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Grade Scale Test 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.

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Now please CLOSE YOUR LAPTOPS and turn off and put away your cell phones. Sample Problems Page Link (Dr. Bruce Johnston )

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Section 4.1A Solving 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.

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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.)

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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.

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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 true Second 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. Example

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Determine whether the given point is a solution of the following system point: (4, 2) system: 2x – 5y = -2 and 3x + 4y = 4 Plug the values into the equations First equation: 24 - 52 = 8 – 10 = -2 true Second equation: 34 + 42 = 12 + 8 = 20 4 false Since the point (4, 2) produces a true statement in only one equation, it is NOT a solution. Example

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Problem from today’s homework:

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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.)

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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)

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Note: Graph Paper Click 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/GraphPaper 2.htm

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Solve the following system of equations by graphing. 2x – y = 6 and x + 3y = 10 Example x y First, 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)

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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 true Second equation, 4 + 3(2) = 4 + 6 = 10 true The point (4, 2) checks, so it is the solution of the system. Example (cont.)

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Problem from today’s homework:

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Solve the following system of equations by graphing. -x + 3y = 6 and 3x – 9y = 9 Example x y First, 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.

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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) Example (cont.) y = x + 2 (divide both sides by 3) Second equation, 3x – 9y = 9 -9y = -3x + 9 (subtract 3x from both sides) y = x – 1 (divide both sides by –9) 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.

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Solve the following system of equations by graphing. x = 3y – 1 and 2x – 6y = -2 Example x y First, 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.

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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) Example (cont.) 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). y = x + (divide both sides by 3) y = x + (divide both sides by -6)

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Watch 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 values for x that also produce integer values for y. Solving the first equation for y gives y = 2x +2, which can be graphed by starting at the y-intercept of (0,2) and then using the slope 2/1 to produce a second point.

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A second method that can be used to solve systems of equations is called the substitution method. To use this method, you solve one equation for one of the variables, then substitute the new form of the equation into the other equation for the solved variable. This gives an equation with only one variable, which then can be solved using the methods used in Chapter 2. Substitution Method

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Solving a system of linear equations in two variables by the substitution method: 1)Solve one of the equations for one of the two variables (if this is not already done.) 2)Substitute the expression from step 1 into the other equation. 3)Solve the new equation for that one remaining variable. 4)Substitute the value found in step 3 into either of the original equations containing both variables. 5)Check the proposed solution in the original equations.

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Solve the following system of equations using the substitution method. y = 4x and 4x + y = 32 Since the first equation is already solved for y (y = 4x), substitute 4x into the second equation in place of y. 4x + y = 32 4x + 4x = 32 (replace y with result from first equation) 8x = 32 (simplify left side) x = 4 (divide both sides by 8) Example

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Substitute the solution (x = 4) into either equation and solve for the other variable. y = 4xor4x + y = 32 y = 4(4)4(4) + y = 32 y = 1616 + y = 32 y = 16 The solution to the system of equations is (4, 16). Make sure you check your answer in BOTH ORIGINAL EQUATIONS to make sure it is a true solution. This is especially important on tests and quizzes when you don’t have the “check answer” button, so practice it in the homework assignments. Example (cont.)

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Problem from today’s homework:

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Note: This problem can also be solved just as easily by the elimination method which we will be covering in the next lecture.

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Problem from today’s homework: Note: This problem can also be solved (perhaps more easily!) by the elimination method with we will be covering in the next lecture.

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Solve the following system of equations using the substitution method. y = 2x – 5 and 8x – 4y = 20 Since the first equation is already solved for y, substitute this value into the second equation. 8x – 4y = 20 8x – 4(2x – 5) = 20 (replace y with result from first equation) 8x – 8x + 20 = 20 (use distributive property) 20 = 20 (simplify left side) Example

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When you get a result, like the one on the previous slide, that is obviously true for any value of the replacements for the variables, this indicates that the two equations actually represent the same line (also called coincident lines). There are an infinite number of solutions for this system. Any solution of one equation would automatically be a solution of the other equation. Q: How could you confirm this answer (no solution/parallel lines) if this was a question on a test or quiz? A: Solve both equations for y and compare the slopes (they should be the same) and the y-intercepts (they should also be the same). Example (cont.)

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Solve the following system of equations using the substitution method 3x – y = 4 and 6x – 2y = 4 Solve the first equation for y. 3x – y = 4 -y = -3x + 4 (subtract 3x from both sides) y = 3x – 4 (multiply both sides by –1) Substitute this value for y into the second equation. 6x – 2y = 4 6x – 2(3x – 4) = 4 (replace y with the result from the first equation) 6x – 6x + 8 = 4 (use distributive property) 8 = 4 (simplify the left side) Example

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When you get a result, like the one on the previous slide, that is never true for any value of the replacements for the variables, this indicates that the two equations actually are parallel and never intersect. There is no solution to this system. Q: How could you confirm this answer (no solution/parallel lines) if this was a question on a test or quiz? A: Solve both equations for y and compare the slopes (they should be the same) and the y-intercepts (they should be different). Example (cont.)

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Problem from today’s homework:

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Reminder: This homework assignment on Section 4.1A is due at the start of next class period.

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You may now OPEN your LAPTOPS and begin working on the homework assignment.

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