SOLUTION: Can someone help me on this problem? The length of a rectangle is 2 cm more than twice its width. If the perimeter of the rectangle is 52 cm, find the dimensions of the recta

Algebra ->  Equations -> SOLUTION: Can someone help me on this problem? The length of a rectangle is 2 cm more than twice its width. If the perimeter of the rectangle is 52 cm, find the dimensions of the recta      Log On


   

Question 54802: Can someone help me on this problem?
The length of a rectangle is 2 cm more than twice its width. If the perimeter of the rectangle is 52 cm, find the dimensions of the rectangle.
Thanks, Ashley
Answer by Cintchr(481) About Me  (Show Source):
You can put this solution on YOUR website!
+P=2%28l%2Bw%29+
+l+=+2%2Bw+
Substitute the 2nd equation into the first
+P+=2%282%2Bw%2Bw%29+
add like terms
+P+=+2%282+%2B+2w%29+
Distribute the 2
+P+=+4+%2B+4w+
By your problem the Perimiter is 52
+52+=+4+%2B+4w+
subtract 4 from both sides
+48+=+4w+
divvide by 4
+12+=+w+
Substitute 12 back into the 2nd equation
+l+=+2%2Bw+
+l+=+2%2B12+
+l+=+14+
Dimensions of the rectangle are 12 X 14