SOLUTION: I'm having trouble with this question: "Given the length of a rectangle is one less than four times the width. The perimeter of the rectangle is 88. Find the dimensions of the rect

Algebra ->  Equations -> SOLUTION: I'm having trouble with this question: "Given the length of a rectangle is one less than four times the width. The perimeter of the rectangle is 88. Find the dimensions of the rect      Log On


   

Question 917467: I'm having trouble with this question: "Given the length of a rectangle is one less than four times the width. The perimeter of the rectangle is 88. Find the dimensions of the rectangle." So I put it into an equation of 4x-1=88. I worked out the equation, but then I was left with 89/4. It became a decimal of 22.25, but I have to find the dimensions of the rectangle which would be finding the length and width I suppose which is not the same size. Can someone please help me of solving this problem?
Answer by rothauserc(4718) About Me  (Show Source):
You can put this solution on YOUR website!
let x be the width of the rectangle, then
length = 4x -1
Perimeter(P) = 2l + 2w
P = 2(4x-1) + 2x
now the perimeter is 88, therefore
88 = 8x-2 +2x
10x = 90
x = 9
therefore
width = 9 and length = 35