Test 1 Results Average class score after partial credit Commonly missed questions Grade Scale 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 Any questions on the 4 1A homework problems Now please CLOSE YOUR LAPTOPS and turn off and put away your cell phones Sample Problems Page Link Dr Bruce Johnston Section 4 1 B Solving Systems of Equations by Elimination In addition to the graphing and substitution methods you learned in the last section a third method that can be used to solve systems of equations is called the addition or elimination method This method is probably the one you will use most often so pay special attention to these next few slides To use this method you multiply one or both equations by numbers that will allow you to add the two equations together and eliminate one of the variables From that point on the rest of the solution follows the same steps as the substitution method solving for the one remaining variable then plugging that back in to the original equations to get the value of the other variable Solving a system of linear equations by the addition or elimination method 1 Rewrite each equation in standard form eliminating fraction coefficients 2 If necessary multiply one or both equations by a number so that the coefficients of a chosen variable are opposites 3 Add the equations 4 Find the value of the remaining variable by solving the equation from step 3 5 Find the value of the second variable by substituting the value found in step 4 into either original equation 6 Check the proposed solution in the original equations ALWAYS do this since it s very easy to make one of those annoying arithmetic mistakes in these kinds of problems Solve the following system of equations using the elimination method 6x 3y 3 and 4x 5y 9 Multiply both sides of the first equation by 5 and the second equation by 3 First equation 5 6x 3y 5 3 30x 15y 15 use the distributive property Second equation 3 4x 5y 3 9 12x 15y 27 use the distributive property Combine the two resulting equations eliminating the variable y 30x 15y 15 12x 15y 27 42x 42 x 1 divide both sides by 42 Substitute the value for x into one of the original equations 6x 3y 3 6 1 3y 3 replace the x value in the first equation 6 3y 3 simplify the left side 3y 3 6 3 add 6 to both sides and simplify y 1 divide both sides by 3 Our computations have produced the point 1 1 Check the point in the original equations First equation 6x 3y 3 6 1 3 1 3 Second equation 4x 5y 9 4 1 5 1 9 true true The solution of the system is 1 1 Problem from today s homework Problem from yesterday s homework Note This problem can also be solved by the elimination method What would be the first step in solving this problem with the substitution method What would be your first step in using the elimination method on this problem Which method is easier to use on this problem Solve the following system of equations using the elimination method 2 1 3 x y 3 4 2 1 1 x y 2 2 4 First multiply both sides of the equations by a number that will clear the fractions out of the equations Multiply both sides of each equation by 12 Note you don t have to multiply each equation by the same number but in this case it will be convenient to do so First equation 2 1 3 x y 3 4 2 1 2 12 x 4 3 3 y 12 2 8 x 3 y 18 multiply both sides by 12 simplify both sides Second equation 1 1 x y 2 2 4 1 1 12 x y 12 2 4 2 6 x 3 y 24 multiply both sides by 12 simplify both sides Combine the two equations 8x 3y 18 6x 3y 24 14x 42 x 3 divide both sides by 14 Substitute the value for x into one of the original equations For ease I ll use one that has been cleared of fractions 8x 3y 18 8 3 3y 18 24 3y 18 3y 18 24 6 y 2 Our computations have produced the point 3 2 Check the point in the original equations Note Here you should use the original equations before any modifications even though they involve fractions to detect any computational errors that you might have made First equation Second equation 2 1 3 x y 3 4 2 1 1 x y 2 2 4 2 1 3 3 2 3 4 2 1 1 3 2 2 2 4 1 3 2 2 2 3 1 2 2 2 true The solution is the point 3 2 true Problem from today s homework Problem from yesterday s homework Note This problem can also be solved by the elimination method What would be the first step in solving this problem with the substitution method What would be your first step in using the elimination method on this problem Which method is easier to use on this problem There are three types of answers you will encounter in the Section 4 1 homework problems corresponding to the three different ways two lines can intersect 1 Intersection in a single point Answer is an ordered pair The two lines have different slopes 2 No common intersection parallel lines Answer N for No Solution The lines have the same slope different y intercepts all variables drop out by elimination leaving a false statement such as 0 3 3 The two equations represent the same line so the Intersection is all the points on the line Answer I for Infinitely many solutions Lines have same slope AND y intercept all variables drop out by elimination leaving a true statement such as 0 0 When using the elimination addition method to combine two equations be alert for results like 5 5 which is always true thus indicating that there are infinitely many solutions since the two equations represent the same line or 0 6 which is never true thus indicating that there are no solutions since the two equations represent parallel lines Problem from today s homework Problem from today s homework What if you end up with y 0 when solving a system Example 2x 5y 5 2x 2y 5 3y 0 y 0 3 0 Now plug y 0 back into one of the two original equations 2x 5 0 5 2x 0 5 2x 5 x 5 2 So the solution is 5 2 0 a single ordered pair Now check that ordered pair in the second equation 2x 2y 5 2 5 2 2 0 5 0 5 checks Recap of methods for solving a system of linear equations 1 Graphing method Not usually the method of choice for getting an …