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Published byKristina Thomas Modified over 9 years ago

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**Please open your laptops, log in to the MyMathLab course web site, and open Daily Quiz 16.**

IMPORTANT NOTE: If you have time left out of your five minutes after you finish the five problems on this quiz, use it to check your answers before you submit the quiz! A scientific calculator may be used on this quiz. Remember to turn in your answer sheet to the TA when the quiz time is up.

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Please pass your completed graphing worksheets (HW 18) to the center aisle so the TA can collect them. The worksheets will be graded tonight and returned to you tomorrow. There will be a graphing question similar to one on the worksheet on Wednesday’s daily in-class quiz.

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**Section 4.1 Solving Systems of Equations in Two Variables by Graphing**

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: The lines intersect in a single point (so the answer is one ordered pair). The lines don’t intersect at all, i.e. they are parallel (so the answer is “no solution”.) 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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**Plug the values into the equations.**

Example 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 = true Second equation: 2(-3) + 10(1) = = true Since the point (-3, 1) produces a true statement in both equations, it is a solution.

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**Plug the values into the equations**

Example 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: 2•4 - 5•2 = 8 – 10 = true Second equation: 3•4 + 4•2 = = 20 false Since the point (4, 2) produces a true statement in only one equation, it is NOT a solution.

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**Problem from today’s homework:**

(try this one in your notebook) x = 3 and y = 5 works in the first equation, but not in the second one.

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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 for solving systems of equations that we will be studying in this chapter. The other two methods we will be using are: Substitution method (Section 4.2) Addition or elimination method (Section 4.3)

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Note: Graph Paper Click on the “Announcements” 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.)

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**Example: Solve the following system of equations by graphing: 2x – y = 6 and x + 3y = 10 **

First, graph 2x – y = 6. Solving for y gives y = 2x – 6. Plot the y-intercept of -6, then use the slope of 2 to go up two, over one. (4, 2) (1, 3) Second, graph x + 3y = 10. Solving for y gives y = -1/3x + 10/3. The y-intercept is a fraction, so let x = 1; then y = -1/3 + 10/3 = 9/ 3 = 3. Plot (1,3), then use the slope of -1/3 to go down one, over three. (1, -4), (0, -6) The lines APPEAR to intersect at (4, 2).

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**The point (4, 2) checks, so it is the solution of the system.**

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

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**Problem from today’s homework:**

(try this one in your notebook) Solution: Graph the 2 lines. They appear to intersect at (3,3) Now check x = 3, y = 3 back into BOTH equations to make sure they both give true statements.

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**Solve the following system of equations by graphing.**

Example Solve the following system of equations by graphing. -x + 3y = 6 and 3x – 9y = 9 x y (3,3) First, graph -x + 3y = 6. Solving for y gives y = 1/3x + 2. Plot the y-intercept of 2, then use the slope of 1/3 to go up one, over three. (0, 2) (3, 0) (0, -1) Next, graph 3x - 9y = 9. Solving for y gives y = 1/3x - 1. Plot the y-intercept of -1, then use the slope of 1/3 to go up one, over three. The lines APPEAR to be parallel.

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Example (cont.) Although the lines appear to be parallel, you still need to check that they have the same slope. You can do this by solving both equations for y, as we did on the previous slide. 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.**

Example x y Solve the following system of equations by graphing. x = 3y – 1 and 2x – 6y = -2 (5, 2) (-1, 0) (2, 1) (-4, -1) (7, -2) First, graph x = 3y – 1. Second, graph 2x – 6y = -2. The lines APPEAR to be identical.

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Example (cont.) 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 (add 1 to both sides) y = x (divide both sides by 3) Second equation, 2x – 6y = -2 -6y = -2x – (subtract 2x from both sides) y = x (divide both sides by -6) 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).

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**Watch out for graphing problems when the y-intercept is not an integer:**

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. 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.

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**CLOSE Please YOUR LAPTOPS, and get out your note-taking materials.**

and turn off and put away your cell phones, and get out your note-taking materials.

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**The assignment on this material (HW 19) **

REMINDER: The assignment on this material (HW 19) is due at the start of the next class session. Lab hours in 203: Mondays through Thursdays 8:00 a.m. to 7:30 p.m. Please remember to sign in on the Math 110 clipboard by the front door of the lab

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**and begin working on the homework assignment.**

You may now OPEN your LAPTOPS and begin working on the homework assignment. We expect all students to stay in the classroom to work on your homework till the end of the 55-minute class period. If you have already finished the homework assignment for today’s section, you should work ahead on the next one or work on the next practice quiz/test.

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