Systems of Linear Equations: Elimination by Addition
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1 OpenStax-CNX module: m Systems of Linear Equations: Elimination by Addition Wade Ellis Denny Burzynski This work is produced by OpenStax-CNX and licensed under the Creative Commons Attribution License 3.0 Abstract This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr. Beginning with the graphical solution of systems, this chapter includes an interpretation of independent, inconsistent, and dependent systems and examples to illustrate the applications for these systems. The substitution method and the addition method of solving a system by elimination are explained, noting when to use each method. The ve-step method is again used to illustrate the solutions of value and rate problems (coin and mixture problems), using drawings that correspond to the actual situation. Objectives of this module: know the properties used in the addition method, be able to use the addition method to solve a system of linear equations, know what to expect when using the addition method with a system that consists of parallel or coincident lines. 1 Overview The Properties Used in the Addition Method The Addition Method Addition and Parallel or Coincident Lines 2 The Properties Used in the Addition Method Another method of solving a system of two linear equations in two variables is called the method of elimination by addition. It is similar to the method of elimination by substitution in that the process eliminates one equation and one variable. The method of elimination by addition makes use of the following two properties. 1. If A, B, and C are algebraic expressions such that A = B C = D A + C = B + D and then Version 1.4: Jun 1, :40 pm
2 OpenStax-CNX module: m ax + ( ax) = 0 Property 1 states that if we add the left sides of two equations together and the right sides of the same two equations together, the resulting sums will be equal. We call this adding equations. Property 2 states that the sum of two opposites is zero. 3 The Addition Method To solve a system of two linear equations in two variables by addition, 1. Write, if necessary, both equations in general form, ax + by = c. 2. If necessary, multiply one or both equations by factors that will produce opposite coecients for one of the variables. 3. Add the equations to eliminate one equation and one variable. 4. Solve the equation obtained in step Do one of the following: (a) Substitute the value obtained in step 4 into either of the original equations and solve to obtain the value of the other variable, or (b) Repeat steps 1-5 for the other variable. 6. Check the solutions in both equations. 7. Write the solution as an ordered pair. The addition method works well when the coecient of one of the variables is 1 or a number other than 1. 4 Sample Set A Example 1 Solve x y = 2 (1) 3x + y = 14 (2) Step 1: Both equations appear in the proper form. Step 2: The coecients of y are already opposites, 1 and 1, so there is no need for a multiplication. Step 3: Add the equations. x y = 2 3x + y = 14 4x+0=16 Step 4: Solve the equation 4x = 16. 4x = 16 x = 4 The problem is not solved yet; we still need the value of y. Step 5: Substitute x = 4 into either of the original equations. We will use equation 1.
3 OpenStax-CNX module: m y = 2 Solve for y. y = 2 y = 2 We now have x = 4, y = 2. Step 6: Substitute x = 4 and y = 2 into both the original equations for a check. (1) x y = 2 (2) 3x + y = = 2 Is this correct? 3 (4) + 2 = 14 Is this correct? 2 = 2 Yes, this is correct = 14 Is this correct? 14 = 14 Yes, this is correct. Step 7: The solution is (4, 2). The two lines of this system intersect at (4, 2). 5 Practice Set A Solve each system by addition. Exercise 1 (Solution on p. 11.) x + y = 6 2x y = 0 Exercise 2 (Solution on p. 11.) x + 6y = 8 x 2y = 0 6 Sample Set B Solve the following systems using the addition method. Example 2 Solve 6a 5b = 14 (1) 2a + 2b = 10 (2) Step 1: The equations are already in the proper form, ax + by = c. Step 2: If we multiply equation (2) by 3, the coecients of a will be opposites and become 0 upon addition, thus eliminating a. 6a 5b = 14 3 (2a + 2b) = 3 (10) 6a 5b = 14 6a 6b = 30
4 OpenStax-CNX module: m Step 3: Add the equations. 6a 5b = 14 6a 6b = b=44 Step 4: Solve the equation 11b = b = 44 b = 4 Step 5: Substituteb = 4 into either of the original equations. We will use equation 2. 2a + 2b = 10 2a + 2 ( 4) = 10 Solve for a. 2a 8 = 10 2a = 2 a = 1 We now have a = 1 and b = 4. Step 6: Substitute a = 1 and b = 4 into both the original equations for a check. (1) 6a 5b = 14 (2) 2a + 2b = 10 6 ( 1) 5 ( 4) = 14 Is this correct? 2 ( 1) + 2 ( 4) = 10 Is this correct? = 14 Is this correct? 2 8 = 10 Is this correct? Step 7: The solution is ( 1, 4). 14 = 14 Yes, this is correct. 10 = 10 Yes, this is correct. Example 3 3x + 2y = 4 Solve 4x = 5y + 10 (1) (2) Step 1: Rewrite the system in the proper form. 3x + 2y = 4 4x 5y = 10 (1) (2) Step 2: Since the coecients of y already have opposite signs, we will eliminate y. Multiply equation (1) by 5, the coecient of y in equation 2. Multiply equation (2) by 2, the coecient of y in equation 1. 5 (3x + 2y) = 5 ( 4) 2 (4x 5y) = 2 (10) 15x + 10y = 20 8x 10y = 20 Step 3: Add the equations.
