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STEYNING & DISTRICT U3A Discovering Mathematics Session 4

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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 subtract7y = 28and 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 – 2ySubstitute x in b) 4(7 – 2y) – 3y = 6 or 28 – 8y – 3y = 6or 28 – 11y = 6 By transposing ; 22 = 11y or y = 2 Replacing the value of y in a);x + 4 = 7 or x = 3

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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

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Applying the Principle i)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 – 4substitute4j – 4 = 3j + 1 So 4j – 3j = 1 + 4j = 5&p = 3x5 + 1p = 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 – 18substitute 9j – 18 = 6j + 3 So 9j – 6j = 3 – 18, ie 3j = 21 & j = 7p = 6x7 + 3p = 45

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Quadratic Equations If possible, quadratic equations should be solved by factorisation. Eg i)x 2 + 6x + 8 = 0 is of the form of ax 2 + 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) = x 2 + 8x + x + 8 = 0Wrong! Try (x + 4)(x + 2) = x 2 + 4x + 2x + 8 = 0 x 2 + 6x + 8 = 0 Correct. So x+4=0 or x+2=0 & x = -4 or -2 ii) 2x 2 - 7x - 15 = 0 Factors of 15 are 5, 3, 15, 1 Try (2x + 5)(x – 3) = 2x 2 + 5x - 6x - 15 = 0 Wrong! Try (2x + 3)(x – 5) = 2x 2 + 3x - 10x - 15 = 0 2x 2 - 7x -15 = 0 Correct. & 2x+3=0 or x-5=0 so x = -3/2 or +5

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Quadratic Equations 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 : ax 2 + bx + c = 0 Divide throughout by a, we get ; x 2 +bx/a + c/a = 0 Rearranging this we get ; x 2 +bx/a = -c/a Add (b/2a) 2 to both sides, we have. x 2 +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 + b 2 /4a 2 = (-4ac + b 2 )/4a 2 If we now take the square root of each side, we have ; x + b/2a = +- √(-4ac + b 2 )/2a So x = [-b +- √( b 2 - 4ac )]/2a

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

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Solving Quadratic Equations by graphical means. X X2X x X 2 –x To solve the equation x 2 –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 x 2 +2x–3x -6 = 0 x 2 –x -6 = 0 X-Values

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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 1 st June 2013 “Ministers misusing statistics to mislead voters must pay”.

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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..

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Statistics (continued 2) Tabular Form Bar Charts

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Statistics (continued 3) Pie Charts Sales 1st Qtr8.2 2nd Qtr3.2 3rd Qtr1.4 4th Qtr1.2

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Statistics (continued 4) Line Graphs

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