+8 The length of a rectangle exceedss its width by 8 inches. The perimeter is 80 inches. Find the length and width of the rectangle
+118608 Hint,
If you let the width be x units long. Then how long will the length be?
If the perimeter is 80 what will length + width be?
+118608 Notreallymath has asked if this is correct:
2a+2a+8=80
No, Not quite. 
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If a is the width then
The the length will be a+8
This means that the perimeter will be a+a+ a+8 + a+8 = 4a+16
so
4a+16 = 80
But what I was getting at is that the perimeter is 2*length + 2*width
so a HALF of the perimeter is length+ width
so
x + x+8 = 40
2x+8=40
2x = 32
x = 16
so the width is 16 and the length is ??
You can do it either way though. 
Let length=x+8
Width= x
Apply the perimeter formula for rectangle which is:
2*(Length+Width)
we already know that length=x+8 and width is x
so,
2*(x+8+x)= 2*(2x+8) = 4x+16
well the perimeter as the question says is 80 inches
so set,
4x+16=80
subtract 16 from both sides
4x=64
divide by 4
x=16
Well width is x so it is 16
length= 16+8=24
You can make sure by subistuting length and width in the perimeter formula
so
2*(24+16)=80
2*(40)=80
yes indeed these values work.
