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Revision: 1.Solving Equations

Revision: 1.Solving Equations

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Revision: 1.Solving Equations

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  1. 1.3 ALGEBRA Relationships between tables, equations or Graphs4 Credits (External)AS 91028Gamma Mathematics - Ex 12,13,14 and 15 Page 254 Work book – Chapter 3 Page73

  2. Revision: 1.Solving Equations To solve equations you must: • Get the unknown to one side of the equal sign by itself • Do the same operation to both sides of the equals sign • Start each step on a new line with all the equals signs lined up

  3. 5x = 42 x + 5 = 7 x = x = 7 - 5 x = 2 = -3 x - 7 = 7 x = 7 + 7 x = -3 x 2 x = -6 x = 14 Examples: Solve

  4. Example: Solve = 7 6x + 5 = 17 - 3 = 9 6x = 17 - 5 x + 3 = 7 x 2 = 9 + 3 6x = 12 x + 3 = 14 x = 2 x = 14 - 3 = 12 x = 11 x = 12 x 5 x = 60 In an equation with more than one operation, the opposite operations must be applied in reverse order

  5. 2. Solving Inequations Linear inequations have an inequality symbol instead of an equals sign: < less than < less than or equal to > greater than > greater than or equal to

  6. To solve inequations follow the same procedure as solving equations. When multiplying or dividing by a negative number swap the inequality sign around.

  7. -5x - 2 < -22 3x + 5 > -7 -5x < -22 + 2 3x > -7 - 5 -5x < -20 3x > -12 x >4 x > -4 Dividing by a negative – swap sign!! Note: always check your answer by putting putting in a number to check that the equation is true!! Examples: Solve

  8. -7 – 3x < -1 -2x - 4 > 10 -3x < 6 -2x > 14 x > -2 x < -7 More examples:

  9. 3. Substitution Substitution means replacing a symbol with a value. Remember to follow the rules of BEDMAS. Examples: Calculate the value of these expressions 7x – 1 when x = 2 7  2 – 1 = 14 – 1 = 13

  10. when f = 2 and g = 6 = 4 5x2 - 3x + 2 when x = -3 5x(-3)2 - 3  -3 + 2 = 56 = 45 - -9 + 2

  11. Page 258 Exercise 12.01

  12. Note 1: Linear Patterns Linear Patterns are sequences of numbers where the difference between successive terms is always the same. The general rule for a LINEAR pattern is always in the form: Term y = dx + c x is the position of the term in the sequence d is the difference c is a constant

  13. Example 1:Find the rule that generates the sequence Term # (x) 1 2 3 4 5 6 7 ……. e.g. 4, 7, 10, 13, 16, 19, 22……. +3 This sequence has a common difference of _______ Each Term = 3 x Term # + C C = constant term C is found by substituting ANY term and term # into the formula e.g. 4= 3 x1+ C C = 4 - 3 Rule to find each term is y = 3x + 1 C = 1

  14. e.g. 2 Identify if the following relations are linear. -3 2 -3 4 -3 8 -3 16 32 -3 Common difference = -3 Therefore, Linear No Common difference Therefore, Not Linear

  15. e.g. Find the differences, and if the result is linear find the rule. Common difference = -3 Therefore, Linear (X) Each Term = −3 x Term # + C 16 = −3 x 1 + C -3 -3 16 = −3 + C -3 19 = C -3 -3 Each Term = −3x + 19 Rule: y = -3x + 19 Check that your formula works !

  16. NOW YOU TRY THIS ONE -8, -6, -4, -2, 0, 2…… Find the differences, and if the result is linear find the rule. Common difference = Therefore, +2 linear (X) Each Term = 2 x Term # + C -8 = 2 x 1 + C 2 2 -8 = 2 + C 2 -10 = C 2 2 Each Term = 2x − 10 Rule: y = 2x - 10 Check that your formula works !

  17. NOW YOU TRY THIS ONE 6, 11, 16, 21, 26, 31…… Find the differences, and if the result is linear find the rule. Common difference = Therefore, 5 linear (X) Each Term = 5 x Term # + C 6 = 5 x 1 + C 5 5 6 = 5 + C 5 IWB Mathematics Pg 269-273 Ex. 12.02 # 3 -12 1 = C 5 5 Each Term = 5x + 1 Rule: y = 5x + 1 HMWKpg 74 Ex A Check that your formula works !

