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    O-Level Additional Math

    Scheduled Pinned Locked Moved Secondary Schools - Academic Support
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    • I Offline
      iFruit
      last edited by

      (x+1)/(x-1) is of the form


      D = d.q + r where D = x+1, q =1, d = x-1, r =2

      whereas (-x-1)/(1-x) is of the form

      -D = -d.q -r ==> (-D) = (-d).q + (-r) where (-D) = (-x-1), q = 1, (-d) = (1-x) and -r = -2


      Because the divisor is changed to -ve sign, the remainder also has to change to -ve sign so that dividend also changes sign

      HTH.

      SKT:
      Can anyone help to clear my DD's doubt for the following:

      (x+1)/(x-1) = (-x-1)/(1-x), but the remainder of (x+1)/(x-1) is 2, and the remainder of (-x-1)/(1-x) is -2. Why are they not the same?

      TIA.

      1 Reply Last reply Reply Quote 0
      • B Offline
        benorito
        last edited by

        Hi,


        Can you kindly help with this question :
        Given that 280 and a number y have a LCM of 6160 and a HCF of 40, find the number y.

        Thanks in advance!

        1 Reply Last reply Reply Quote 0
        • I Offline
          iFruit
          last edited by

          benorito:
          Hi,


          Can you kindly help with this question :
          Given that 280 and a number y have a LCM of 6160 and a HCF of 40, find the number y.

          Thanks in advance!
          You need to know that product of two numbers = HCF x LCM of those two numbers


          280 * y = 6160 * 40

          y = (6160 * 40)/280 = 880

          1 Reply Last reply Reply Quote 0
          • B Offline
            benorito
            last edited by

            iFruit:


            You need to know that product of two numbers = HCF x LCM of those two numbers


            280 * y = 6160 * 40

            y = (6160 * 40)/280 = 880
            Thank you !!

            1 Reply Last reply Reply Quote 0
            • S Offline
              SKT
              last edited by

              Hi,

              Given that x and y satisfy the simultaneous equations
              mx + (m-1)y = 10,
              (m-2)x + 3my = 20.

              (a) If the equations have no unique solution, find the values of m.
              (b) If the equations have no solutions, find the value of m.

              TIA.

              1 Reply Last reply Reply Quote 0
              • I Offline
                iFruit
                last edited by

                SKT:
                Hi,

                Given that x and y satisfy the simultaneous equations
                mx + (m-1)y = 10,
                (m-2)x + 3my = 20.

                (a) If the equations have no unique solution, find the values of m.
                (b) If the equations have no solutions, find the value of m.

                TIA.
                Using Cramer’s rule

                x = [30m - 20(m-1)]/ [3m² – (m² -3m +2)] = (10m+20)/ (2m² + 3m -2)

                y = [20m -10(m-2)] / (2m² + 3m -2) = (10m+20)/ (2m² + 3m -2)


                Recall that equation has
                1) infinite solutions when (10m+20) = 0 and (2m²+ 3m -2)=0
                2) No solutions when (10m+20) # 0 and (2m² + 3m -2)=0

                (2m² + 3m -2)= (2m-1)(m+2) =0----> m=-2 or m=1/2


                When m=-2, 10m+20 = 0,
                when m=1/2, 10m+20 = 25


                So infinite solutions when m=-2
                No solutions when m=1/2

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                • A Offline
                  atutor2001
                  last edited by

                  SKT:
                  Hi,

                  Given that x and y satisfy the simultaneous equations
                  mx + (m-1)y = 10,
                  (m-2)x + 3my = 20.

                  (a) If the equations have no unique solution, find the values of m.
                  (b) If the equations have no solutions, find the value of m.

                  TIA.
                  Below is an ELEMENTARY approach

                  For linear equation : y = mx +c

                  (a) 2 lines will have no unique solution if they are the same line i.e. both lines have the same gradient, m and the same y-intercept, c

                  mx + (m-1)y = 10 that is: y = -mx/(m-1) +10/(m-1)
                  (m-2)x + 3my = 20 that is: y = -(m-2)/(3m) + 20/(3m)

                  If the gradients are the same :

                  -mx/(m-1) = -(m-2)/(3m)
                  3m² = m²-2m-m+2
                  2m²+3m-2 = 0
                  (2m-1)(m+2) = 0
                  m = 1/2 or -2

                  If the y-intercepts are the same :

                  10/(m-1) = 20/(3m)
                  30m = 20m - 20
                  m = -2

                  Therefore, there is no unique solution if m = -2

                  (b) 2 lines will have no solution if they are parallel i.e. same gradient but different y-intercept.

                  Therefore, there is no solution if m = 1/2

                  1 Reply Last reply Reply Quote 0
                  • S Offline
                    SKT
                    last edited by

                    Hi,


                    However, the answer provided for (a) is 1/2, -2, wonder what's the meaning of \"no unique solution\". Are the two lines parallel or collinear or both?

                    TIA

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                    • I Offline
                      iFruit
                      last edited by

                      SKT:
                      Hi,


                      However, the answer provided for (a) is 1/2, -2, wonder what's the meaning of \"no unique solution\". Are the two lines parallel or collinear or both?

                      TIA
                      That is right.

                      No Unique solution means, either no solution or infinite solutions.

                      when m=-2, there are infinite solutions (when it's the same line)

                      so both 1/2 and -2 are answers for (a).

                      1 Reply Last reply Reply Quote 0
                      • A Offline
                        atutor2001
                        last edited by

                        SKT:
                        Hi,


                        However, the answer provided for (a) is 1/2, -2, wonder what's the meaning of \"no unique solution\". Are the two lines parallel or collinear or both?

                        TIA
                        NO UNIQUE solution means these 2 lines are collinear (2 lines overlapping each other, so every point on the lines is a solution or can say there are infinite number of solutions)
                        So answer for (a) is -2 ONLY

                        NO solution means these 2 lines are parallel but NOT collinear (these 2 lines will never meet which is why there is no solution)
                        So answer for (b) is 1/2

                        1 Reply Last reply Reply Quote 0

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