Chapter 10

SYSTEM OF LINEAR EQUATIONS AND WORD PROBLEMS

10.1 Systems of Two Linear Equations

         Those section aims to:

1.      state and illustrate the three methods of solving two linear equations;

2.      sketch the graphs of consistent, inconsistent, and dependent equations; and

3.      interpret graphically the solutions of two simultaneous linear equations.

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Consider two linear equations in two unknowns x and y

                                a₁x + b₁y =c₁

                                a₂x + b₂y = c₂

This system of two linear equations in two unknowns is called the simultaneous linear equations in two unknowns. A pair of numbers for x and y which satisfies both equations is called simultaneous solution of the given equations.

Three Methods of Solving Systems of Two Linear Equations

1. Elimination Method

a.)    To eliminate one variable, the coefficients in that unknown must be numerically equal. If the signs of the equal coefficients are unlike, add the equations; if like, subtract them.

Example of unlike signs in one variable with equal numerical coefficients;

                      Add             x + 2y = 7

                       (+)            3x – 2y = 5

                                       4x          = 12  or  x = 3

Substituting x = 3 in any of the two given equations, we get 3 + 2y = 7, or y = 2. Thus the simultaneous solution is x = 3, y = 2 or (3, 2).

Example of like signs in one variable but equal numerical but equal numerical coefficients:

                  Subtract         4x  –  y    = 1

                       (y)             4x + 3y   =  13

                   _________________________

                                            - 4y      = -12 or y = 3

Substituting y = 3 in any of the two given equations, we get x = 1

b)      If the coefficients in one variable are not numerically equal, multiply the given equations by constants so that the resulting equations would have the same numerical coefficients in that variable. For example,

                                                 (1)

3x – 2y  =  8                        3x – 2y  = 8

                              (2)

5x – y    = 11                      10x – 2y = 22

Subtracting the last two equations, we get

                                 -7x  =  - 14   or   x = 2.

Substituting x  = 2 in any of the two given equations, we obtain y = -1.

2. Substitution Method

Select one equation, preferably one whose coefficients in one unknown is the least. Then solve for the unknown in terms of the other unknown and substitute this value in the other equations. For example, given the system

                        3x – 2y  =  8

                        5x  - y   =   11

Consider the second equation because y has the smaller coefficient. Then solving for y in terms of x and substituting in the other equations, we have

                                    y  =  5x – 11

                           3x – 2 (5x – 11)  =  8

                                x = 2, y = 11

            Example 10.1.1 Solve the system:

                                    2x² – 3y  = 7

                                      7x + y  =  13

            Solution.  It is more convenient to eliminate y, i.e. multiplying the second equation  by 3 gives

                                    2x – 3y  = 7

                        (+)      21x + 3y = 39

                    _______________________

                                   23x = 46  or  x = 2

Substituting x = 2 in the first equation gives 2(2) – 3y  = 7, or y = -1.

Summary of steps using the substitution method in solving two equations in two unknowns.

1.      Solve one equation for x in terms of y or for y in terms of x.

2.      Substitute this expression for that variable into the other equation.

3.      Solve the resulting equation in one variable.

4.      Substitute the solution from step 3 into either original equation to find the value of the other variable.

5.      Check the solution in both of the given equations.

3. Graphical Method

Geometrically, the graph of a pair of linear equations is intersecting lines, parallel lines, or coinciding lines.

a.       Consistent equations have a unique solution and the graph is a pair of intersecting lines. The point of intersection of the lines is the solution set of the given pair of equations. The set of equations given in the previous two methods (elimination and substitution) are consistent equations.

b.      Inconsistent equations have no solution and their graph is a pair of parallel lines. A pair of inconsistent equations is of the form

ax + by = c₁

ax + by = c₂

                where c₁ ≠ c_2. The following is a set of inconsistent equations:

                                                           2x – 3y = 5

                                                           2x – 3y = 7

c.       Dependent equations have infinitely many solutions and the graph is a pair of coinciding lines. Basically, any two given dependent equations are always reducible to a single equation. The following is a set of dependent equations:

x – 3y  = 5

2x – 6y  = 10

Fig. 10.1.1. Illustrates the graphs of the three types of systems of linear  equations in two unknowns.

