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Elementary Mathematics

Elementary Mathematics. Brush-Up. Why ?. Situation: Lot of different backgrounds in PUB, Many of those bgs: non-mathematical… Observation: People start doubting “obvious” things!. Program ?. Focus is everything , so:

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Elementary Mathematics

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  1. Elementary Mathematics Brush-Up

  2. Why ? Situation: • Lot of different backgrounds in PUB, • Many of those bgs: non-mathematical… Observation: • People start doubting “obvious” things!

  3. Program ? Focus is everything, so: • Not: real numbers, (rational) exponents, multiplying and factoring algebraic equations, rational expressions, solving polynomial equations, … (see 1st 15 pages of booklet) • Functions: numerical, algebraic & graphical viewpoint (with special attention to linear functions) • Systems of 2 equations in 2 unknowns: graphical & algebraic solution (elimination)

  4. Functions Definitions: • A real-valued function f of a real variable is a rule that assigns to each real number x in a specified set of numbers, called the domain of f, a single real number f(x). • The variable x is called the independent variable. If y = f(x) we call y the dependent variable. • A function can be specified: - numerically - algebraically - graphically

  5. Functions Numerically Specified Functions: • Suppose that the function f is specified by the following table: • Then, f(0) is the value of the function when x = 0. From the table, we obtain: • And so on…

  6. Functions Algebraically Specified Functions: • Suppose that the function f is specified by f(x) = 3x2- 4x + 1. Then

  7. Functions Graphically Specified Functions: • The graph of the function f is the set of all points (x, f(x)) in the xy-plane, where we restrict the values of x to lie in the domain of f. • To obtain the graph of a function, plot points of the form (x, f(x)) for several values of x in the domain of f. The shape of the entire graph can usually be inferred from sufficiently many points.

  8. Functions Graphically Specified Functions: • Example: to sketch the graph of the functionf(x) = x2 with domain the set of all real numbers, first choose some values of x in the domain and compute the corresponding y-coordinates: • Plotting these points gives the picture on the left, suggesting the graph on the right:

  9. Functions Linear Functions: • A linear function is one that can be written in the form: where m and b are fixed numbers. (the names m and b are traditional) • A linear function is one whose graph is a straight line (hence the term "linear"). • The graph of the above example looks like this:

  10. Functions Linear Functions: y = mx + b • Example: y = 3x – 1 • Numerically, b is the value of y when x = 0 • Graphically, b is the y-intercept of the graph • Numerically, y increases by m units for every 1-unit increase of x • Geometrically, the graph rises by m units for every 1-unit move to the right; m is the slope of the line

  11. Functions Linear Functions: y = mx + b Exercices: (1’) • y = 2x – 4 • y = x + 3 • y = -2x + 4 • 1 – y/6 = x/2 Graph ???

  12. Functions Linear Functions: y = mx + b Finding the equation of a linear function, when given: • Slope & y-intercept  immediately !!! • Example: Slope = 4 (= m!) y-Intercept = -3 (= b!)  y = 4x – 3 !

  13. Functions Linear Functions: y = mx + b Finding the equation of a linear function, when given: • Slope & y-intercept  immediately !!! • Slope & Point (x1, y1)  Slope-Intercept Formula:  Direct Formula: y = y1 + m(x - x1) Example: (1,2) & slope 3 y = 2 + 3(x-1) = 2 + 3x – 3 = 3x – 1 y = mx + b, where b = y1 – mx1

  14. Functions Linear Functions: y = mx + b Finding the equation of a linear function, when given: • Slope & y-intercept  immediately !!! • Slope & Point (x1, y1)  Slope-Intercept Formula  Direct Formula • Points (x1, y1) & (x2, y2)  calculate slope & intercept: m = (y2 – y1)/(x2 – x1) & b = y1 – mx1 (DF) Example: (0,1) & (2,-3) m = (-3 – 1)/(2 – 0) = -2 b = 1 – (-2)0 = -3 – (-2)2 = 1  y = -2x + 1

  15. Functions Linear Functions: y = mx + b Exercises: (1’) • y-intercept = -4 & slope = 9 • Slope = 1/3 & point (0,-1) • Point (2,1) & slope = -1 • Points (-2,1) and (-1,2) Equation ???

