8.1 Solving Systems of Linear Equations by Graphing

(covered in MTH 098)

A system of linear equations, also called a linear system, is two or more linear equations with the same variables. Solve the system means find values of the variables which make all of the equations true. One methods of solving a linear system is by graphing, where each of the equations is graphed on the same set of axes. The point of intersection of the graphs is the solution to the system (it is the point where both equations are true.)

Problem: Solve the system of equations by graphing: 3x + 5y = 6 and 2x – 2y = 4.

Solution:

(0, 2) is the solution to the system.

 

One drawback to solving systems by graphing is that if the solution to the system has fractional coefficients, it can be difficult to read the point accurately from the graph.

Problem: Solve the system of equations by graphing: 3x + 5y = 8 and 2x – 2y = 5.

Solution:

  is the solution to the system.

8.2 Solving Systems of Linear Equations by the Substitution Method

Objective 1: The substitution method of solving linear equations is useful when one of the equations is already (or can quickly be) solved for one of the variables. The expression for that variable is then substituted into the other equation. This equation should now have only one unknown and can be solved using the methods for solving equations having one unknown (chapter 2).

 

Problem: Solve the system by substitution: 2x + 7y = -12 and x = -2y

Solution: Since the second equation is already solved for x, take the equivalent of x (-2y)

and substitute it into the first equation in the place of x.

2(-2y) + 7y = -12

-4y + 7y = -12

3y = -12

y = -4 x = -2y x = -2(-4) x = 8

Since we are looking for the values of x and y that satisfy both equations, be sure to find both. The solution may be written either as an ordered pair, (8, -4) or as x = 8, y = -4.

***NOTE: It is standard practice to write the variables alphabetically when writing the solution to a system as an ordered pair.

 

Problem: Solve the system by substitution: 2x + 7y = -12 and x = 3 -2y

Solution: Since the second equation is already solved for x, take the equivalent of x (3 -2y)

and substitute it into the first equation in the place of x.

2(3 – 2y) + 7y = -12

6 – 4y + 7y = -12

6 + 3y = -12

3y = -18

y = -6 x = 3 – 2(-6) x = 3 + 12 x = 15

The solution to the system is (15, -6).

 

Problem: Solve the system by substitution: x + 4y = -1 and 2x – 5y = 11

Solution: Solve either equation for either variable. An accepted practice is to solve for the variable which has the smallest coefficient, so in this case, solve the first equation for x.

x + 4y = -1 x = -4y – 1 Substitute this expression for x (-4y – 1) into the second equation.

2(-4y – 1) – 5y = 11

-8y – 2 – 5y = 11

-13y – 2 = 11

-13y = 13

y = - 1 x = -4(- 1) – 1 = 4 - 1 x = 3

The solution to the system is (3, -1).

Objective 2: From previous work, we know that sometimes systems of equations have no solution or that they have an infinite number of solutions.

 

Problem: Solve the system by substitution: y = 8x + 4 and 16x – 2y = 8

Solution: Since the first equation is already solved for y, substitute 8x + 4 into the second equation in place of y.

                                        16x – 2(8x + 4) = 8

                                        16x – 16x – 8 = 8

-8 = 8 This statement is false. This means that there is no solution to the system – there is no (x, y) which satisfies both equations. Geometrically, the lines are parallel. The system is inconsistent.

 

Problem: Solve the system by substitution: -x – 3y = 7 and 4x + 12y = -28

Solution: From the first equation: x = -3y – 7

Substitute into the second equation: 4(-3y – 7) + 12y = -28

-12y – 28 + 12y = -28

-28 = -28 TRUE

Since the statement –28 = -28 is TRUE, the system is said to have an infinite number of solutions. Any (x, y) which satisfies one of the equations will also satisfy the other equation. Geometrically, the equations represent the same line.

Objective 3: When working with systems of equations involving fractions or decimals, consider eliminating the fractions or decimals as the first step.

 

Problem: Solve the system by substitution:

Solution: Multiply the first equation by 6: 3x + 2y = -2

Multiply the second equation by 2: x + 4y = -14

The system now looks like this: 3x + 2y = -2 and x + 4y = -14.

