratic equation
It can be re arranged into standard form

Y^2 + 5.5Y + 6Y - 882 = 0

Y^2 + 11.5Y - 882 = 0

Solve this for Y using the Quadratic formulae
then calculate the value of X using the other equations.
Guest
 

Re: Dimensions - Mat

Postby Guest » Sat Jul 18, 2015 7:37 pm

Y^2 + 11.5Y - 882 = 0

Quadratic formula:

A = 1
B = 11.5
C = -882

Y = -B + - sq. rt. B^2 - 4AC / 2A

Y = 11.5 + - 11.5^2 - 4(1) (-882)/2(1)

Y = -11.5 + - 132.75 - 3528 / 2

Y = 11.5 + - - 3395.25 / 2

Y = 11.5 + - 58.268 / 2

Y = 11.5 + - 29.13 = 17.63 = 18

Y = 11.5 - 29.63 = - 40.63

Y = 18

882/18 = 49 = X

49 - 6 = 43

18 + 5.5 = 23.5
Guest
 

Re: Dimensions - Mat

Postby Guest » Sat Jul 18, 2015 8:13 pm

Y^2 + 11.5Y - 882 = 0

Quadratic formula:

A = 1
B = 11.5
C = -882

Y = -B + - sq. rt. B^2 - 4AC / 2A.....It should be all over 2A and sqrt of all of (B^2 - 4AC)

as below....
Y = (-B + - sq. rt. (B^2 - 4AC)) / 2A
Guest
 

Re: Dimensions - Mat

Postby Guest » Sat Jul 18, 2015 10:05 pm

Y = (-B + - sq. rt. (B^2 - 4AC)) / 2A

Y = (-11.5 + - sq. rt. (11.5^2 - 4(1)(-882) / 2

Y = -11.5 + - 113116.5 / 2

Y = -11.5 + - 336.32 / 2

Y = -11.5 + - 168

Y = 156.50, - 179.50

X = 882/156.60 = 5.6

That can't be right either. I don't know.
Guest
 

Re: Dimensions - Mat

Postby Guest » Sun Jul 19, 2015 6:23 am

Y = (-11.5 + - sq. rt. (11.5^2 - 4(1)(-882) / 2
make sure you calculate the "bits correctly" and follow the bits through each step. remember its all over 2A
I have put in extra brackets to try and show the "bits" more clearly....I suppose I should use tex.
Y = [ (-11.5 )+ - {sq. rt. ((11.5^2) -( 4*1*(-882)) } ] / 2
Guest
 

Re: Dimensions - Mat

Postby Guest » Sun Jul 19, 2015 9:48 am

"Y = [ (-11.5 )+ - {sq. rt. ((11.5^2) -( 4*1*(-882)) } ] / 2"

Y = (-11.5) + - {sq. rt. (132.25 + 3528)} /2

Y =( -11.5) + - {3660.25} / 2

Y = (-11.5) + - {60.5} / 2

Y = -11.5 + 60.5 = 49 / 2 = 24.50

Y = -11.5 - 60.5 = -72 / 2 = -36

X = 882 / 24.50 = 36

24.50 * 36 = 882 sq. ft.

36 - 6 = 30

Y = 24.50 + 5.5 = 30

30 * 30 = 900 sq. ft.
Guest
 

Re: Dimensions - Mat

Postby Guest » Sun Jul 19, 2015 3:21 pm

Yes OK

So original size was 24.5 x 36 = 882

Square size was 30 x 30 = 900
Guest
 

Re: Dimensions - Mat

Postby Guest » Sun Jul 19, 2015 4:32 pm

Thanks for the hints.
Guest
 

Re: Dimensions - Mat

Postby oliver1 » Wed Nov 21, 2018 1:05 am

Hello, Guest,


Y^2 + 11.5Y - 882 = 0

Quadratic formula:

A = 1
B = 11.5
C = -882

Y = -B + - sq. rt. B^2 - 4AC / 2A

Y = 11.5 + - 11.5^2 - 4(1) (-882)/2(1)

Y = -11.5 + - 132.75 - 3528 / 2

Y = 11.5 + - - 3395.25 / 2

Y = 11.5 + - 58.268 / 2

Y = 11.5 + - 29.13 = 17.63 = 18

Y = 11.5 - 29.63 = - 40.63

Y = 18

882/18 = 49 = X

49 - 6 = 43

18 + 5.5 = 23.5


Thanks for sharing this formula I use this.

oliver1
 
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Re: Dimensions - Mat

Postby Guest » Thu Nov 22, 2018 12:54 am

You're welcome, oliver1.
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