5 OpenStax-CNX module: m x + 10y = 20 8x 10y = 20 23x+0=0 Step 4: Solve the equation 23x = 0 23x = 0 x = 0 Step 5: Substitute x = 0 into either of the original equations. We will use equation 1. 3x + 2y = 4 3 (0) + 2y = 4 Solve for y y = 4 y = 2 We now have x = 0 and y = 2. Step 6: Substitution will show that these values check. Step 7: The solution is (0, 2). 7 Practice Set B Solve each of the following systems using the addition method. Exercise 3 (Solution on p. 11.) 3x + y = 1 5x + y = 3 Exercise 4 (Solution on p. 11.) x + 4y = 1 x 2y = 5 Exercise 5 (Solution on p. 11.) 2x + 3y = 10 x + 2y = 2 Exercise 6 (Solution on p. 11.) 5x 3y = 1 8x 6y = 4 Exercise 7 (Solution on p. 11.) 3x 5y = 9 4x + 8y = 12
6 OpenStax-CNX module: m Addition And Parallel Or Coincident Lines When the lines of a system are parallel or coincident, the method of elimination produces results identical to that of the method of elimination by substitution. Addition and Parallel Lines If computations eliminate all variables and produce a contradiction, the two lines of the system are parallel and the system is called inconsistent. Addition and Coincident Lines If computations eliminate all variables and produce an identity, the two lines of the system are coincident and the system is called dependent. 9 Sample Set C Example 4 Solve 2x y = 1 (1) 4x 2y = 4 (2) Step 1: The equations are in the proper form. Step 2: We can eliminate x by multiplying equation (1) by 2. 2 (2x y) = 2 (1) 4x 2y = 4 4x + 2y = 2 4x 2y = 4 Step 3: Add the equations. 4x + 2y = 2 4x 2y = = 2 0 = 2 This is false and is therefore a contradiction. The lines of this system are parallel. This system is inconsistent. Example 5 Solve 4x + 8y = 8 (1) 3x + 6y = 6 (2) Step 1: The equations are in the proper form. Step 2: We can eliminate x by multiplying equation (1) by 3 and equation (2) by 4. 3 (4x + 8y) = 3 (8) 4 (3x + 6y) = 4 (6) 12x 24y = 24 12x + 24y = 24 Step 3: Add the equations.
