S3TN Solving Fractional Equations and Others

Solving Fractional Quadratic Equations

Cross-Multiplying =>

            a = c     =>      x bd  a  =   c   x bd    =>   ad = bc

            b    d                        b       d

[[ The Equation must be ax²+ bx + c = 0 before factorising ]]

Example 

solve    6   = x + 3

          x + 4

            6      =    x + 3 

         x + 4

 6 = (x + 3)(x + 4)         (Step1:  Cross-Multiply)

   6 = x²+ 7x + 12          (Step2: Expand to form a quadratic equation)

   x²+ 7x + 12 – 6 = 0

   x²+ 7x  + 6 = 0

  (x + 1)(x + 6) = 0         (Step3: Factorise and Solve for x)

   x + 1 = 0 or x + 6 = 0

   x = -1 or x = -6

Example

Solve      1       +    2       = 5

            x -2        x – 3 

              (x – 3)  + 2 x (x – 2)   = 5        (Step1: LCM the denominator)

        (x - 3)(x -2)  (x – 2)(x – 3)

            x – 3  + 2x – 4  =  5                   (Step 2 : Combine to single fraction)

              (x – 2)(x -3)

         x + 2x – 3 – 4 = 5        

           (x – 2)(x -3)

               3x – 7       = 5

         (x – 2)(x – 3)

          3x – 7 = 5(x – 2)(x – 3).              (Step 3 : Cross-Multiply and Expand)

          3x – 7 = 5(x²– 5x + 6)

          3x – 7 = 5x²– 5x + 6

          5x²– 5x + 6 – 3x + 7 = 0. 

          5x²-8x + 13 = 0                           (Step 4: Factorise and Solve for x)                                       

          (5x + 13)(x + 1) = 0                                           5x   \ /    13

     5x + 13 = 0 or x + 1 = 0                                           x    / \     1  

           x = -13/5 or x = -1

Solving Equations by equating Coefficients

Example

Find the value of A, B and C

2x⁴ - 13x³ + 19x² + 5x + 1 = (x - 4)(Ax + 1)(x² + Bx + 1) + C

Tips : 
1. Such questions are usually solved by equating the coefficient values
2. If there are more than 2 brackets, expands in pairs
3. Solve the unit value first, and then the highest powers of X

Step 1 - Expand brackets (if needed)

   2x⁴ - 13x³ + 19x² + 5x + 1 = (x - 4)(Ax + 1)(x² + Bx + 1) + C

   2x⁴ - 13x³ + 19x² + 5x + 1 = (x - 4)(Ax + 1)(x² + Bx + 1) + C [Expands 1 pair]

                    (x - 4) (Ax + 1) = (Ax² + (1 - 4)x - 4) (x² + Bx + 1) + C

  2x⁴ - 13x³ + 19x² + 5x + 1 = (Ax² + (1 - 4A)x - 4)(x² + Bx + 1) + C

Step 2 - extract  and solve from comparing coefficients 

  Extract coefficient  of x4.   [=> x² x x² , x³ x x , x⁴ x unit value ]

      2x⁴ - 13x³ + 19x² + 5x + 1 = (Ax² + (1 - 4A)x - 4)(x² + Bx + 1) + C

                   2x⁴  =  Ax² x x²

    A = 2  [Equating x4 coefficient]

    

  Extract and equating coefficient of unit value

    1 = (-4 x 1) + C

C = 1 + 4

C = 5

Step 3 - Substitute values and find remaining unknown

     2x⁴ - 13x³ + 19x² + 5x + 1 = (2x² -7x - 4)(x² + Bx + 1) + 5

  Extract and equating x coefficients

5x = -7x x 1 - 4Bx

             5 = -7 - 4B

  4B = -12

                B = -3 

A = 2, B = -3, C = 5