The petimiterof a rectangle is 110cm. the length is 1cm more than twice the width. write two equations that would be used to solve the system

Step-by-step explanation:

The perimeter is the sum of the four side lengths of the rectangle. If L and W represent the length and width, respectively, then there are two sides of length L and two sides of length W. This means the perimeter is ...

  2L +2W, or 2(L+W)

The problem statement tells us the perimeter is 110 cm, so if all lengths are in cm, the equation for perimeter can be ...

  2(L+W) = 110 . . . . equation for the perimeter

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The problem statement also gives a relationship between length and width. Twice the width will be represented by 2W, so one more cm than that will be (2W+1). We are told that is the same as the length:

  2W+1 = L . . . . . equation relating length and width

Solution (not required by the problem statement)

Substituting the second equation into the first, you get

  2(2W+1 +W) = 110

  3W+1 = 55 . . . . . divide by 2

  3W = 54 . . . . . . .subtract 1

  W = 18 . . . . . . . . .divide by 3

  L = 2·18+1 = 37