Consistent Equations                              Inconsistent Equations                      Dependent Equations

       x + 2y  = 7                                                   2x – 3y  = 5                                       x – 3y  = 5

      4x + 3y = 8                                                   2x – 3y  = 7                                    2x – 6y  = 10

Example 10.1.2  Solve the following systems of linear equations by elimination method.

a.       4x – 5y = 14

7x + 3y  = 1                       Ans.  x = 1,  y = -2

                        b.  3x + 2y – 2  = 0                  Ans. x = -2, y = 4

                             4x – y + 12  = 1

c. 6  -  2  = 1

    x      y                                   Ans. x = 2, y = 1

    4  +  10 = 12

                            x        y

Example 10.1.3. Solve the systems of equations.

                        a.  2x – 5y  = 1                      b.  x / 2 -  y/3  =  -2

                            3x  +  4y  = 13                        x/4  +  y/6  = 3

Solution. a. To eliminate x, multiply the first equation by 3 and the second by 2 and subtract, i.e.

                                 6x  - 15y  = 3

                                 6x  + 8y  =  26

                         ________________

                            -23y  =  -23

                                  y = 1

Substituting y = 1 in any of the two given equations gives

                                6x – 15  = 3,  or x = 3.

Thus the solution is (3, 1).

b.      It is more convenient to get rid of the denominators.  Multiplying the first equation by 6 and the second by 12 gives

3x – 2y  =  -12

                         (+)      3x  +  2y  = 36

                     ______________________

                                          6x  =  24

                                            x  =  4

Substituting   = 4 in any of the two given equation gives

                          3(4) – 2y = -12 or y = 12

Example 10.1.4 Solve the system

a.    1  +  1  = 5  ;  4  + 3  = 3                                    b.  x + y   =  3   ;   x  +  y   =  18

       x      y      6     x     y                                                3     6               5       4        5

Solutions.

 .    1  +  1  = 5  ;          Multiply by 3          3  +  3     5           Subtracting              1  =  1

       x      y      6                                          x      y      2                                          x      2, or = 2

   4  +  3  = 3                                                     4  +  3  = 3                                    

   x       y                                                            x       y

    Substituting x = 2 in any of the 2 given equations gives   4  +  3  = 3, y = 3

2            y

b.  x   +  y  = 3            x ½                          x  +  y  = 3

     3       6                                                   6      12    3

    x    +  y  = 18           x ½                        x  +  y   =  6

   5         4      5                                          15    12      5

Subtracting the last set of equations, we have

          LCD  =  30  x   -   x   = 3  -  6   5x – 2x  = 45  - 36, or x = 3

            6       15    2      5

Substituting x = 3 in any of the 2 given equation gives y = 12.

Example 10.1.5  Solve graphically the systems/

a.  x  + y  = 5                   Ans. x = 2

    2x – y  = 1                           y = 3

b.  x + y  = -1                    Ans. x = ½

    3x – y = 3                             y = 3/2

Example 10.1.6 State whether the system is consistent, inconsistent, or dependent. State the solutions of consistent equations.

1a.   2x  = y – 5                        Ans.  dependent

        y/2  = x + 5/2

   b.   2x  = y – 5                      Ans.  inconsistent

        y/2  =  x  + 5/2

2a.  4x  - y  = 5                        Ans.  inconsistent

          4x  =  7 + y                   

   b.  4x – y  = 7                       Ans.  dependent

                x  = (7 + y) / 4

3a.  4x + y  = 7                        Ans. consistent x = 2, y = -1

        x  - 4y  = 6

b. 3x  + y  = 5                          Ans. consistent x + 1, y +2

   2x  + 5y  = 12