Chapter 1 Linear Equations and GraphsThe Cartesian Coordinate SystemSlide 3Linear Equations in Two VariablesLinear Equations in Two Variables (continued)Using Intercepts to Graph a LineSlide 7Using a Graphing CalculatorSlide 9Special CasesSolutionsSlope of a LineSlope-Intercept FormFind the Slope and Intercept from the Equation of a LineSlide 15Point-Slope FormExampleSlide 18ApplicationSlide 20Supply and DemandSupply and Demand ExampleSupply and Demand Example (continued)Slide 24Slide 25Chapter 1Linear Equations and GraphsSection 2Graphs and Lines2The Cartesian Coordinate SystemThe Cartesian coordinate system was named after René Descartes. It consists of two real number lines, the horizontal axis (x-axis) and the vertical axis (y-axis) which meet in a right angle at a point called the origin. The two number lines divide the plane into four areas called quadrants. The quadrants are numbered using Roman numerals as shown on the next slide. Each point in the plane corresponds to one and only one ordered pair of numbers (x,y). Two ordered pairs are shown.3 xyIIIIII IV(3,1)(–1,–1)The Cartesian Coordinate System(continued)Two points, (–1,–1) and (3,1), are plotted. Four quadrants are as labeled.4Linear Equations in Two VariablesA linear equation in two variables is an equation that can be written in the standard form Ax + By = C, where A, B, and C are constants (A and B not both 0), and x and y are variables.A solution of an equation in two variables is an ordered pair of real numbers that satisfy the equation. For example, (4,3) is a solution of 3x - 2y = 6.The solution set of an equation in two variables is the set of all solutions of the equation.The graph of an equation is the graph of its solution set.5Linear Equations in Two Variables (continued)If A is not equal to zero and B is not equal to zero, thenAx + By = C can be written asThis is known as slope-intercept form.If A = 0 and B is not equal to zero, then the graph is a horizontal lineIf A is not equal to zero and B = 0, then the graph is a vertical line A Cy x mx bB B=- + = +CyB=CxA=6Using Intercepts to Graph a LineGraph 2x – 6y = 12.7Using Intercepts to Graph a LineGraph 2x – 6y = 12.x y0 –2 y-intercept6 0 x-intercept3 –1 check point8Using a Graphing CalculatorGraph 2x – 6y = 12 on a graphing calculator and find the intercepts.9Using a Graphing CalculatorGraph 2x – 6y = 12 on a graphing calculator and find the intercepts.Solution: First, we solve the equation for y.2x – 6y = 12 Subtract 2x from each side. –6y = –2x + 12 Divide both sides by –6 y = (1/3)x – 2Now we enter the right side of this equation in a calculator, enter values for the window variables, and graph the line.10Special CasesThe graph of x = k is the graph of a vertical line k units from the y-axis. The graph of y = k is the graph of the horizontal line k units from the x-axis. Examples:1. Graph x = –72. Graph y = 311Solutionsx = –7y = 412Slope of a Line Slope of a line:rise runoo( )2 2,x y( )1 1,x yrunrisexxyym 121213Slope-Intercept Form The equation y = mx+b is called the slope-intercept form of an equation of a line. The letter m represents the slope and b represents the y-intercept.14Find the Slope and Intercept from the Equation of a Line Example: Find the slope and y intercept of the line whose equation is 5x – 2y = 10.15Find the Slope and Intercept from the Equation of a Line 5 2 102 5 105 10 552 2 2x yy xxy x- =- =- +-= + = -- -Example: Find the slope and y intercept of the line whose equation is 5x – 2y = 10. Solution: Solve the equation for y in terms of x. Identify the coefficient of x as the slope and the y intercept as the constant term. Therefore: the slope is 5/2 and the y intercept is –5.16Point-Slope Form 1 1( )y y m x x- = -2 12 1y ymx x-=-Cross-multiply and substitute the more general x for x2where m is the slope and (x1, y1) is a given point.It is derived from the definition of the slope of a line: The point-slope form of the equation of a line is17ExampleFind the equation of the line through the points (–5, 7) and (4, 16).18Example199)5(4716mNow use the point-slope form with m = 1 and (x1, x2) = (4, 16). (We could just as well have used (–5, 7)).Find the equation of the line through the points (–5, 7) and (4, 16).Solution: 12164)4(116xxyxy19Application Office equipment was purchased for $20,000 and will have a scrap value of $2,000 after 10 years. If its value is depreciated linearly, find the linear equation that relates value (V) in dollars to time (t) in years:20Application Office equipment was purchased for $20,000 and will have a scrap value of $2,000 after 10 years. If its value is depreciated linearly, find the linear equation that relates value (V) in dollars to time (t) in years: Solution: When t = 0, V = 20,000 and when t = 10, V = 2,000. Thus, we have two ordered pairs (0, 20,000) and (10, 2000). We find the slope of the line using the slope formula. The y intercept is already known (when t = 0, V = 20,000, so the y intercept is 20,000). The slope is (2000 – 20,000)/(10 – 0) = –1,800. Therefore, our equation is V(t) = –1,800t + 20,000.21Supply and DemandIn a free competitive market, the price of a product is determined by the relationship between supply and demand. The price tends to stabilize at the point of intersection of the demand and supply equations.This point of intersection is called the equilibrium point.The corresponding price is called the equilibrium price.The common value of supply and demand is called the equilibrium quantity.22Supply and DemandExampleYear SupplyMil buDemandMil buPrice$/bu2002 340 270 2.222003 370 250 2.72Use the barley market data in the following table to find: (a) A linear supply equation of the form p = mx + b (b) A linear demand equation of the form p = mx + b (c) The equilibrium point.23Supply and DemandExample (continued)(a) To find a supply equation in the form p = mx + b, we must first find two points of the form (x, p) on the supply line. From the table, (340, 2.22) and (370, 2.72) are two such points. The slope of the line is 2.72 2.22 0.50.0167370 340 30m-= = =-Now use the point-slope form