Need Help on two maths questions

• Ahm97sic

  12 Apr 08, 10:01

  Question 1

   

  Given that x^2 - y^2 = 37. Find the values of x and y.

   

  Teacher says that this question is on simultaneous equations but there is only one question.

   

  Teacher also says cannot use guess and check method to find the values of x and y.

   

  Question 2

   

  Given that

   

  37/13 = 2 + 1/(x + (1/(y + 1/z)))

   

  Find the values of x, y and z.

   

  Teacher says that cannot use guess and check method to find the values of x, y and z.

• ^tamago^

  12 Apr 08, 10:22

  x²-y² = 37
  (x-y)(x+y) = 37

  Since 37 is a prime number, the 2 factors can only be 1 and 37.

  x-y = -37 ----- (1)
  x+y = -1 -----(2)

  (1)+(2): 2x = -38, x = -19, y = 18

  x-y = -1 ----- (3)
  x+y = -37 -----(4)

  (3)+(4): 2x = -38, x = -19, y = -18

  x-y = 1 ----- (5)
  x+y = 37 ----- (6)

  (5)+(6): 2x = 38, x = 19, y = 18

  x-y = 37 ----- (7)
  x+y = 1 ----- (8)

  (7)+(8): 2x = 38, x = 19, y = -18

  So the solution pairs are (18,19), (18,-19), (-18,19) and (-18,-19).

• ^tamago^

  12 Apr 08, 10:44

  There should be a slightly more elegant method to solve Question 1, but I forgot.

  For Question 2, x=1, y=5 and z=2.

  (yz+1)/[x(yz+1)+z] = 11/13
  yz+1 = 11 ----(1)
  11x+z = 13 ----(2)

  For x, y and z are integers, the only solution is x=1, y=5 and z=2.

• Ahm97sic

  12 Apr 08, 12:03

  Tamago, thanks for solving question 1 but I do not understand your answer for question 2.

  Why is the only solution for x is 1, y is 5 and z is 2.

  Teacher has given hint that this question can be solved by using reciprocal concept BUT how ?

   

• FirePig

  12 Apr 08, 12:37

  Originally posted by ^tamago^:

  x²-y² = 37
  (x-y)(x+y) = 37

  Since 37 is a prime number, the 2 factors can only be 1 and 37.

  x-y = -37 ----- (1)
  x+y = -1 -----(2)

  (1)+(2): 2x = -38, x = -19, y = 18

  x-y = -1 ----- (3)
  x+y = -37 -----(4)

  (3)+(4): 2x = -38, x = -19, y = -18

  x-y = 1 ----- (5)
  x+y = 37 ----- (6)

  (5)+(6): 2x = 38, x = 19, y = 18

  x-y = 37 ----- (7)
  x+y = 1 ----- (8)

  (7)+(8): 2x = 38, x = 19, y = -18

  So the solution pairs are (-19,18), (-19,-18), (19,18) , (19,18)

  
  Some minor corrections in red.

• FirePig

  12 Apr 08, 12:54

  Originally posted by ^tamago^:

  There should be a slightly more elegant method to solve Question 1, but I forgot.

  For Question 2, x=1, y=5 and z=2.

  (yz+1)/[x(yz+1)+z] = 11/13
  yz+1 = 11 ----(1)
  11x+z = 13 ----(2)

  For x, y and z are integers, the only solution is x=1, y=5 and z=2.

  
  yz+1 = 11n ----(1)
  11nx+z = 13n ----(2) where n is an integer that can range from -infinity to postive infinity

   

  TS, tell your teacher to go solve for you.

• ^tamago^

  12 Apr 08, 13:01

  Originally posted by FirePig:

  
  Some minor corrections in red.

  thanks.

• eagle

  12 Apr 08, 13:30

  Originally posted by ^tamago^:

  x²-y² = 37
  (x-y)(x+y) = 37

  Since 37 is a prime number, the 2 factors can only be 1 and 37.

  x-y = -37 ----- (1)
  x+y = -1 -----(2)

  (1)+(2): 2x = -38, x = -19, y = 18

  x-y = -1 ----- (3)
  x+y = -37 -----(4)

  (3)+(4): 2x = -38, x = -19, y = -18

  x-y = 1 ----- (5)
  x+y = 37 ----- (6)

  (5)+(6): 2x = 38, x = 19, y = 18

  x-y = 37 ----- (7)
  x+y = 1 ----- (8)

  (7)+(8): 2x = 38, x = 19, y = -18

  So the solution pairs are (18,19), (18,-19), (-18,19) and (-18,-19).

  There's no limitations that x and y must be integers

  x and y, when plotted out, should be a hyperbola curve.

