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Practice set a

Solve each system by addition.

{ x + y = 6 2 x y = 0

( 2 , 4 )

{ x + 6 y = 8 x 2 y = 0

( 4 , 2 )

Sample set b

Solve the following systems using the addition method.

Solve { 6 a 5 b = 14 ( 1 ) 2 a + 2 b = 10 ( 2 )

Step 1: The equations are already in the proper form, a x + b y = c .

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

       { 6 a 5 b = 14 3 ( 2 a + 2 b ) = 3 ( 10 ) { 6 a 5 b = 14 6 a 6 b = 30

Step 3:  Add the equations.

       6 a 5 b = 14 6 a 6 b = 30 0 11 b = 44

Step 4:  Solve the equation 11 b = 44.

       11 b = 44
        b = 4

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

       2 a + 2 b = 10 2 a + 2 ( 4 ) = 10 Solve for  a . 2 a 8 = 10 2 a = 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 ) 6 a 5 b = 14 ( 2 ) 2 a + 2 b = 10 6 ( 1 ) 5 ( 4 ) = 14 Is this correct? 2 ( 1 ) + 2 ( 4 ) = 10 Is this correct? 6 + 20 = 14 Is this correct? 2 8 = 10 Is this correct? 14 = 14 Yes, this is correct . 10 = 10 Yes, this is correct .

Step 7:  The solution is ( 1 , 4 ) .

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

Step 1:  Rewrite the system in the proper form.

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

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

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

Step 3:  Add the equations.

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

Step 4:  Solve the equation 23 x = 0

       23 x = 0

       x = 0

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

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

Practice set b

Solve each of the following systems using the addition method.

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

( 1 , 2 )

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

( 3 , 1 )

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

( 2 , 2 )

{ 5 x 3 y = 1 8 x 6 y = 4

( 1 , 2 )

{ 3 x 5 y = 9 4 x + 8 y = 12

( 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

Solve { 2 x y = 1 ( 1 ) 4 x 2 y = 4 ( 2 )

Step 1: The equations are in the proper form.

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

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

Step 3:  Add the equations.

       4 x + 2 y = 2 4 x 2 y = 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.

Solve  { 4 x + 8 y = 8 ( 1 ) 3 x + 6 y = 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 ( 4 x + 8 y ) = 3 ( 8 ) 4 ( 3 x + 6 y ) = 4 ( 6 ) { 12 x 24 y = 24 12 x + 24 y = 24

Step 3:  Add the equations.

       12 x 24 y = 24 12 x + 24 y = 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.

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

inconsistent

{ 4 x 28 y = 4 x 7 y = 1

dependent

Exercises

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

{ x + y = 11 x y = 1

( 5 , 6 )

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

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

( 2 , 2 )

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

{ 3 x + 4 y = 24 3 x 7 y = 42

( 0 , 6 )

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

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

( 2 , 2 )

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

{ x + y = 4 x y = 4

dependent

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

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

( 1 , 1 )

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

{ 3 x + y = 4 5 x 2 y = 14

( 2 , 2 )

{ 5 x 3 y = 20 x + 6 y = 4

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

( 4 , 3 )

{ x 11 y = 17 2 x 22 y = 4

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

( 1 , 6 )

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

{ 3 x 4 y = 2 9 x 12 y = 6

dependent

{ 3 x 5 y = 28 4 x 2 y = 20

{ 6 x 3 y = 3 10 x 7 y = 3

( 1 , 1 )

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

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

inconsistent

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

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

dependent

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

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

inconsistent

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

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

( 0 , 4 )

{ 8 x 3 y = 25 4 x 5 y = 5

{ 10 x 4 y = 72 9 x + 5 y = 39

( 258 7 , 519 7 )

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

{ 25 x 32 y = 14 50 x + 64 y = 28

dependent

Exercises for review

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

( [link] ) Simplify 8 + 3 50 .

17 2

( [link] ) Solve the radical equation 2 x + 3 + 5 = 8.

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

( 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.

( [link] ) Solve using the substitution method: { 3 x 4 y = 11 5 x + y = 3

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Source:  OpenStax, Algebra i for the community college. OpenStax CNX. Dec 19, 2014 Download for free at http://legacy.cnx.org/content/col11598/1.3
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