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11.2 Elimination by substitution

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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 when the substitution method works best, be able to use the substitution method to solve a system of linear equations, know what to expect when using substitution with a system that consists of parallel lines.

Overview

  • When Substitution Works Best
  • The Substitution Method
  • Substitution and Parallel Lines
  • Substitution and Coincident Lines

When substitution works best

We know how to solve a linear equation in one variable. We shall now study a method for solving a system of two linear equations in two variables by transforming the two equations in two variables into one equation in one variable.

To make this transformation, we need to eliminate one equation and one variable. We can make this elimination by substitution .

When substitution works best

The substitution method works best when either of these conditions exists:
  1. One of the variables has a coefficient of 1 ,   or
  2. One of the variables can be made to have a coefficient of 1 without introducing fractions.

The substitution method

The substitution method

To solve a system of two linear equations in two variables,
  1. Solve one of the equations for one of the variables.
  2. Substitute the expression for the variable chosen in step 1 into the other equation.
  3. Solve the resulting equation in one variable.
  4. Substitute the value obtained in step 3 into the equation obtained in step 1 and solve to obtain the value of the other variable.
  5. Check the solution in both equations.
  6. Write the solution as an ordered pair.

Sample set a

Solve the system { 2 x + 3 y = 14 3 x + y = 7 ( 1 ) ( 2 )

Step 1:  Since the coefficient of y in equation 2 is 1, we will solve equation 2 for y .

       y = − 3 x + 7

Step 2:  Substitute the expression − 3 x + 7 for y in equation 1.

       2 x + 3 ( − 3 x + 7 ) = 14

Step 3:  Solve the equation obtained in step 2.
      2 x + 3 ( − 3 x + 7 ) = 14 2 x − 9 x + 21 = 14 − 7 x + 21 = 14 − 7 x = − 7 x = 1
Step 4:  Substitute x = 1 into the equation obtained in step 1 ,   y = − 3 x + 7.
      y = − 3 ( 1 ) + 7 y = − 3 + 7 y = 4
 We now have x = 1 and y = 4.

Step 5:  Substitute x = 1 , y = 4 into each of the original equations for a check.
( 1 ) 2 x + 3 y = 14 ( 2 ) 3 x + y = 7 2 ( 1 ) + 3 ( 4 ) = 14 Is this correct? 3 ( 1 ) + ( 4 ) = 7 Is this correct? 2 + 12 = 14 Is this correct? 3 + 4 = 7 Is this correct? 14 = 14 Yes, this is correct . 7 = 7 Yes, this is correct .

Step 6:  The solution is ( 1 , 4 ) . The point ( 1 , 4 ) is the point of intersection of the two lines of the system.

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Practice set a

Slove the system { 5 x − 8 y = 18 4 x +     y =       7

The point ( 2 , − 1 ) is the point of intersection of the two lines.

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Substitution and parallel lines

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

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Source:  OpenStax, Elementary algebra. OpenStax CNX. May 08, 2009 Download for free at http://cnx.org/content/col10614/1.3
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