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Elimination by Addition

Module by: Wade Ellis, Denny Burzynski. E-mail the authors

Summary: 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 five-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.

Overview

  • The Properties Used in the Addition Method
  • The Addition Method
  • Addition and Parallel or Coincident Lines

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 A , B B , and C C are algebraic expressions such that

    A = B C = D A+C = B+D and then A = B C = D A+C = B+D and then
  2. ax+( ax )=0 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.

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. ax+by=c.
  2. If necessary, multiply one or both equations by factors that will produce opposite coefficients for one of the variables.
  3. Add the equations to eliminate one equation and one variable.
  4. Solve the equation obtained in step 3.
  5. 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 coefficient of one of the variables is 1 or a number other than 1.

Sample Set A

Example 1

Solve  { xy=2 ( 1 ) 3x+y=14 ( 2 ) { xy=2 ( 1 ) 3x+y=14 ( 2 )

Step 1:  Both equations appear in the proper form.

Step 2:  The coefficients of y y are already opposites, 1 and 1, 1, so there is no need for a multiplication.

Step 3:  Add the equations.

      xy=2 3x+y=14 4x+0=16 xy=2 3x+y=14 4x+0=16

Step 4:  Solve the equation 4x=16. 4x=16.

      4x=16 4x=16

      x=4 x=4

 The problem is not solved yet; we still need the value of y y .

Step 5:  Substitute x=4 x=4 into either of the original equations. We will use equation 1.

      4y = 2 Solve for y. y = 2 y = 2 4y = 2 Solve for y. y = 2 y = 2

 We now have x=4,y=2. x=4,y=2.

Step 6:  Substitute x=4 x=4 and y=2 y=2 into both the original equations for a check.

       (1) xy = 2 (2) 3x+y = 14 42 = 2 Is this correct? 3(4)+2 = 14 Is this correct? 2 = 2 Yes, this is correct. 12+2 = 14 Is this correct? 14 = 14 Yes, this is correct. (1) xy = 2 (2) 3x+y = 14 42 = 2 Is this correct? 3(4)+2 = 14 Is this correct? 2 = 2 Yes, this is correct. 12+2 = 14 Is this correct? 14 = 14 Yes, this is correct.

Step 7:  The solution is ( 4,2 ). ( 4,2 ).

The two lines of this system intersect at ( 4,2 ). ( 4,2 ).

Practice Set A

Solve each system by addition.

Exercise 1

{ x+y=6 2xy=0 { x+y=6 2xy=0

Solution

( 2,4 ) ( 2,4 )

Exercise 2

{ x+6y=8 x2y=0 { x+6y=8 x2y=0

Solution

( 4,2 ) ( 4,2 )

Sample Set B

Solve the following systems using the addition method.

Example 2

Solve { 6a5b=14 ( 1 ) 2a+2b=10 ( 2 ) { 6a5b=14 ( 1 ) 2a+2b=10 ( 2 )

Step 1: The equations are already in the proper form, ax+by=c. ax+by=c.

Step 2: If we multiply equation (2) by —3, the coefficients of a a will be opposites and become 0 upon addition, thus eliminating a a .

       { 6a5b=14 3( 2a+2b )=3( 10 ) { 6a5b=14 6a6b=30 { 6a5b=14 3( 2a+2b )=3( 10 ) { 6a5b=14 6a6b=30

Step 3:  Add the equations.

       6a5b=14 6a6b=30 011b=44 6a5b=14 6a6b=30 011b=44

Step 4:  Solve the equation 11b=44. 11b=44.

       11b=44 11b=44
        b=4 b=4

Step 5:  Substitute b=4 b=4 into either of the original equations. We will use equation 2.

       2a+2b = 10 2a+2( 4 ) = 10 Solve for a. 2a8 = 10 2a = 2 a = 1 2a+2b = 10 2a+2( 4 ) = 10 Solve for a. 2a8 = 10 2a = 2 a = 1

 We now have a=1 a=1 and b=4. b=4.

Step 6:  Substitute a=1 a=1 and b=4 b=4 into both the original equations for a check.

       (1) 6a5b = 14 (2) 2a+2b = 10 6(1)5(4) = 14 Is this correct? 2(1)+2(4) = 10 Is this correct? 6+20 = 14 Is this correct? 28 = 10 Is this correct? 14 = 14 Yes, this is correct. 10 = 10 Yes, this is correct. (1) 6a5b = 14 (2) 2a+2b = 10 6(1)5(4) = 14 Is this correct? 2(1)+2(4) = 10 Is this correct? 6+20 = 14 Is this correct? 28 = 10 Is this correct? 14 = 14 Yes, this is correct. 10 = 10 Yes, this is correct.

Step 7:  The solution is ( 1,4 ). ( 1,4 ).

