Perimeter of Rectangle

jane

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Joined
Jul 26, 2007
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16
Would you please let me know if I am doing this correct? Thanks

the width is x/2x-5
the length is 8/2x-5

P=2l+2w

p=2(8/2x-5)+2(x/2x-5)

p=(16/2x-5)+(2x/2x-5)

p=2x+16/2x-5
 

Subhotosh Khan

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Jun 18, 2007
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jane said:
Would you please let me know if I am doing this correct? Thanks

the width is x/2x-5
the length is 8/2x-5

P=2l+2w

p=2(8/2x-5)+2(x/2x-5)

p=(16/2x-5)+(2x/2x-5)

p=2x+16/2x-5

Correct but use parentheses to group your operations

the width is x/(2x-5)
the length is 8/(2x-5)

P = 2l + 2w

p = 2[8/(2x-5)] + 2[x/(2x-5)]

p = 16/(2x-5) + 2x/(2x-5)

p = (2x+16)/(2x-5) = 2(x+8)/(2x-5)
 

Denis

Senior Member
Joined
Feb 17, 2004
Messages
1,495
jane, much easier and less confusing if you leave the doubling till the end:

x / (2x - 5) + 8 / (2x - 5)

= (x + 8) / (2x - 5) ; now multiply by 2:

= 2(x + 8) / (2x - 5)
 

jane

New member
Joined
Jul 26, 2007
Messages
16
Thanks.
 
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