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

N621-G23 Cartesian coordinates in two dimensions y = ax + b

This is called a cartesian graph. 

(1) A Cartesian coordinate graph: two axes ("axes" plural of "axis"):

                           

• Horizontal axis is called the x-axis. 
• Vertical one is the y-axis. 

(2) Coordinates are points on a graph 

       Format: (x, y)

(3) The centre point of the graph is called the origin (0, 0). 

      It is the zero point of both the x-axis and the y-axis. 

Plotting a coordinate: 

Step1: locate on x-axis

Step2: locate on y-axis

Step3: "Move"/plot for both to "meet"

Example: 

     Plot coordinate (5,10) on the graph                          

Practice on same graph:

     Plot coordinates (-5, 5), (10,0), (0,5)

Linear functions y = ax + b

A function that produces a straight line when on a graph with a constant rate of change.

A linear (straight line) equation has a standard form:  

                            y = ax + b 

Example: 

     y = 2x +c 

     y = -x + 5 

     y = 4

     x = 3 are all linear equations.

Equation of the Line and its Coordinates

On the graph, the equation of the line y = mx+c = the LINE drawn.
-    y and x are the variables, and a and b are constant values (a number)

-    a is the slope (rate of change) 

-    b crosses the y-axis called the y-intercept with coordinate (0,b)

-  Coordinates are points on the line that satisfies/can be substituted into the equation.

    => the coordinates (x,y) can be substituted into the equation.

                  Equation of the line y = 2x + 8

Examples

To test if the following coordinates are on the line y = 2x + 8. 

a. (-4,0),        b. (-3,2)        c. (5, 5) 

Equation of the line is y = 2x + 8

(a) Substitute (-4,0) into y = 2x + 8 

                        =>  0 = 2(-4) +8

      (-0, -4) is on the line.

(b) Substitute (-3,2) into y = 2x + 8 

                        =>  2 = 2(-3) +8

      (-3, 2) is on the line.

(c) Substitute (5,5) into y = 2x + 8

                  => 5 ≠ 2(5) + 8

(5, 5) is a coordinate on the graph but not on the line.

Graph Below : y = 2x + 8, and the coordinates (-3,2), (-4,0) and (5,5).

Name the other coordinate on the line (satisfy the equation).

=> (1,10)

If the line passes through a coordinate (x,y), => can be used for the (x,y) of the equation : y = mx + c

Practice: What are the coordinates of (x₁, y₁), (x₂, y₂), (x₃, y₃)?

What is c (the y-intercept)?

c = y-intercept 

=> where the equation(line) intercept/‘cuts’ the y-axis. 

     Coordinate : (0, y), (0, c). 

         

         Line cuts y-axis at 5          

         c = 5

Finding the value of c from the equation

Equation of a straight line:

              y = mx + c 

c= the y-intercept; coordinate (0,c)

Example:

(1)       y = 3x + 5  => c = 5, (0,5)

(2)       y = -x – 3   => c = -3 (0,-3)

(3)       2y = 5x + 6      (must change equation of format: y = mx + c)

             y = 5x + 6

                        2

               = 5x + 3,  => c = 3 (0, 3)

                   2

(4)       -3y = -2x + 9    (must change equation format to: y = mx + c)

            2x - 9 = 3y

             3y = 2x - 9

                          3

                  = 2/3 x - 3,  => c = -3 (0, -3)

What is m?

                      y = mx + c

Gradient(m) is the ‘steepness’ or slope of a graph. 

Formula: 

                    Gradient = y₂ – y₁

                                       x₂ – x₁

We need 2 sets of coordinates (x, y) on the line to find the gradient(m). 

Example: 

Find the gradients of the following graphs.

(x₁, y₁) = (0, -2), (x₂, y₂) = (5,8)                  (x₁, y₁) = (-5, 7), (x₂, y₂) = (1,-5)              

   Gradient = -2 – 8                                                   Gradient = 7 – (-5) 

                        0 - 5                                                                        -5 - 1 

                   = -10/-5                                                                     =12/-6

                   = 2                                                                             = -2

 

Observation:

1. The gradient/"slope" is positive (2) for  (0, -2), (5,8) as it is  "ascending" from left to right.

  Slope Upward => positive gradient(+m)

2. The gradient/"slope" is -2 for (-5,7), (1-5) as it is "descending" from left to right.

          Slope downward => -m

3.  When calculating the value of m, the (x,y) must be of the same "set order" =>

y₁ - y₂ or y₂ - y₁  NOT  y₁ - y₂

                x₁ - x₂     x₂ - x₁            x₂ - x₁

Positive and negative gradient equation

To find the Value of m from the equation

Equation of a straight line:

            y = mx + c 

Example:

(1)       y = 3x + 5 , m = 3

(2)       y = -x – 3, => y = (-1)x - 3, m = -1

(3)       2y = 5x + 6    (must change equation of format: y = mx + c)

             y = 5x + 6

                        2

               = 5x + 3, m = 5/2

                  2                   

(4)       -3y = 2x + 9  (must change equation of format: y = mx + c)

             -2x - 9 = 3y

                   3y = -2x - 9

                               3

                        = -2/3 x - 3,  m = -2/3

Equation Of Vertical And Horizontal Lines

(1) x = numeric value 

      -> a vertical line graph

Example: 

                        x=5

                     

         Equation of the line : x = 5

(2) y = numerical value 

         => a horizontal line graph. 

