# SOLUTION: find the value of x and y in the parallelogram PQRS the diagonals of the parallelogram are QS and RP,and they bisect at point T. QT=3x, RT=y+3, PT=2x, and ST=2y I have done nu

Algebra ->  Algebra  -> Parallelograms -> SOLUTION: find the value of x and y in the parallelogram PQRS the diagonals of the parallelogram are QS and RP,and they bisect at point T. QT=3x, RT=y+3, PT=2x, and ST=2y I have done nu      Log On

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 Geometry: Parallelograms Solvers Lessons Answers archive Quiz In Depth

 Question 115500This question is from textbook Prentice Hall Mathematics : find the value of x and y in the parallelogram PQRS the diagonals of the parallelogram are QS and RP,and they bisect at point T. QT=3x, RT=y+3, PT=2x, and ST=2y I have done number 18, and I have been working on this one for probably half an hour. I just don't know how to get the answer from the information given. Thank you!This question is from textbook Prentice Hall Mathematics Answer by bucky(2189)   (Show Source): You can put this solution on YOUR website!If I understand your problem correctly, the most critical information you need to use is that the diagonals bisect each other. From this you know that: . QT = ST and RT = PT . For the top equation you can substitute 3x for QT and 2y for ST to make the equation: . 3x = 2y . For the bottom equation you can substitute y + 3 for RT and 2x for PT and it becomes: . y + 3 = 2x . So you have the two equations: . 3x = 2y and y + 3 = 2x . Let's solve them by substitution. Solve the bottom equation for y by subtracting 3 from both sides to end up with: . y = 2x - 3 . Now go to the top equation [3x = 2y] and substitute 2x - 3 for y to get: . 3x = 2(2x - 3) . Multiply out the right side: . 3x = 4x - 6 . get rid of the 6 on the right side by adding 6 to both sides and you have: . 3x + 6 = 4x . Subtract 3x from both sides and you finally get to: . 6 = x . Now that you know x = 6 you can return to either of the two original equations, substitute 6 for x, and solve for y. Let's return to the equation 2y = 3x and substitute 6 for x to get: . 2y = 3*6 . Multiply out the right side: . 2y = 18 . Solve for y by dividing both sides by 2 to get: . y = 9 . In summary, we've found that x = 6 and y = 9 . Let's check. From the geometry we know PT should equal RT. PT is 2x which is 2*6 or 12. RT is y + 3 which is 9 + 3 = 12. Line PR is bisected at T because both PT and RT equal 12. . Also from the geometry we know that QT should equal ST. QT = 3x = 3*6 = 18. And ST = 2y = 2*9 = 18. This means that diagonal QS is bisected at T because both QT and ST equal 18. . Hope this helps you to understand the problem and how to get the answers for x and y. .