Example 3

Solve  { 3x+2y=4 4x=5y+10 (1) (2) { 3x+2y=4 4x=5y+10 (1) (2)

Step 1:  Rewrite the system in the proper form.

       { 3x+2y=4 4x5y=10 ( 1 ) ( 2 ) { 3x+2y=4 4x5y=10 ( 1 ) ( 2 )

Step 2:  Since the coefficients of y y already have opposite signs, we will eliminate y y .
     Multiply equation (1) by 5, the coefficient of y y in equation 2.
     Multiply equation (2) by 2, the coefficient of y y in equation 1.

       { 5( 3x+2y )=5( 4 ) 2( 4x5y )=2( 10 ) { 15x+10y=20 8x10y=20 { 5( 3x+2y )=5( 4 ) 2( 4x5y )=2( 10 ) { 15x+10y=20 8x10y=20

Step 3:  Add the equations.

       15x+10y=20 8x10y=20 23x+0=0 15x+10y=20 8x10y=20 23x+0=0

Step 4:  Solve the equation 23x=0 23x=0

       23x=0 23x=0

       x=0 x=0

Step 5:  Substitute x=0 x=0 into either of the original equations. We will use equation 1.

       3x+2y = 4 3( 0 )+2y = 4 Solve for y. 0+2y = 4 y = 2 3x+2y = 4 3( 0 )+2y = 4 Solve for y. 0+2y = 4 y = 2

 We now have x=0 x=0 and y=2. y=2.

Step 6:  Substitution will show that these values check.

Step 7:  The solution is ( 0,2 ). ( 0,2 ).

Practice Set B

Solve each of the following systems using the addition method.

Exercise 3

{ 3x+y=1 5x+y=3 { 3x+y=1 5x+y=3

Solution

( 1,2 ) ( 1,2 )

Exercise 4

{ x+4y=1 x2y=5 { x+4y=1 x2y=5

Solution

( 3,1 ) ( 3,1 )

Exercise 5

{ 2x+3y=10 x+2y=2 { 2x+3y=10 x+2y=2

Solution

( 2,2 ) ( 2,2 )

Exercise 6

{ 5x3y=1 8x6y=4 { 5x3y=1 8x6y=4

Solution

( 1,2 ) ( 1,2 )

Exercise 7

{ 3x5y=9 4x+8y=12 { 3x5y=9 4x+8y=12

Solution

( 3,0 ) ( 3,0 )

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.

Sample Set C

Example 4

Solve { 2xy=1 ( 1 ) 4x2y=4 ( 2 ) { 2xy=1 ( 1 ) 4x2y=4 ( 2 )

Step 1: The equations are in the proper form.

Step 2: We can eliminate x x by multiplying equation (1) by –2.

       { 2( 2xy )=2( 1 ) 4x2y=4 { 4x+2y=2 4x2y=4 { 2( 2xy )=2( 1 ) 4x2y=4 { 4x+2y=2 4x2y=4

Step 3:  Add the equations.

       4x+2y=2 4x2y=4 0+0=2 0=2 4x+2y=2 4x2y=4 0+0=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 ) { 4x+8y=8 ( 1 ) 3x+6y=6 ( 2 )

Step 1:  The equations are in the proper form.

Step 2:  We can eliminate x x by multiplying equation (1) by –3 and equation (2) by 4.

       { 3( 4x+8y )=3( 8 ) 4( 3x+6y )=4( 6 ) { 12x24y=24 12x+24y=24 { 3( 4x+8y )=3( 8 ) 4( 3x+6y )=4( 6 ) { 12x24y=24 12x+24y=24

Step 3:  Add the equations.

       12x24y=24 12x+24y=24 0+0=0 0=0 12x24y=24 12x+24y=24 0+0=0 0=0

 This is true and is an identity. The lines of this system are coincident.

 This system is dependent.

Practice Set C

Solve each of the following systems using the addition method.

Exercise 8

{ x+2y=6 6x+12y=1 { x+2y=6 6x+12y=1

Solution

inconsistent

Exercise 9

{ 4x28y=4 x7y=1 { 4x28y=4 x7y=1

Solution

dependent

Exercises

For the following problems, solve the systems using elimination by addition.

