SOLUTION: The perimter of a rectangle is 66 inches. The length exceeds the width by 27 inches. Find the length and width. The length is ?? inches The width is ?? inches Please HELP.

Algebra ->  Coordinate Systems and Linear Equations  -> Lessons -> SOLUTION: The perimter of a rectangle is 66 inches. The length exceeds the width by 27 inches. Find the length and width. The length is ?? inches The width is ?? inches Please HELP.      Log On


   

Question 147955: The perimter of a rectangle is 66 inches. The length exceeds the width by 27 inches. Find the length and width.
The length is ?? inches
The width is ?? inches
Please HELP.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Let x=length and y=width

Remember, the perimeter formula is P=2x%2B2y. Since the perimeter is 66 inches, this means that P=66. So the equation is now 66=2x%2B2y.


Also, because the "length exceeds the width by 27 inches", this tells us that x=y%2B27


66=2x%2B2y Start with the first equation.


66=2%28y%2B27%29%2B2y Plug in x=y%2B27


66=2y%2B54%2B2y Distribute.


66=4y%2B54 Combine like terms on the right side.


0=4y%2B54-66 Subtract 66 from both sides.


-4y=54-66 Subtract 4y from both sides.


-4y=-12 Combine like terms on the right side.


y=%28-12%29%2F%28-4%29 Divide both sides by -4 to isolate y.


y=3 Reduce.

So the width is 3 inches.


x=y%2B27 Go back to the second equation.


x=3%2B27 Plug in y=3


x=30 Add.


So the length is 30 inches


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Answer:


So the length is 30 inches and the width is 3 inches