  18. Using Rules for Linear Sequences If the rule for a linear sequence is known, then the values of terms or the term number can be found algebraically. Examples: The rule for a linear sequence is t = 4n -2 Find the 7th term of the sequence t = 4 x 7 – 2 =26 Which term has a value of 74? 74 = 4n -2 76 = 4n n = 19

  19. What is the first term in the sequence that has a value over 151? 4n – 2 > 151 4n > 153 n > 38.25 39th term is the first term greater than 151 IWB Mathematics Pg 269-273 Ex. 12.02 HMWKpg 74-82 Ex B

  20. StarterIdentify if the following relations are linear. y = 2x – 3 xy = -13 y = x(x + 2) y = 2 Linear y1 = 2x1 - 3 Not Linear x = -13y-1 Not Linear y1 = x2 + 2x1 Linear y1 = 2

  21. Note 2: Graphs of linear functions • The graph of a linear function is a straight line. • Plot points by: • Draw a table of x values. • Substitute the x value into the formula and solve for y. • Plot the pair of co-ordinates onto the axes In a linear equation, x and y are only raised to the power of 1. (i.e. Not x2, x3, x-1 etc) There is a common difference between ordered pairs.

  22. Plot points from the table on an axis. (x, y) Up/Down the y-axis Across the x-axis

  23. Note 2: Graphs of linear functions • Drawing up a table of x values. • Substitute the x value into the formula and solve for y. • Plot the pair of co-ordinates onto the axes egDraw the graph of y = 2x – 1 IWB Mathematics Pg 295 Ex. 13.01 -5 -3 -1 1 3 5 When x = -2 y = 2 -2 – 1 = -5 When x = -1 y = 2 -1 – 1 = -3 HMWKpg 78 Ex C and D

  24. Vertical Change rise Horizontal Change run Note 3: Gradients • The gradient of a graph tells us how steep the graph is. • It is given as a whole number or a fraction. • The top number is the change in y, and the bottom number is the change in x. • Gradient =

  25. Note 3: Gradients Special gradients to remember A horizontal line has a gradient of 0 (the rise = 0) Gradient = 0 A vertical line has a gradient that is undefined. (the run = 0) Gradient = undefined A line that is going up hill is positive. Gradient = positive A line that is going down hill is negative. Gradient = negative

  26. Note 3: Gradients To draw lines with a certain gradient: 1.) Pick a starting point. 2.) Count the number of squares Up (if rise is +) Down (if rise is -) 3.) Count the number of squares Right (if run is +) Left (if run is -) If there is no run, (i.e. the gradient is a whole number) it is always 1. 4.) Mark the endpoint.

  27. Note 3: Gradients e.g. Draw lines with the following gradients. a = b = 3

  28. Note 3: Gradients To find the gradients of a straight line: Pick 2 points on the line that go through a corner of a grid line. The points marked here are the only ones that can be used as they are the only ones that go through the corners of the grid. Count the rise from one point to the other, then count the run. Simplify where possible. =

  29. Note 3: Gradients b= e.g. Find the gradients of the following lines. undefined = GAMMAPg 298 Ex. 13.02 HMWKpg 83 ExE

  30. Note 4: Straight Line Graphs The general formula for a straight line is y = mx + c There are a couple of straight lines that do not fit this general formula: y = number gives a horizontal line x = number gives a vertical line y-int gradient

  31. Note 4: Straight Line Graphs e.g. y = 2 x = 4

  32. Note 4: Straight Line Graphs To draw straight line graphs using y = mx + c • Mark the y-intercept (given by c). If there is no c, the y-intercept is 0. • From the y-intercept, count the gradient (given by m) Put a point and then count the gradient again. If there is no m, it is always 1. • Rule the straight line between the three points. • Put arrows on both ends of the line and label it.

  33. Note 4: Straight Line Graphs y y = • y = 4x - 2 • y = -2x e.g.Draw the following straight lines. • y = • y = 4x – 2 • y = x + 1 • y = -2x x GAMMAPg 306 Ex. 13.03 HMWKpg 86 Ex F • y = x + 1

  34. Note 4: Straight Line Graphs Using Intercept-Intercept Method To find the x-intercept, set y = 0. To find the y-intercept, set x = 0. Join the two intercepts to make a straight line. e.g.Find the x and y intercepts of the line y = -4x + 12 Find y-intercept, set x = 0 y = -4 (0) + 12 y = 12 therefore the y intercept is (0, 12) Find x-intercept, set y = 0 0 = -4x + 12 4x= 12 x = 3 therefore the x intercept is (3, 0)

  35. Note 4: Straight Line Graphs Using Intercept-Intercept Method Draw the graph of 2x – y = 6 Find x-int, Set y = 0 2x – 0 = 6 2x = 6 x = 3 therefore the x-intercept is (3, 0) Find y-int, Set x = 0 2(0) – y = 6 -y = 6 y = -6 therefore the y-intercept is (0, -6)

  36. x-intercept is (3, 0) y-intercept is (0, -6)

