lecture2Lecturer.pdf - Lecture 2 Unit 2 Linear Equations Outline 1 Linear equations and functions 2 Simultaneous linear equations Two variables in two

lecture2Lecturer.pdf - Lecture 2 Unit 2 Linear Equations...

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Lecture 2: Unit 2, Linear Equations 5/2/2016
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Outline 1 Linear equations and functions 2 Simultaneous linear equations Two variables in two equations Three variables in three equations 3 Economic applications Demand and supply analysis National Income IS-LM analysis
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Objectives Manipulate and solve linear equations Sketch the graph of a linear function Solve systems of simultaneous linear equations
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Literature Renshaw, ch. 3
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General form of linear equation Consider ax + b = c , in which x is the only variable, while a , b and c are constants and/or parameters The solution to this equation is x = c - b a Linear function: Constant relationship between x and y
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The function y = ax + b Consider the linear function y = ax + b In this function: y is the dependent variable x is the independent variable a is the slope Slope: Shows the change in y for one-unit change in x , i.e. a = Δ y Δ x y -intercept: Shows the value of y if x is 0 x -intercept: The x -intercept shows the value of x if y is 0, i.e. x = - b / a
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Inverse functions and Implicit vs. Explicit functions Inverse function: if y = ax + b then x = y - b a is the inverse function Note that for the inverse function, the slope is 1 / a while the x -intercept is - b / a Explicit function: y = ax + b . It’s clear which variable is the dependent, and which is the independent Implicit function: ax + by - c = 0. It’s not clear which variable is the dependent, and which one the independent
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Simultaneous equations: 2 by 2 The solution to systems of linear equations involves finding values for all of the variables that satisfy all of the equations in the system simultaneously. Consider the following example: Solve for x and y in the following system of equations 8 x + 4 y = 12 - 2 x + y = 9 There are two ways to solve this problem: Substitution Elimination
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Solution via substitution Take the second equation, and make y the subject: y = 2 x +9 Now, plug this expression for y into the first equation: 8 x + 4(2 x + 9) = 12 Next, simplify: 8 x + 8 x + 36 = 12 16 x = - 24 x = - 24 16 = - 3 2 = - 1 . 5 Plug this solution into the expression for y : y = 2( - 1 . 5) + 9 = 6
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Solution via elimination Again, solve for x and y given 8 x + 4 y = 12 - 2 x + y = 9 First, decide which variable you want to get rid of (let’s choose y ) Now, multiply first equation by 1; multiply the second equation by 4 The system then becomes: 8 x + 4 y = 12 - 8 x + 4 y = 36 Note that both equations contain the term 4 y
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Solution via elimination 2 Now subtract the second equation from first equation : 8 x + 4 y - ( - 8 x + 4 y ) = 12 - 36 This yields 16 x = - 24 x = - 1 . 5 Now plug x = - 1 . 5 into any of the equations above (let’s choose the first equation here): - 2(1 . 5) + y = 9 y = 6 Note that these answers are the same (as they should be!)
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Simultaneous equations: 3 variables in 3 equations We can again solve for these using either elimination or substitution There are just a few more steps to follow Consider the following system of equations: x + y - z = - 4 2 x - y - z = 0 3 x - 4 y - z = 0
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Substitution Rewrite the first equation, making z the subject: x + y + 4 = z .
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