Answered

AB = x + 3
BC = x + 4
CD = 2x - 1
AD = 3x - y

How do you solve for x and y
ABCD is a kite​

AB X 3BC X 4CD 2x 1 AD 3x YHow Do You Solve For X And YABCD Is A Kite class=

Answer :

Answer:

Step-by-step explanation:

IF it is a SYMMETRIC kite

            CB = CD

         x + 4 = 2x - 1

x - x + 4 + 1 = 2x - x - 1 + 1

                5 = x

  AB = AD

x + 3 = 3x - y

5 + 3 = 3(5) - y

     8 = 15 - y

      y = 7

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