  16. Functions Linear Functions: y = mx + b Exercises: (1’) • y-intercept = -4 & slope = 9  y = 9x – 4 • Slope = 1/3 & point (0,-1)  y = (1/3)x – 1 • Point (2,1) & slope = -1  y = -x + 3 • Points (-2,1) and (-1,2)  y = x + 3

  17. Systems of 2 Equations in 2 ?’s Definitions: • A linear equation in two unknowns x and y is an equation of the form ax + by = c, where a, b, and c are numbers, and where a and b are both not zero. • Examples: 4x + 5y = 0 This has a = 4, b = 5, c =0 x – y = 11 This has a = 1, b = -1, c = 11

  18. Systems of 2 Equations in 2 ?’s Definitions: • A system of linear equations is just a collection of these beasts. To solve a system of linear equations means to find a solution (or solutions) (x, y) that simultaneously satisfies all of the equations in the system. • Examples: 4x + 5y = 40 x – y = 1 -x + 2y = 22 5x + 7y = 3 { {

  19. Systems of 2 Equations in 2 ?’s Graphical Solution: • To locate solutions to a system of two equations in two unknowns, plot the graphs, and locate the intersection points (if any). • Examples: 2x + y = 4 2x – y = 2  solution = (1,5;1) {

  20. Systems of 2 Equations in 2 ?’s Graphical Solution: • Examples: x – 2y = -2 x – 2y = 2  no solution ! (the lines are parallel) {

  21. Systems of 2 Equations in 2 ?’s Graphical Solution: • Examples: x – 2y = -2 -2x + 4y = 4  infinitely many solutions ! (both equations are represented by the same line) General solution: x = 2y – 2 y is arbitrary { {

  22. Systems of 2 Equations in 2 ?’s Algebraical Solution: • Problem with graphical approach: only gives approximate solutions! • Now: “method of elimination” = algebraic way of obtaining the exact solution(s) of a system of equations in two unknowns by manipulating the equations in such a way as to eliminate of the variables (x or y).

  23. Systems of 2 Equations in 2 ?’s Algebraical Solution (“Elimination”): • Examples: 2x + 3y = 4 (1) 2x + 3y = 4 (1) x – 3y = 2 (2) 3x + 0y = 6 (1) + (2) 22 + 3y = 4 x = 2 y = 0 x = 2 { {  {  { 

  24. Systems of 2 Equations in 2 ?’s Algebraical Solution (“Elimination”): • Examples: 2x + 3y = 3 (1) 6x + 9y = 9 (1)  3 3x – 2y = -2 (2) -6x + 4y = 4 (2)  (-2) 0x + 13y = 13  y = 1 6x + 9  1= 9 x = 0 y = 2 { {  {  { 

  25. Systems of 2 Equations in 2 ?’s Algebraical Solution (“Elimination”): • Examples: 3x – 5y = 1 (1) 6x – 10y = 2 (1)  2 -6x + 10y = -2 (2) -6x + 10y = -2 (2) 0x + 0y = 0  ?!? ! Infinitely many solutions! x = (5/3)y + (1/3) y is arbitrary { {  { General solution:

  26. Systems of 2 Equations in 2 ?’s Algebraical Solution (“Elimination”): • Examples: 3x – 5y = 1 (1) 6x – 10y = 2 (1)  2 -6x + 10y = 2 (2) -6x + 10y = 2 (2) 0x + 0y = 4  ?!? ! NO solutions ! (cf. graphically: 2 parallel lines  no intersection…) { { 

  27. Systems of 2 Equations in 2 ?’s Algebraical Solution (“Elimination”): • Exercise: (1’) 4x – 3y = 2 (1) x + y = 1 (2) { … 

  28. Systems of 2 Equations in 2 ?’s Algebraical Solution (“Elimination”): • Exercise: (1’) 4x – 3y = 2 (1) 4x –3y = 2 (1) x + y = 1 (2) 3x + 3y = 3 (2)  3 7x + 0y = 5  x = 5/7 4(5/7) - 3y = 2 x = 5/7 x = 5/7 x = 5/7 (20/7) – (21/7)y = (14/7) 21y = 20 – 14 y = 6/21 { {  { { { {

  29. Thank you !!! For further assistance: Filip Goeman Operations & Technology Management Centre (1st floor) filip.goeman@vlerick.be

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