Solve the second equation for x: x = -4y – 14

Substitute this expression for x into the first equation: 3(-4y – 14) + 2y = -2

-12y – 42 + 2y = -2

-10y – 42 = -2

-10y = 40

y = -4 x = 2

(2, -4) is the solution to the system.

ASSIGNMENT PAGE 537-8 problems 3 – 19 (odd), 25, 29, 31

 

8.3 Solving Systems of Linear Equations by Elimination

A drawback to solving linear systems by substitution is that, for some systems, the algebra involved can get very messy.

Example: Solve the system by substitution: 5x + 7y = -9 and –7x + 16y = 20

Solve the first equation for x:

Substitute into the second equation:

Solve this equation for y:

    • A third method to solve linear systems is the elimination method , sometimes called the addition method. To solve a system by elimination, add the equations or multiples of the equations together so that one of the variables is eliminated.

    Problem: Solve the system by elimination: 3x – y = 7

                                                                            2x + y = 3

    Solution: Since the coefficients of the y terms are opposites (additive inverses), adding the equations together as they are will eliminate y.

    3x – y = 7

    2x + y = 3

    5x = 10 x = 2 and 3(2) – y = 7 y = -1

    ***OR….. x = 2 and 2(2) + y = 3 4 + y = 3 y = -1

    (2, -1) is the solution to the system.

    ***NOTE: Once the value of one of the variables has been found, this value may be "plugged-in" to either of the original equations to find the value of the other variable.

    Problem: Solve the system by elimination: x + 2 = - 3y

                                    2x = y + 10

    Solution: First, write both equations in standard form (Ax + By = C)

    x + 3y = -2

    2x – y = 10

    Next, decide which variable to eliminate. In this problem, I am eliminating y, so the coefficients of the y terms need to be opposites. (I need 3

    and –3.) Multiply one or both equations by a constant so that the variable to be eliminated "adds out".

    Multiply equation two by 3.

    x + 3y = -2 x + 3y = -2

    3(2x – y = 10) 6x - 3y = 30

    Add these equations. The sum will contain only one variable.

    x + 3y = -2

    6x - 3y = 30

    7x = 28

    Solve for the remaining variable. x = 4

    Substitute this answer back into either of the original equations to find the value of the other variable that satisfies the system.

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

    The solution to the system is (4, -2).

     

    Problem: Solve the system by elimination: 4x – 5y = -18

                                                                            3x + 2y = -2

    Solution (by eliminating y): 4x – 5y = -18 2(4x – 5y = -18)

    3x + 2y = - 2 5(3x + 2y = - 2)

    8x – 10y = -36

    15x + 10y = -10

    23x = -46 and x = -2 y = 2

    The solution to the system is (-2, 2).

    Solution (by eliminating x): 4x – 5y = -18 3(4x – 5y = -18)

    3x + 2y = - 2 -4(3x + 2y = - 2)

    12x – 15y = -54

    -12x – 8y = 8

    -23y = -46 and y = 2 x = -2

     

    Problem: Solve by elimination: 3y = 8 + 4x

    6x = 9 – 2y

    Solution: 4x – 3y = -8 2(4x – 3y = -8) 8x – 6y = -16

    6x + 2y = 9 3(6x + 2y = 9) 18x +6y = 27

    26x = 11 x =

    x = y =

    The solution to the system is

     

    Problem: Solve by elimination: 3x + y = -7

    6x + 2y = 5

    Solution: -2(3x + y = -7) -6x – 2y = 14

    6x + 2y = 5 6x + 2y = 5

    0 = 19 The system is inconsistent.

    The system has no solution.

    The lines are parallel.

     

    Problem: Solve by elimination: 2x + 5y = 1

    -4x - 10y = -2

    Solution: 2(2x + 5y = 1) 4x + 10y = -2

    -4x - 10y = -2 -4x – 10y = -2

                     0 = 0 The system has an infinite number of solutions. (The equations represent the same line.)

     

    ASSIGNMENT PAGE 544 problems 5 – 37 (every other odd)

    OR PAGE 546 problems 5 – 24 (odd)

    ***NOTE: Linear systems may be solved by using any of the 3 methods. Both graphing

                    and substitution have drawbacks. Generally, elimination is simpler to use.

                    Good advice would be to pick a method and use it on all of the problems.