Grade ScalePowerPoint PresentationSection 4.1A Solving Systems of Equations in Two VariablesWhat does this look like on a graph?Slide 5Slide 6Slide 7Problem from today’s homework:Slide 9Slide 10Note: Graph PaperSlide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Watch out for graphing problems #5 and #6, when the y-intercept is not an integer:Slide 20Slide 21Slide 22Slide 23Slide 24Slide 25Slide 26Slide 27Slide 28Slide 29Slide 30Slide 31Reminder:Slide 33Grade ScaleTest 1 Results:•Average class score after partial credit: __________•Commonly missed questions: #_________________•If you got less than 70% on Test 1, make sure to go over your quiz with me or a TA sometime today or tomorrow to help you prepare for tomorrow’s test.Now pleaseCLOSE YOUR LAPTOPSand turn off and put away your cell phones.Sample Problems Page Link(Dr. Bruce Johnston)Section 4.1ASolving Systems of Equations in Two Variables•A system of linear equations consists of two or more linear equations.•This section focuses on only two equations at a time.•The solution of a system of linear equations in two variables is any ordered pair that solves both of the linear equations.What does this look like on a graph?The SOLUTION to a system of two linear equations is the intersection (if any) of the two lines.There are only three possible solution scenarios:1. The lines intersect in a single point (so the answer is one ordered pair).2. The lines don’t intersect at all, i.e. they are parallel (so the answer is “no solution”.)3. The two lines are identical, i.e. coincident, so there are infinitely many solutions (all of the points that fall on that line.)To be a SOLUTION of a system of equations, an ordered pair must result in true statements for BOTH equations when the values for x & y are plugged into them. If either one (or both) gives a false statement, the ordered pair is NOT a solution of the system.Determine whether the given point is a solution of the following system.point: (-3, 1)system: x – y = -4 and 2x + 10y = 4•Plug the values into the equations.First equation: -3 – 1 = -4 trueSecond equation: 2(-3) + 10(1) = -6 + 10 = 4 true•Since the point (-3, 1) produces a true statement in both equations, it is a solution.Determine whether the given point is a solution of the following systempoint: (4, 2)system: 2x – 5y = -2 and 3x + 4y = 4•Plug the values into the equationsFirst equation: 2•4 - 5•2 = 8 – 10 = -2 trueSecond equation: 3•4 + 4•2 = 12 + 8 = 20  4 false•Since the point (4, 2) produces a true statement in only one equation, it is NOT a solution.Problem from today’s homework:•Note that our chances of guessing the right coordinates for a solution just by looking at the two equations are not very good.•Since a solution of a system of equations is a solution common to both equations, it would also be a point common to the graphs of both equations.•So one way to find the solution of a system of 2 linear equations is to graph the equations and see where the lines intersect.•You can use any of the techniques from Chapter 3 to graph the two lines (e.g. solving each equation for y and using the slope and intercept, or making a table of x- and y-values for each equation and plotting the ordered pairs.)Graphing is the first of three methods that we will be studying in this section to find a solution for a system of equations. The other two methods we will be using are:•Substitution method (also covered today)•Addition or elimination method (in Sec 4.1B)Note: Graph PaperClick on the “News Bulletins” button to find a site that allows you to print free graph paper.• If you want to be able to draw accurate graphs but you don't want to buy a whole pack of graph paper for one assignment, go to this web site and print a couple pages of graph paper for free. (You don’t have to do this – graphing by hand on plain paper is fine, but sometimes it’s easier to see the solutions if you can plot your points carefully on real graph paper instead of a hand-drawn graph grid.) http://www.printfree.com/Office_forms/GraphPaper2.htmSolve the following system of equations by graphing.2x – y = 6 andx + 3y = 10xyFirst, graph 2x – y = 6.(0, -6)(3, 0)(6, 6)Second, graph x + 3y = 10.(1, 3)(-2, 4)(-5, 5)The lines APPEAR to intersect at (4, 2).(4, 2)•Although the solution to the system of equations appears to be (4, 2), you still need to check the answer by substituting x = 4 and y = 2 into the two equations.First equation,2(4) – 2 = 8 – 2 = 6 trueSecond equation,4 + 3(2) = 4 + 6 = 10 true•The point (4, 2) checks, so it is the solution of the system.Problem from today’s homework:Solve the following system of equations by graphing.-x + 3y = 6 and3x – 9y = 9xyFirst, graph -x + 3y = 6.(-6, 0)(0, 2)(6, 4)Second, graph 3x – 9y = 9.(0, -1)(6, 1)(3, 0)The lines APPEAR to be parallel.•Although the lines appear to be parallel, you still need to check that they have the same slope. You can do this by solving for y.First equation,-x + 3y = 6 3y = x + 6 (add x to both sides)31 y = x + 2 (divide both sides by 3)Second equation,3x – 9y = 9 -9y = -3x + 9 (subtract 3x from both sides)31y = x – 1 (divide both sides by –9)31• Both lines have a slope of , since they have different y-intercepts they are parallel and do not intersect. Hence, there is no solution to the system.Solve the following system of equations by graphing.x = 3y – 1 and2x – 6y = -2xyFirst, graph x = 3y – 1.(-1, 0)(5, 2)(7, -2)Second, graph 2x – 6y = -2.(-4, -1)(2, 1)The lines APPEAR to be identical.•Although the lines appear to be identical, you still need to check that they are identical equations. You can do this by solving for y.First equation, x = 3y – 1 3y = x + 1 (add 1 to both sides)Second equation,2x – 6y = -2 -6y = -2x – 2 (subtract 2x from both sides)• The two equations are identical, so the graphs must be identical. There are an infinite number of solutions to the system (all the points on the line y = 1/3 x + 1/3).31 y = x + (divide both sides by 3)3131 y = x + (divide both sides by -6)31Watch out for graphing problems #5 and #6, when the y-intercept is not an integer:Solving the second equation for y gives y = -1/3 x + 13/3.The graphing tool will not allow you to plot the y-intercept (0,13/3). To use the graphing tool, you must plot two points by selecting integer