# SOLUTION: The length of a rectangle is 4 inches more than twice the width. The perimeter is 68 cm. Find the dimensions of the rectange. (Round to two decimal places.)

Algebra ->  Algebra  -> Rectangles -> SOLUTION: The length of a rectangle is 4 inches more than twice the width. The perimeter is 68 cm. Find the dimensions of the rectange. (Round to two decimal places.)      Log On

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 Geometry: Rectangles Solvers Lessons Answers archive Quiz In Depth

 Question 341706: The length of a rectangle is 4 inches more than twice the width. The perimeter is 68 cm. Find the dimensions of the rectange. (Round to two decimal places.)Answer by Jeannie1012(8)   (Show Source): You can put this solution on YOUR website!To solve the perimeter of a rectangle you will use the formula 2W + 2L = perimter First: Draw a picture of a rectangle. Label the width w since you don't know its value. Now for the length you know that it is 4 inches more than 2 times the width or 2w + 4 so we know that: width = W Length = 2W + 4 perimeter = 68 STEP 2: plug the values you have into the formula. So 2L + 2w = perimeter becomes 2(2w+4) + 2w = 68 STEP 3: Use distribution to solve and you get 4w + 8 + 2w = 68 STEP 4: combine like terms and you get 6W + 8 = 68 STEP 5: subtract 8 from both sides to get 6w alone and you get 6w = 60 STEP 6: Divide by 6 to get the w alone, don't for get what you do to one side you have to do to the other so 6w/6 = w and 60 /6 = 10 and we now know that W = 10 now you can check your work by plugging 10 into the equation anywhere there is a w. so for length 2w + 4 become 2(10) + 4 = 20 + 4 L = 24 and W = 10 so 2L + 2W = 68 becomes 2(24) +2(10) = 68 48 + 20 = 68 68 = 68 so it checks.