The length of a rectangle is five inches less than twice it’s width. If the rectangle has a perimeter of 80 inches, find the length and width
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Cassidy V. asked • 10/13/15
The length of a rectangle is five inches less than twice its width. If the rectangle has a perimeter of 80 inches, find the length and width
I don't understand how to turn this word problem into an equation. I really need help. Thank you very much!
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Lorie T. answered • 10/13/15
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Thanks for your question! Let's step through it by breaking down the sentence.
But first we need to know the formula to find the perimeter of a rectangle. It is:
2*length + 2*width = perimeter
two times the length plus two times the width equals the perimeter
--- Based on that formula, let's use "2x" to represent the width
--- Next, in the sentence it says "thelength... is five inches less than twice the width"; we can show that as:
twice its five inches
width less
2* (2x - 5)
---- to simplify this, we have to use the algebraic concept of distribution which says to multiply everything in the parenthesis by "2". When we do, we get:
(4x - 10)
--- Now let's write a math "sentence" to match the word sentence above:
the length the width the perimeter
(4x - 10) + 2x = 80from here we will solve for "x" (aka the width) first:
--- we do so by combining the "x"s in the equation
(4x - 10) + 2x = 80
--- let's rearrange things to see it easier
4x + 2x - 10 = 80
----- we can now sum (add) the "x" numbers to get
6x - 10 = 80
----- we now reduce the equation by adding "10" to each side (as the left side has "- 10" showing)
6x - 10 + 10 = 80 + 10-- this leaves us with
6x = 90
----- to further solve for "x" we will need to divide both sides by "6" (as currently we have 6*"x", aka 6x)
6x = 90
6 = 6
x = 15 - width of the rectangle
We now know that the width of the rectangle is 15. Now let's find the length!
To do that we will need to use our little handy equation we built above:
the length the width the perimeter
2*(2x - 5) + 2*x = 80
----- from that formula we need "2*(2x - 5) and can use our new width number of 15 to substitute in for "x":
the length
2*(2x - 5) = ?
2*(2(15) - 5) = ?
2*(30 - 5) = ?
2*(25) = ?
50 = ? = the length
We can check our numbers by putting them all together:
the length the width the perimeter
50 + 30 = 80