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Substitution and parallel lines

If computations eliminate all the variables and produce a contradiction, the two lines of a system are parallel, and the system is called inconsistent.

Sample set b

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

Step 1:  Solve equation 1 for y .
      2 x y = 1 y = 2 x + 1 y = 2 x 1

Step 2:  Substitute the expression 2 x 1 for y into equation 2.
      4 x 2 ( 2 x 1 ) = 4

Step 3:  Solve the equation obtained in step 2.
      4 x 2 ( 2 x 1 ) = 4 4 x 4 x + 2 = 4 2 4

Computations have eliminated all the variables and produce a contradiction. These lines are parallel.
A graph of two parallel lines. One line is labeled with the equation two x minus y is equal to one and passes through the points one, one, and zero, negative one. A second line is labeled with the equation four x minus two y is equal to four and passes through the points one, zero, and zero, negative two.
This system is inconsistent.

Practice set b

Slove the system { 7 x 3 y = 2 14 x 6 y = 1

Substitution produces 4 1 , or 1 2 2 , a contradiction. These lines are parallel and the system is inconsistent.

Substitution and coincident lines

The following rule alerts us to the fact that the two lines of a system are coincident.

Substitution and coincident lines

If computations eliminate all the variables and produce an identity, the two lines of a system are coincident and the system is called dependent.

Sample set c

Solve the system { 4 x + 8 y = 8 3 x + 6 y = 6 ( 1 ) ( 2 )

Step 1:  Divide equation 1 by 4 and solve for x .
      4 x + 8 y = 8 x + 2 y = 2 x = 2 y + 2

Step 2:  Substitute the expression 2 y + 2 for x in equation 2.
      3 ( 2 y + 2 ) + 6 y = 6

Step 3:  Solve the equation obtained in step 2.
      3 ( 2 y + 2 ) + 6 y = 6 6 y + 6 + 6 y = 6 6 = 6

Computations have eliminated all the variables and produced an identity. These lines are coincident.
A graph of two coincident lines. The line is labeled with the equation x plus two y is equal to two and a second label with the equation three x plus six y is equal to six. The lines pass through the points zero, one and two, zero. Since the lines are coincident, they have the same graph.
This system is dependent.

Practice set c

Solve the system { 4 x + 3 y = 1 8 x 6 y = 2

Computations produce 2 = 2 , an identity. These lines are coincident and the system is dependent.

Systems in which a coefficient of one of the variables is not 1 or cannot be made to be 1 without introducing fractions are not well suited for the substitution method. The problem in Sample Set D illustrates this “messy” situation.

Sample set d

Solve the system { 3 x + 2 y = 1 4 x 3 y = 3 ( 1 ) ( 2 )

Step 1:  We will solve equation ( 1 ) for y .
      3 x + 2 y = 1 2 y = 3 x + 1 y = 3 2 x + 1 2

Step 2:  Substitute the expression 3 2 x + 1 2 for y in equation ( 2 ) .
      4 x 3 ( 3 2 x + 1 2 ) = 3

Step 3:  Solve the equation obtained in step 2.
      4 x 3 ( 3 2 x + 1 2 ) = 3 Multiply both sides by the LCD ,  2 . 4 x + 9 2 x 3 2 = 3 8 x + 9 x 3 = 6 17 x 3 = 6 17 x = 9 x = 9 17

Step 4:  Substitute x = 9 17 into the equation obtained in step 1 , y = 3 2 x + 1 2 .
      y = 3 2 ( 9 17 ) + 1 2
      y = 27 34 + 17 34 = 10 34 = 5 17
     We now have x = 9 17 and y = 5 17 .

Step 5:  Substitution will show that these values of x and y check.

Step 6:  The solution is ( 9 17 , 5 17 ) .

Practice set d

Solve the system { 9 x 5 y = 4 2 x + 7 y = 9

These lines intersect at the point ( 1 , 1 ) .

Exercises

For the following problems, solve the systems by substitution.

{ 3 x + 2 y = 9 y = 3 x + 6

( 1 , 3 )

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

{ 2 x + 2 y = 0 x = 3 y 4

( 1 , 1 )

{ 3 x + 5 y = 9 x = 4 y 14

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

( 2 , 2 )

{ 4 x + y = 7 2 x + 5 y = 9

{ 6 x 6 = 18 x + 3 y = 3

( 4 , 1 3 )

{ x y = 5 2 x + y = 5

{ 5 x + y = 4 10 x 2 y = 8

Dependent (same line)

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

{ 4 x 2 y = 8 6 x + 3 y = 0

( 1 , 2 )

{ 2 x + 3 y = 12 2 x + 4 y = 18

{ 3 x 9 y = 6 6 x 18 y = 5

inconsistent (parallel lines)

{ x + 4 y = 8 3 x 12 y = 10

{ x + y = 6 x y = 4

( 1 , 5 )

{ 2 x + y = 0 x 3 y = 0

{ 4 x 2 y = 7 y = 4

( 15 4 , 4 )

{ x + 6 y = 11 x = 1

{ 2 x 4 y = 10 3 x = 5 y + 12

( 1 , 3 )

{ y + 7 x + 4 = 0 x = 7 y + 28

{ x + 4 y = 0 x + 2 3 y = 10 3

( 4 , 1 )

{ x = 24 5 y x 5 4 y = 3 2

{ x = 11 6 y 3 x + 18 y = 33

inconsistent (parallel lines)

{ 2 x + 1 3 y = 4 3 x + 6 y = 39

{ 4 5 x + 1 2 y = 3 10 1 3 x + 1 2 y = 1 6

( 1 , 1 )

{ x 1 3 y = 8 3 3 x + y = 1

Exercises for review

( [link] ) Find the quotient: x 2 x 12 x 2 2 x 15 ÷ x 2 3 x 10 x 2 2 x 8 .

( x 4 ) 2 ( x 5 ) 2

( [link] ) Find the difference: x + 2 x 2 + 5 x + 6 x + 1 x 2 + 4 x + 3 .

( [link] ) Simplify 81 x 8 y 5 z 4 .

9 x 4 y 2 z 2 y

( [link] ) Use the quadratic formula to solve 2 x 2 + 2 x 3 = 0.

( [link] ) Solve by graphing { x y = 1 2 x + y = 5
An xy coordinate plane with gridlines labeled negative five and five with increments of one unit for both axes.

( 2 , 1 )
A graph of two lines intersecting at a point with coordinates negative two, one. One of the lines is passing through a point with coordinates zero, five and the other line is passing through a point with coordinates zero, negative one.

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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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