Exercise 10

{ x+y=11 xy=1 { x+y=11 xy=1

Solution

( 5,6 ) ( 5,6 )

Exercise 11

{ x+3y=13 x3y=11 { x+3y=13 x3y=11

Exercise 12

{ 3x5y=4 4x+5y=2 { 3x5y=4 4x+5y=2

Solution

( 2,2 ) ( 2,2 )

Exercise 13

{ 2x7y=1 5x+7y=22 { 2x7y=1 5x+7y=22

Exercise 14

{ 3x+4y=24 3x7y=42 { 3x+4y=24 3x7y=42

Solution

( 0,6 ) ( 0,6 )

Exercise 15

{ 8x+5y=3 9x5y=71 { 8x+5y=3 9x5y=71

Exercise 16

{ x+2y=6 x+3y=4 { x+2y=6 x+3y=4

Solution

( 2,2 ) ( 2,2 )

Exercise 17

{ 4x+y=0 3x+y=0 { 4x+y=0 3x+y=0

Exercise 18

{ x+y=4 xy=4 { x+y=4 xy=4

Solution

dependent

Exercise 19

{ 2x3y=6 2x+3y=6 { 2x3y=6 2x+3y=6

Exercise 20

{ 3x+4y=7 x+5y=6 { 3x+4y=7 x+5y=6

Solution

( 1,1 ) ( 1,1 )

Exercise 21

{ 4x2y=2 7x+4y=26 { 4x2y=2 7x+4y=26

Exercise 22

{ 3x+y=4 5x2y=14 { 3x+y=4 5x2y=14

Solution

( 2,2 ) ( 2,2 )

Exercise 23

{ 5x3y=20 x+6y=4 { 5x3y=20 x+6y=4

Exercise 24

{ 6x+2y=18 x+5y=19 { 6x+2y=18 x+5y=19

Solution

( 4,3 ) ( 4,3 )

Exercise 25

{ x11y=17 2x22y=4 { x11y=17 2x22y=4

Exercise 26

{ 2x+3y=20 3x+2y=15 { 2x+3y=20 3x+2y=15

Solution

( 1,6 ) ( 1,6 )

Exercise 27

{ 5x+2y=4 3x5y=10 { 5x+2y=4 3x5y=10

Exercise 28

{ 3x4y=2 9x12y=6 { 3x4y=2 9x12y=6

Solution

dependent

Exercise 29

{ 3x5y=28 4x2y=20 { 3x5y=28 4x2y=20

Exercise 30

{ 6x3y=3 10x7y=3 { 6x3y=3 10x7y=3

Solution

( 1,1 ) ( 1,1 )

Exercise 31

{ 4x+12y=0 8x+16y=0 { 4x+12y=0 8x+16y=0

Exercise 32

{ 3x+y=1 12x+4y=6 { 3x+y=1 12x+4y=6

Solution

inconsistent

Exercise 33

{ 8x+5y=23 3x3y=12 { 8x+5y=23 3x3y=12

Exercise 34

{ 2x+8y=10 3x+12y=15 { 2x+8y=10 3x+12y=15

Solution

dependent

Exercise 35

{ 4x+6y=8 6x+8y=12 { 4x+6y=8 6x+8y=12

Exercise 36

{ 10x+2y=2 15x3y=3 { 10x+2y=2 15x3y=3

Solution

inconsistent

Exercise 37

{ x+ 3 4 y= 1 2 3 5 x+y= 7 5 { x+ 3 4 y= 1 2 3 5 x+y= 7 5

Exercise 38

{ x+ 1 3 y= 4 3 x+ 1 6 y= 2 3 { x+ 1 3 y= 4 3 x+ 1 6 y= 2 3

Solution

( 0,4 ) ( 0,4 )

Exercise 39

{ 8x3y=25 4x5y=5 { 8x3y=25 4x5y=5

Exercise 40

{ 10x4y=72 9x+5y=39 { 10x4y=72 9x+5y=39

Solution

( 258 7 , 519 7 ) ( 258 7 , 519 7 )

Exercise 41

{ 12x+16y=36 10x+12y=30 { 12x+16y=36 10x+12y=30

Exercise 42

{ 25x32y=14 50x+64y=28 { 25x32y=14 50x+64y=28

Solution

dependent

Exercises For Review

Exercise 43

((Reference)) Simplify and write ( 2 x 3 y 4 ) 5 ( 2x y 6 ) 5 ( 2 x 3 y 4 ) 5 ( 2x y 6 ) 5 so that only positive exponents appear.

Exercise 44

((Reference)) Simplify 8 +3 50 . 8 +3 50 .

Solution

17 2 17 2

Exercise 45

((Reference)) Solve the radical equation 2x+3 +5=8. 2x+3 +5=8.

Exercise 46

((Reference)) Solve by graphing { x+y=4 3xy=0 { x+y=4 3xy=0
An xy coordinate plane with gridlines labeled negative five and five with increments of one unit for both axes.

Solution

( 1,3 ) ( 1,3 )
A graph of two lines intersecting at a point with coordinates negative one, three. One of the lines is passing through a point with coordinates zero, zero and the other line is passing through two points with coordinates zero, four and four, zero.

Exercise 47

((Reference)) Solve using the substitution method: { 3x4y=11 5x+y=3 { 3x4y=11 5x+y=3

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