Example:

                         y=5

                          

   Equation of the line : y = 5

To find intercept of 2 lines

(1) Use simultaneous equation or 

(2) the point of interception of the 2 lines on the graph 

2 linear equations:  y = 2x – 2

                               y = -2x + 4

From the graph, the intersection coordinate is (3/2, 1)

=> x = 3/2 and y = 1

Practice

1.  Draw a simple graph. Plot and label the point A(-2, 1) and B (3, 5).

2.  Find the gradient of the line AB   [17/I/92,2/T]

3.  Find the gradient of the line joining the points A (2, 6) and B(7,3)

4.   Find the gradient of the line joining the points A(2, 1) and B(4, 6)

N63-G23 Quadratic Functions and Graphs

Graph of Quadratic Functions

(1)  Of the form y= ax² + bx + c ( b and c can be zero, but a ≠ 0)

(2) All have a line of symmetry

(3) U-shaped

(4) Turning Point:

      (i) Positive [+ax²] graph - turning point at the bottom (Minimum point)

     (ii) Negative [-ax²] graph - turning points at the top (maximum) 

Plotting the Graph

Example (1) 

(a) Complete the table for y = -x² + x -3 for -2 <= x <=3.

X	-2	-1	0	1	2	3
Y	-9

(b) Plot and draw the graph.

(c) Use the graph to solve the equations 

     (a) –x² + x – 3 = -7

     (b)  –x² + x – 3 = 0

(a) Filling in the table

(a) The table is for the x and y values of equation y = –x² + x – 3

Step 1:  Write the equation, and substitute x value into y = –x² + x – 3.

Starting with x = -2

                        y = –x² + x – 3

                        y = -(-2)² + (-2) – 3

                        y = -(4) – 2 – 3

                           = -4 -2 – 3

                           = -9

Do for other x values:-

            When x = - 1, y = -(-1)² + (-1) – 3 = -5

            When x = 0, y = -(0)² + (0) – 3 = -3

            When x = 1, y = -(1)² + (1) – 3 = -3

            When x = 2, y = -(2)² + (2) – 3 = -5

            When x = 3, y = -(3)² + (3) – 3 = -9

Step 2: Fill in the values 

X	-2	-1	0	1	2	3
Y	-9	-5	-3	-3	-5	-9

(b) Drawing the Graph

(b) Plot and draw the graph.

The graph is plotted by placing and joining the coordinates on the graph.

(c) Solving Equation using the Graph

There are two variables ( x and y) in the equation y = ax² + bx + c. 

<<a, b, c are constant (a specific number) >>

We can solve an equation when either x or y is given. 

When y is given => find x, 

When x is given => find y

When using a graph to solve an equation ax² + bx + c, 

Step 1 : Form an equation such that

    y = ax² + bx + c = k

=> y = k

Step 2 : Draw y = k on the graph

Step 3 : Solve the equation by locating the intersection of y = k and y = ax² + bx + c and find the required value/s. 

(c) Use the graph to solve the equation 

      (i)  –x² + x – 3 = -7

Step 1: Write and form the equation

 (i)         y =  –x² + x – 3 = -7

Step 2: ‘Equate’ both to get equation to plot/draw on graph

           y = -7

Step 3: Draw the line and find the coordinate               

                  Draw y = -7

From the graph, x = -1 2/5 and 2 2/5

(ii) –x² + x – 3 = 0 

        y =  –x² + x – 3 = 0 (Step 1) 

        Y = 0                       (Step 2)

       Draw y = 0               (Step 3)

Since y = 0 and y =  –x² + x – 3 do not "cut" or intersect, the values are undefined => (no intersections of the line and equation).

Example(2)

Use the graph of y = 3x² – 2x + 4 to find

      (a) 3x² – 2x + 4 = 12

                 y =  3x² – 2x + 4 = 12  (Step 1)

                 y = 12                          (Step 2)

Draw the line and find coordinate(Step 3)

          x = -4/3 or x = 2  

(b) 3x² – 2x – 3  = 0

Step 1: Write the equation

      y =  3x² – 2x + 4

      y =  3x² – 2x - 3 = 0  (Step 1)

Step 2: Equate 

[need to ‘change’ equation to 3x² – 2x + 4]

To equate, +7 needs to be "added" 

       y = 3x² – 2x - 3 +7 = 0 + 7

       y= 3x² – 2x + 4 = 7

           y = 7

Step 3: Draw the line y = 7 and find coordinate

            x = -0.72 or x = 1.4

Finding Minimum/Maximum Points

Example:  Find the maximum and minimum value of y = -x² + 6x + 6 and y = x² + 4x + 5

Step 1: Find the value of x when y = 0

  y = -x² + 6x + 6 y = x² + 4x - 5

     When y = 0 Step 1: find the value of x when y = 0

-x² + 6x + 6 =0 x² + 4x - 5 = 0

     (x + 1)(x -7) = 0 (x + 5)(x - 1) = 0  (by factorisation)

      x = -1, or x = 7 x = -5, x = 1

Step 2: Find the mid-point for the values of x

Mid point of x = -1 and x = 7 Mid point of x = -5 and x = 1

   |--|--|--|--|--|--|--|--|-   -|---|---|---|---|---|---|--

  -1 0 1 2  3 4 5 6 7 -5 -4 -3  -2 -1  0   1

From the number line = 3 From the number line = -2

Step 3: Find the value of y

Value of y: 

y = -(3)² +6(3) + 6  y = (-2)² + 4(-2) - 5

  = -9 + 18 + 6     = 4 - 8 - 5

              = 15     = -9

 

Step 4: List the maximum/minimum point

     Mid-point = (3, 15)   Mid-point = (-3, -9)

By graphical method:

Example:  Find the maximum and minimum value of y = -x² + 6x + 6 and y = x² + 4x + 5

Summary

Practice

Using Example(2)

(a)        Find the values of x for 3x² – 2x + 4 = x + 6 

(b)        Find the values of x for 3x² – 3x – 2  = 0