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Section 15.3 Solving Systems of Linear Equations by Elimination
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15.3 solving systems of equations by elimination
1.
Slide - 1Copyright
© 2018, 2014, 2010 Pearson Education Inc.A L W A Y S L E A R N I N G 2 Systems of Linear Equations and Inequalities 15
2.
Slide - 2Copyright
© 2018, 2014, 2010 Pearson Education Inc.A L W A Y S L E A R N I N G 1. Solve linear systems by elimination. 2. Multiply when using the elimination method. 3. Use an alternative method to find the second value in a solution. 4. Solve special systems by elimination. Objectives 15.3 Solving Systems of Linear Equations by Elimination
3.
Slide - 3Copyright
© 2018, 2014, 2010 Pearson Education Inc.A L W A Y S L E A R N I N G Recall if A = B and A + C = B + C. This addition can be taken a step further. Adding equal quantities, rather than the same quantity, to both sides of an equation also results in equal sums. Solve Linear Systems by Elimination If A = B and C = D, then A + C = B + D. Using the addition property to solve systems is called the elimination method. When using this method, the idea is to eliminate one of the variables. To do this, one of the variables in the two equations must have coefficients that are opposites.
4.
Slide - 4Copyright
© 2018, 2014, 2010 Pearson Education Inc.A L W A Y S L E A R N I N G Example Use the elimination method to solve the system. 2x – y = 5 x + y = 7 Each equation in this system is a statement of equality, so the sum of the left sides equals the sum of the right sides. Solve Linear Systems by Elimination + 3x = 12 Notice that y has been eliminated. Now we can solve the result for x. 3 3 x = 4 To find the y-value of the solution, substitute 4 for x into either equation. x + y = 7 4 + y = 7 –4 –4 y = 3 The solution set of the system is {(4, 3)}.
5.
Slide - 5Copyright
© 2018, 2014, 2010 Pearson Education Inc.A L W A Y S L E A R N I N G Solve Linear Systems by Elimination CAUTION A system is not completely solved until values for both x and y are found. Do not stop after finding the value of only one variable. Remember to write the solution set as a set containing an ordered pair.
6.
Slide - 6Copyright
© 2018, 2014, 2010 Pearson Education Inc.A L W A Y S L E A R N I N G Solving a Linear System by Elimination Step 1 Write both equations in standard form Ax + By = C. Step 2 Transform so that the coefficients of one pair of variable terms are opposites. Multiply one or both equations by appropriate numbers so that the sum of the coefficients of either the x- or y-terms is 0. Step 3 Add the new equations to eliminate a variable. The sum should be an equation with just one variable. Step 4 Solve the equation from Step 3 for the remaining variable. Step 5 Find the other value. Substitute the result from Step 4 into either of the original equations and solve for the other variable. Step 6 Check the solution in both of the original equations. Then write the solution set containing an ordered pair. Solve Linear Systems by Elimination
7.
Slide - 7Copyright
© 2018, 2014, 2010 Pearson Education Inc.A L W A Y S L E A R N I N G Example Solve the system. Step 1 Both equations are already in standard form. Multiply When Using the Elimination Method 3x + y = 8 5x – 2y = 6 Step 2 The y-terms have opposite signs, but their coefficients are different. However, if we multiply both sides of the first equation by 2, then the y terms will cancel when we add. 2( ) 6x + 2y = 16 5x – 2y = 6 Step 3 Now when we add, the y terms are eliminated. + 11x = 22 Step 4 We are left with an equation with just x, which we can solve easily. 11 11 x = 2 ( )2 3x + y = 8 5x – 2y = 6
8.
Slide - 8Copyright
© 2018, 2014, 2010 Pearson Education Inc.A L W A Y S L E A R N I N G Example 3 (cont) Solve the system. Step 5 Substitute x = 2 into either of the equations. Multiply When Using the Elimination Method 3x + y = 8 5x – 2y = 6 The solution set is {(2, 2)}. 3(2) + y = 8 6 + y = 8 –6 –6 y = 2 Step 6 Check that (2,2) satisfies both equations. 3(2) + 2 = 8 ? 6 + 2 = 8 ? 8 = 8 5(2) – 2(2) = 6 ? 10 – 4 = 6 ? 6 = 6
9.
Slide - 9Copyright
© 2018, 2014, 2010 Pearson Education Inc.A L W A Y S L E A R N I N G x + 4y = 9 3x + 2y = 5 –3( ) Example Solve the system. We multiply the first equation by –3 to eliminate the x’s. x + 4y = 9 3x = –2y + 5 ( ) · –3 –3x – 12y = –27 3x + 2y = 5+ –10y = –22 10 10 11 5 y Use an Alternative Method to Find a Second Value in a Solution
10.
Slide - 10Copyright
© 2018, 2014, 2010 Pearson Education Inc.A L W A Y S L E A R N I N G Solve the system. Example (cont) Substituting –11/5 for y into one of the given equations would give x, but the arithmetic involved would be messy. Instead, solve for x by starting again with the original equations and eliminating y. Use an Alternative Method to Find a Second Value in a Solution 11 5 y x + 4y = 9 3x + 2y = 5–2( ) ( )·–2 x + 4y = 9 –6x – 4y = –10+ –5x = –1 5 5 1 5 x 1 11 The solution set is , . 5 5
11.
Slide - 11Copyright
© 2018, 2014, 2010 Pearson Education Inc.A L W A Y S L E A R N I N G Example Solve the system by the elimination method. Multiply each side of the first equation by 3; then add the two equations. Solve Special Systems by Elimination A true statement occurs when the equations are equivalent. As before, this indicates that every solution of one equation is also a solution of the other. The solution set is ⅔x – y = 1 –2x + 3y = –3 3( ) ( )3 2x – 3y = 3 –2x + 3y = –3+ 0 = 0 True. 2 3 3 , .x y x y 2 1 3 2 3 3 x y x y
12.
Slide - 12Copyright
© 2018, 2014, 2010 Pearson Education Inc.A L W A Y S L E A R N I N G x + 4y = 3 2x + 8y = 5 –2( ) Example (cont) Solve the system by the elimination method. ( )·–2 –2x – 8y = –6 2x + 8y = 5+ 0 = –1 The false statement 0 = –1 indicates that the solution set is False. 0 . Solve Special Systems by Elimination 4 3 2 8 5 x y x y
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