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

Answer:2(L+W) = 110L = 2W+1Step-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___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