1 / 13

Understanding Simultaneous Equations and Quadratic Solutions

In this session, we explore methods for solving simultaneous equations, including elimination and substitution. We discuss examples using two equations with common unknowns, solving them both graphically and algebraically. Additionally, we delve into quadratic equations, emphasizing factorization, the quadratic formula, and graphical methods to find solutions. We've also touched upon statistics and various data presentation methods, aiming to provide a well-rounded understanding of mathematical principles.

denzel
Télécharger la présentation

Understanding Simultaneous Equations and Quadratic Solutions

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. STEYNING & DISTRICT U3ADiscovering MathematicsSession 4

  2. Simultaneous Equations Two equations with common unknowns can be solved by elimination or by substitution. Eg. i) Eqn a) 10x + 3y = 2 and Eqn. b) 2x + 2y = 6 By inspection we see that multiplying b) by 5 & subtracting a) from b) We have ; 10x + 10y = 30 10x + 3y = 2 subtract 7y = 28 and y = 4 Find x by substituting y in b); 2x + 8 = 6, 2x = -2 and x = -1 ii) Solve a) x + 2y = 7 & b) 4x – 3y = 6 by substitution. From a) x = 7 – 2y Substitute x in b) 4(7 – 2y) – 3y = 6 or 28 – 8y – 3y = 6 or 28 – 11y = 6 By transposing ; 22 = 11y or y = 2 Replacing the value of y in a); x + 4 = 7 or x = 3

  3. Solving Simultaneous Equations with Graphs y = 7 – 2x If x = 0, y = 7 y = 0, x = 3.5 y = 3 + x If x = 0, y = 3 x = 4, y = 7 Y = 3 + x The x & y values where the lines intersect is the solution. In this case x = 1.33 and y = 4.33

  4. Applying the Principle I have a number of pencils (p) to be placed in jars (j). If I put 4 pencils in each jar, I have 1 jar left over. If I put 3 pencils in each jar, I am left with 1 spare pencil. How many jars and pencils? j = p/4 + 1 & p = 3j + 1 4j = p + 4 or p = 4j – 4 substitute 4j – 4 = 3j + 1 So 4j – 3j = 1 + 4 j = 5 & p = 3x5 + 1 p = 16 ii) With 9 pencils in each jar, I have 2 jars empty. With 6 pencils in each jar, I have 3 spare pencils. j = p/9 + 2 & p = 6j + 3 9j = p + 18 or p = 9j – 18 substitute 9j – 18 = 6j + 3 So 9j – 6j = 3 – 18, ie 3j = 21 & j = 7 p = 6x7 + 3 p = 45

  5. Quadratic Equations If possible, quadratic equations should be solved by factorisation. Egi) x2 + 6x + 8 = 0 is of the form of ax2 + bx+ c = 0 this should factorise to (ax + ?)(x + ??) a=1. b=6 and c=8 The factors of b are 6, 1, 3, 2 & of c are 8, 1, 4, 2 Try (x + 8)(x + 1) = x2 + 8x + x + 8 = 0 Wrong! Try (x + 4)(x + 2) = x2 + 4x + 2x + 8 = 0 x2 + 6x + 8 = 0 Correct. So x+4=0 or x+2=0 & x = -4 or -2 ii)2x2 - 7x - 15 = 0 Factors of 15 are 5, 3, 15, 1 Try (2x + 5)(x – 3) = 2x2 + 5x - 6x - 15 = 0 Wrong! Try (2x + 3)(x – 5) = 2x2 + 3x - 10x - 15 = 0 2x2 - 7x -15 = 0 Correct. & 2x+3=0 or x-5=0 so x = -3/2 or +5

  6. At school we were taught a formula for the solution of those quadratic equations which couldn’t be factorised. We can derive it from very basic algebraic principles. The standard equation is : ax2 + bx + c = 0 Divide throughout by a, we get ; x2 +bx/a + c/a = 0 Rearranging this we get ; x2 +bx/a = -c/a Add (b/2a)2 to both sides, we have . x2 +bx/a + (b/2a)2 = -c/a + (b/2a)2 The left side may now be factorised to give ; (x + b/2a)2 = -c/a + b2/4a2 = (-4ac + b2)/4a2 If we now take the square root of each side, we have ; x + b/2a = +- √(-4ac + b2)/2a So x = [-b +- √( b2 - 4ac)]/2a Quadratic Equations

  7. Quadratic Equations (continued) Try; 5x2 +4x – 3 = 0 and x = [-b +- √( b2 - 4ac)]/2a Then x = [-4 +- √( 42 – 4x5x-3)]/2x5 x = [-4 +- √( 16 + 60)]/10 = [-4 +- √(76)]/10 √(76) is somewhere between 8 & 9, ie approx. 8.7, so x = (-4 + 8.7)/10 = 0.47 or -12.7/10 = -1.27

  8. Solving Quadratic Equations by graphical means. To solve the equation x2 –x -6 = 0 Make the equation = y. Plot x against y. Then solutions occur where the curve intersects the x – axis. In this case: x = -2 & +3 ie (x+2)(x-3) = 0 or x2 +2x–3x -6 = 0 x2 –x -6 = 0 X-Values

  9. STATISTICS Who said: “There are three kinds of lies; lies, damned lies and statistics”. The statement was attributed to Disraeli in Mark Twain’s autobiography. (1924) but an anonymous version occurred earlier in The Economic Journal (1892). Guardian Headline 1st June 2013 “Ministers misusing statistics to mislead voters must pay”.

  10. Statistics (continued 1) Statistics is the collection, organization and presentation of data. There are several methods of presenting statistical data, Eg. Tabular Form, Bar Charts, Histograms, Pie Charts, Line Graphs, etc..

  11. Statistics (continued 2) Tabular Form Bar Charts

  12. Statistics (continued 3) Pie Charts Sales 1st Qtr 8.2 2nd Qtr 3.2 3rd Qtr 1.4 4th Qtr 1.2

  13. Statistics (continued 4)Line Graphs

More Related