  37. The cover up method – a quick way of graphing a straight line in the form ax + by = c -x + 2y = 8 Is to cover up the x term to get the y-intercept -x + 2y = 8 And then cover the y term to get the x-intercept -x + 2y = 8 Then plot the 2 points. i.e. y = 4 (0,4) i.e. x = -8 (-8,0)

  38. y -x + 2y = 8 (0,4) (-8,0) x IWB Mathematics EX 13.04 pg 310 HMWK G, H

  39. Starter – Graph to Equation 2 y = -1/2 x + 0 3 y = 1/3x - 5 y = 2x + 3 1 4 y = -4x + 4

  40. Writing Equations for Graphs • y-intercept (c) • Gradient (m) • The equation for the graph is found by replacing m and c into the general format for a line: y = mx + c. • For a vertical line: x = c • For a horizontal line: y = c Where c is a constant By looking at a graph we can write the equation for a line by finding the:

  41. Example: Write the equations for the lines Equation a c = 2 m = -¾ y = -¾x + 2 Equation b x = 5 Equation c c = -6 m = 6/4 = 3/2 y = 3/2x - 6

  42. Starter y 2x - y = 4 y = 0 3x = 6 x = 2 Int: (2, 0) y = 0 2x = 4 x = 2 Int: (2, 0) x = 0 6y = 6 y = 1 Int (0, 1) x = 0 -y = 4 y = -4 Int: (0, -4) x 2x – y = 4 3x + 6y = 6 Draw the following straight lines. 3x + 6y = 6

  43. Note 5: Applications of the Straight Line Graph Many circumstances in nature and life can be graphed using straight line relationships. • The y-intercept represents the value of y initially (when x=0) Example: A fixed administration fee • The gradient represents the rate at which y changes compared with x. Example: An hourly hire charge

  44. Note 5: Applications of the Straight Line Graph e.g. De Vinces specializes in pizza. The approximate relationship between x, the # of pizzas they sell daily and y their daily costs, is given by: y = 10x + 50 Draw a graph showing the number of pizzas they sell on the x-axis, and their daily costs on the y-axis. Use an appropriate scale. (Because the numbers are large, it is best to have the y-axis going up in 20’s).

  45. Cost $ y 140 y = 10x + 50. 120 100 80 60 40 20 x 4 14 2 6 8 10 12 Number of Pizzas sold

  46. $130 5 What are their costs if they sell 8 pizzas? Either read off the graph or substitute into the equation. If their costs are $100, how many pizzas did they sell? Either read off graph or substitute into the equation. Give the y-intercept. What does it represent? Give the gradient. What does it represent? If no pizzas are sold, it still costs them $50 50 Represents the fact that each pizza they make, costs $10 10 Cost $ Number of Pizzas sold

  47. Note 6: Rate of Change If two variables (e.g. distance and time) are directly dependent, the graph of this relation is a straight line. We can calculate the rate of change of one variable compared to the other. rate of change = change in dependent variable change in independent variable e.g. = change in distance change in time Distance (km) The gradient of the graph is also a measure of the rate of change of the dependent variable Time (hrs)

  48. e.g. The graph below shows the value of a section in Auckland. The rate of increase shows two distinct slopes. Find the rate in increase before and after 2004. Value of section in Auckland Before 2004 Rate of change = Δ price / Δ time 400 = 330 – 280 2004 – 1998 380 Price in $000 360 = 50 6 340 = $8 333/yr 320 = 400 – 330 2010 – 2004 After 2004 300 280 = 70 6 1998 2000 2002 2004 2006 2008 2010 = $11 667/yr Year

  49. e.g. An engineer studying the expansion of a bar of metal made the following measurements. At 14°C the length of was 41.4 mm, at 22°C the length was 42.0 mm and at 34°C the length was 42.9 mm. Plot the data on a graph & find the rate of change of length compared to temperature. Expansion of a metal bar 43 Plot points (14,41.4), (22,42) & (34,42.9) mm Rate of change = Δ length / Δ temperature 42 = 42.9 – 41.4 34 – 14 = 1.5 20 41 =0.075 mm/°C 10 20 30 °C

  50. IWBMathematics Ex 13.06 Pg 316 - 321 Work book Ex I Starter km This graph shows the distance a car has travelled after leaving Dunedin. • Find the rate of change for each of the three sections A = B = C = b.) Give the equation that represent section A of the journey D = c.) Using your equation from b) find what time the car is 68 km from Dunedin. 140 120 B 100 100 km/hr C 80 20 km/hr A 60 -65 km/hr 40 20 100t - 400 4 5 6 7 8 9 Time (pm) 4.68 hrs = 4hrs 41mins