SOLUTION: The perimeter of a rectangle is 44 inches. If the length is increased by twice the width, the result is 31 inches. Find the length and width of the rectangle.
x+y=44 and 2x+y=31?
Algebra.Com
Question 1072614: The perimeter of a rectangle is 44 inches. If the length is increased by twice the width, the result is 31 inches. Find the length and width of the rectangle.
x+y=44 and 2x+y=31?
Answer by josgarithmetic(39630) (Show Source): You can put this solution on YOUR website!
44 to 31 is not an increase. Recheck the problem description.
RELATED QUESTIONS
The perimeter of a rectangle is 40m. If the width is increased by twice the length, the... (answered by gwendolyn)
The perimeter of a rectangle is 196 inches. The length exceeds the width by 44 inches.... (answered by rapaljer)
The perimeter of a rectangle is six times the width. If the length were increased by 4... (answered by ankor@dixie-net.com)
The perimeter of a rectangle is six times the width. If the length were increased by 4... (answered by mananth)
The perimeter of a rectangle is 44 inches. If the length of the rectangle is 14 inches,... (answered by josgarithmetic)
The length of a rectangle is 3 inches less than twice the width. If the perimeter of the... (answered by ewatrrr)
The length of a rectangle is 3 inches less than twice the width. If the perimeter of the... (answered by Nigelyul)
The perimeter of a rectangle is 44 inches. If the width of the rectangle is four more... (answered by ikleyn)
The length of a rectangle is 8 inches more than twice its width. If the perimeter of the (answered by Paul)