7 OpenStax-CNX module: m x 24y = 24 12x + 24y = = 0 0 = 0 This is true and is an identity. The lines of this system are coincident. This system is dependent. 10 Practice Set C Solve each of the following systems using the addition method. Exercise 8 (Solution on p. 11.) x + 2y = 6 6x + 12y = 1 Exercise 9 (Solution on p. 11.) 4x 28y = 4 x 7y = 1 11 Exercises For the following problems, solve the systems using elimination by addition. Exercise 10 (Solution on p. 11.) x + y = 11 x y = 1 Exercise 11 x + 3y = 13 x 3y = 11 Exercise 12 (Solution on p. 11.) 3x 5y = 4 4x + 5y = 2 Exercise 13 2x 7y = 1 5x + 7y = 22 Exercise 14 (Solution on p. 11.) 3x + 4y = 24 3x 7y = 42 Exercise 15 8x + 5y = 3 9x 5y = 71
8 OpenStax-CNX module: m Exercise 16 (Solution on p. 11.) x + 2y = 6 x + 3y = 4 Exercise 17 4x + y = 0 3x + y = 0 Exercise 18 (Solution on p. 11.) x + y = 4 x y = 4 Exercise 19 2x 3y = 6 2x + 3y = 6 Exercise 20 (Solution on p. 11.) 3x + 4y = 7 x + 5y = 6 Exercise 21 4x 2y = 2 7x + 4y = 26 Exercise 22 (Solution on p. 11.) 3x + y = 4 5x 2y = 14 Exercise 23 5x 3y = 20 x + 6y = 4 Exercise 24 (Solution on p. 11.) 6x + 2y = 18 x + 5y = 19 Exercise 25 x 11y = 17 2x 22y = 4 Exercise 26 (Solution on p. 11.) 2x + 3y = 20 3x + 2y = 15 Exercise 27 5x + 2y = 4 3x 5y = 10 Exercise 28 (Solution on p. 11.) 3x 4y = 2 9x 12y = 6
9 OpenStax-CNX module: m Exercise 29 3x 5y = 28 4x 2y = 20 Exercise 30 (Solution on p. 11.) 6x 3y = 3 10x 7y = 3 Exercise 31 4x + 12y = 0 8x + 16y = 0 Exercise 32 (Solution on p. 11.) 3x + y = 1 12x + 4y = 6 Exercise 33 8x + 5y = 23 3x 3y = 12 Exercise 34 (Solution on p. 11.) 2x + 8y = 10 3x + 12y = 15 Exercise 35 4x + 6y = 8 6x + 8y = 12 Exercise 36 (Solution on p. 11.) 10x + 2y = 2 15x 3y = 3 Exercise 37 x y = x + y = 7 5 Exercise 38 (Solution on p. 11.) x y = 4 3 x y = 2 3 Exercise 39 8x 3y = 25 4x 5y = 5 Exercise 40 (Solution on p. 11.) 10x 4y = 72 9x + 5y = 39 Exercise 41 12x + 16y = 36 10x + 12y = 30
10 OpenStax-CNX module: m Exercise 42 (Solution on p. 12.) 25x 32y = 14 50x + 64y = Exercises For Review Exercise 43 ( here 1 ) Simplify and write ( 2x 3 y 4) 5( 2xy 6 ) 5 so that only positive exponents appear. Exercise 44 (Solution on p. 12.) ( here 2 ) Simplify Exercise 45 ( here 3 ) Solve the radical equation 2x = 8. Exercise 46 (Solution on p. 12.) ( here 4 ) Solve by graphing x + y = 4 3x y = 0 Exercise 47 ( here 5 ) Solve using the substitution method: 3x 4y = 11 5x + y = 3 1 "Basic Operations with Real Numbers: Multiplication and Division of Signed Numbers" < 2 "Roots, Radicals, and Square Root Equations: Addition and Subtraction of Square Root Expressions" < 3 "Roots, Radicals, and Square Root Equations: Square Root Equations with Applications" < 4 "Systems of Linear Equations: Solutions by Graphing" < 5 "Systems of Linear Equations: Elimination by Substitution" <
11 OpenStax-CNX module: m Solutions to Exercises in this Module Solution to Exercise (p. 3) (2, 4) Solution to Exercise (p. 3) ( 4, 2) Solution to Exercise (p. 5) (1, 2) Solution to Exercise (p. 5) ( 3, 1) Solution to Exercise (p. 5) ( 2, 2) Solution to Exercise (p. 5) ( 1, 2) Solution to Exercise (p. 5) (3, 0) Solution to Exercise (p. 7) inconsistent Solution to Exercise (p. 7) dependent Solution to Exercise (p. 7) (5, 6) Solution to Exercise (p. 7) (2, 2) Solution to Exercise (p. 7) (0, 6) Solution to Exercise (p. 7) (2, 2) Solution to Exercise (p. 8) dependent Solution to Exercise (p. 8) (1, 1) Solution to Exercise (p. 8) ( 2, 2) Solution to Exercise (p. 8) ( 4, 3) Solution to Exercise (p. 8) ( 1, 6) Solution to Exercise (p. 8) dependent Solution to Exercise (p. 9) (1, 1) Solution to Exercise (p. 9) inconsistent Solution to Exercise (p. 9) dependent Solution to Exercise (p. 9) inconsistent Solution to Exercise (p. 9) (0, 4)
12 OpenStax-CNX module: m Solution ( to ) Exercise (p. 9) 258 7, Solution to Exercise (p. 9) dependent Solution to Exercise (p. 10) 17 2 Solution to Exercise (p. 10) (1, 3)
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