• gunner77

  12 Apr 08, 13:33

  so pro

  the mod are really good and efficient

• FirePig

  12 Apr 08, 13:34

  Originally posted by eagle:

  There's no limitations that x and y must be integers

  x and y, when plotted out, should be a hyperbola curve.

  
  oh ya hor wtf. The solution is the hyperbola curve in itself...

• ^tamago^

  12 Apr 08, 13:36

  Originally posted by eagle:

  There's no limitations that x and y must be integers

  x and y, when plotted out, should be a hyperbola curve.

  
  yeah but when i see "guess and check" it sounds so P6.

• FirePig

  12 Apr 08, 13:46

  For both questions, just plot the graph and you get the answers.

• Ahm97sic

  12 Apr 08, 17:37

  Dear Firepig,

  my teacher's solution is

  LHS = 37/13

         = 2 + 11/13

         = 2 + 1/(13/11) using reciprocal

         = 2 + 1/(1 + 2/11)

         = 2 + 1/(1 + (1/(11/2))) using 

                                          reciprocal

         = 2 + 1/(1 + (1/(5 + 1/2)))

  LHS = RHS

  By Comparison,

  37/13 = 2 + 1/(x + (1/(y + 1/z)))

   

  2 + 1/(1 + (1/(5 + 1/2)))

   

  = 2 + 1/(x + (1/(y + 1/z)))

   

  So, x = 1, y = 5 and z = 2

   

   

• ^tamago^

  12 Apr 08, 20:32

  This must be some advanced question for P6. :(
• weewee

  12 Apr 08, 20:40

  Originally posted by Ahm97sic:

  Dear Firepig,

  my teacher's solution is

  LHS = 37/13

         = 2 + 11/13

         = 2 + 1/(13/11) using reciprocal

         = 2 + 1/(1 + 2/11)

         = 2 + 1/(1 + (1/(11/2))) using 

                                          reciprocal

         = 2 + 1/(1 + (1/(5 + 1/2)))

  LHS = RHS

  By Comparison,

  37/13 = 2 + 1/(x + (1/(y + 1/z)))

   

  2 + 1/(1 + (1/(5 + 1/2)))

   

  = 2 + 1/(x + (1/(y + 1/z)))

   

  So, x = 1, y = 5 and z = 2

   

   

  
  LIke what eagle said, the problem never stated it must have integers solutions.

  So essentially for the last part (or any other part), you can just put in any rational number y and z such that y + 1/z = 11/2 eg z =1, y=9/2. or z=4, y=21/4

  The question didnt stated that all the x, y and z must be integer so your teacher's answer isnt full.

  So let me ask you a question now.

  Find the values of x and y when x = y + 1/2?

  You will realise the answer is infinite no of x and y as long as x and y differ by half. In fact x and y doesnt even need to an integer or even rational numbers.

   

   

• FirePig

  12 Apr 08, 23:31

  Originally posted by Ahm97sic:

  Dear Firepig,

  my teacher's solution is

  LHS = 37/13

         = 2 + 11/13

         = 2 + 1/(13/11) using reciprocal

         = 2 + 1/(1 + 2/11)

         = 2 + 1/(1 + (1/(11/2))) using 

                                          reciprocal

         = 2 + 1/(1 + (1/(5 + 1/2)))

  LHS = RHS

  By Comparison,

  37/13 = 2 + 1/(x + (1/(y + 1/z)))

   

  2 + 1/(1 + (1/(5 + 1/2)))

   

  = 2 + 1/(x + (1/(y + 1/z)))

   

  So, x = 1, y = 5 and z = 2

   

   

  
  Your teacher's solution is only half-right.

  This is an underdetermined system with more unknown variables x y z than equations. Naturally, the no of solutions is usually infinite. The solution is the curve itself.

   

• eagle

  12 Apr 08, 23:35

  which school teacher is that?

• ^tamago^

  12 Apr 08, 23:57

  Ppl 11 years old this year nia. Dun bully him.
• FirePig

  13 Apr 08, 15:46

  Originally posted by ^tamago^:

  Ppl 11 years old this year nia. Dun bully him.

  
  Impressive. Pri 5 or 6 only and already asking this kind of questions.

  Anyway, for this question, even if you are just looking for integers solution, you must prove that x = 1, y = 5 and z = 2 is the only set of integer solution. (Even though its quite obvious that y+1/z=fraction can only yield one set of integers solution or no integer solution at all. eg y+1/z=7/9 has no integer y and z solution.) [obvious doesnt mean it will be easy to prove)

  Ts, since your teacher is just a primary sch teacher, he/she can be excused for not thinking rigorously enough.

• ^tamago^

  13 Apr 08, 19:58

  You know why I say P6? If you tutor P6, you will know "guess and check". :lol: