Chapter 3 Systems of Linear Equations. § 3.1 Systems of Linear Equations in Two Variables.

Chapter 3

Systems of Linear Equations

§ 3.1

Systems of Linear Equations in Two Variables

Blitzer, Intermediate Algebra, 5e – Slide #3 Section 3.1

Systems of Equations

We know that an equation of the form Ax + By = C is a line when graphed.

Two such equations is called a system of linear equations.

A solution of a system of linear equations is an ordered pair that satisfies both equations in the system.

For example, the ordered pair (2,1) satisfies the system

3x + 2y = 8 4x – 3y = 5



Since two lines may intersect in exactly one point, may not intersect at all, or may intersect in every point; it follows that a system of linear equations will have exactly one solution, will have no solution, or will have infinitely many solutions.



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Determine whether (3,2) is a solution of the system

285

1063

yx

yx

102633

10129

Because 3 is the x-coordinate and 2 is the y-coordinate of the point(3,2), we replace x with 3 and y with 2.

Since the result is false, (3,2) is NOT a solution for the system. Also, I need not check the other equation since the first one failed.

1063 yx

103

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?

false



Solve Systems of Two Linear Equations in Two Variables, x and y, by Graphing

1) Graph the first equation.

2) Graph the second equation on the same set of axes.

3) If the lines representing the two graphs intersect at a point, determine the coordinates of this point of intersection. The ordered pair is the solution to the system.

4) Check the solution in both equations.

NOTE: In order for this method to be useful, you must graph the lines very accurately.



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Solve by graphing:

33

44

yx

yx

44 yx

44 xy

1) Graph the first equation. I first rewrite the equation in slope-intercept form.

Now I can graph the equation.

m = -4 = -4/1, b = 4



2) Graph the second equation on the same set of axes. I first rewrite the equation in slope-intercept form.

33 yx

33 xy

33 xy

CONTINUECONTINUEDD

Now I can graph the equation.

m = 3 = 3/1, b = -3



3) Determine the coordinates of the intersection point. This ordered pair is the system’s solution. Using the graph below, it appears that the solution is the point (1,0). We won’t know for sure until after we check this potential solution in the next step.

CONTINUECONTINUEDD



4) Check the solution in both equations.

44 yx

4014

404

CONTINUECONTINUEDD

44

33 yx

3013

303

33

Because both equations are satisfied, (1,0) is the solution and {(1,0)} is the solution set.

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truetrue


Substitution Method

Solving Linear Systems by Substitution1) Solve either of the equations for one variable in terms of the other. (If one of the equations is already in this form, you can skip this step)

2) Substitute the expression found in step 1 into the other equation. This will result in an equation in one variable! (Hence, it’s called the “Substitution Method”)

3) Solve the equation containing one variable. (the resultant equation from step 2)

4) Back-Substitute the value found in step 3 into one of the original equations. Simplify and find the value of the remaining variable.

5) Check the proposed solution in both of the system’s given equations.


Substitution Method

EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Solve by the substitution method:

33

44

yx

yx

44 yx44 xy

1) Solve either of the equations for one variable in terms of the other. We’ll isolate the variable y from the first equation.

Solve for y by subtracting 4x from both sides


Substitution Method

33 yx

3443 xx

3443 xx

347 x

2) Substitute the expression from step 1 into the other equation. -4x + 4 is the ‘expression from step 1’ and ‘the other equation’ is 3x – y = 3. Therefore:

Replace y with -4x+4

CONTINUECONTINUEDD

77 x1x

DistributeAdd like termsAdd 4 to both sidesDivide both sides by 7

3) Solve the resulting equation containing one variable.


Substitution Method

4) Back-substitute the obtained value into one of the original equations. We back-substitute 1 for x into one of the original equations to find y. Let’s use the first equation.

CONTINUECONTINUEDD

Replace x with 1

Multiply

Subtract 4 from both sides

44 yx

414 y

44 y

0y

Therefore, the potential solution is (1,0).


Substitution Method

5) Check. Now we will show that (1,0) is a solution for both of the original equations.

CONTINUECONTINUEDD

44 yx

4014

404

44

33 yx

3013

303

33

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?

truetrue



Addition (Elimination) Method

Solving Linear Systems by Addition1) If necessary, rewrite both equations in the form Ax + By = C.

2) If necessary, multiply either equation or both equations by appropriate nonzero numbers so that the sum of the x-coefficients or the sum of the y-coefficients is 0.

3) Add the equations in step 2. The sum should be an equation in one variable.

4) Solve the equation in one variable (the result of step 3).

5) Back-substitute the value obtained in step 4 into either of the given equations and solve for the other variable.

6) Check the solution in both of the original equations.

NOTE: As you now know, there is more than one method to solve a system of equations. The reason for learning more than one method is because sometimes one method will be preferable or easier to use over another method.



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Solve by the addition method:

1) Rewrite both equations in the form Ax + By = C. We first arrange the system so that variable terms appear on the left and constants appear on the right. We obtain

Add 2y to both sides

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

4x + 2y = 4

3x - y = 3 Add 3x to both sides



2) If necessary, multiply either equation or both equations by appropriate numbers so that the sum of the x-coefficients is 0. We can eliminate the y’s by multiplying the second equation by 2. Or we can eliminate the x’s by multiplying the first equation by -3 and the second equation by 4. Let’s use the first method.

Multiply by 2

4x + 2y = 4

3x - y = 3

CONTINUECONTINUEDD

4x + 2y = 4

6x - 2y = 6

No Change



3) Add the equations.

CONTINUECONTINUEDD

4x + 2y = 4

6x - 2y = 6

Add: 10x + 0y = 10

4) Solve the equation in one variable. Now I solve the equation 10x = 10.

10x = 10

10x = 10

x = 1



5) Back-substitute and find the value of the other variable. Now we will use one of the original equations and replace x with 1 to determine y. I’ll use the second equation.

CONTINUECONTINUEDD

-y = -3x + 3

-y = -3(1) + 3

-y = -3 + 3

-y = 0

y = 0

Replace x with 1

Multiply

Add

Multiply by -1

Therefore, the potential solution is (1,0).



6) Check. I now check the potential solution (1,0) in both original equations.

CONTINUECONTINUEDD

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truetrue

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

4(1) = -2(0) + 4

4 = 0 + 4

4 = 4

-(0) = -3(1) + 3

0 = -3 + 3

0 = 0



Solving Systems of Equations

Comparing Solution MethodsMethod Advantages Disadvantages

Graphing

You can see the solutions. If the solutions do not involve integers or are too large to be seen on the graph, it’s impossible to tell exactly what the solutions are.

Substitution

Gives exact solutions. Easy to use if a variable is on one side by itself.

Solutions cannot be seen. Introduces extensive work with fractions when no variable has a coefficient of 1 or -1.

Addition

Gives exact solutions. Easy to use if a variable has a coefficient of 1 or -1.

Solutions cannot be seen.



The Number of Solutions to a System of Two Linear Equations

Number of Solutions What This Means Graphically

Graphical Examples

Exactly one ordered-pair solution

The two lines intersect at one point.

No solution The two lines are parallel.

Infinitely Many Solutions

The two lines are identical.



NOTE: It is extremely helpful to understand these relationships as well as any other relationship between an equation and it’s graph.

NOTE: To determine that a system has exactly one solution, solve the system using one of the methods. A single solution will occur as in the previous examples.

NOTE: To determine that a system has no solution, solve the system using one of the methods. Eventually, you’ll get an obviously false statement, like 3 = 4.

NOTE: To determine that a system has infinitely many solutions, solve the system using one of the methods. Eventually, you’ll get an obviously true statement, like -2 = -2.



EXAMPLEEXAMPLE

At a price of p dollars per ticket, the number of tickets to a rock concert that can be sold is given by the demand model N = -25p + 7800. At a price of p dollars per ticket, the number of tickets that the concert’s promoters are willing to make available is given by the supply model N = 5p + 6000.

(a) How many tickets can be sold and supplied for $50 per ticket?

(b) Find the ticket price at which supply and demand are equal. At this price, how many tickets will be supplied and sold?



SOLUTIONSOLUTION

CONTINUECONTINUEDD

The number of tickets that can be sold for $50 per ticket is found using the demand model:

(a) How many tickets can be sold and supplied for $50 per ticket?

N = -25(50) + 7800 = -1250 + 7800 = 6550 tickets sold.

The number of tickets that can be supplied for $50 per ticket is found using the supply model:

N = 5(50) + 6000 = 250 + 6000 = 6250 tickets supplied.



CONTINUECONTINUEDD

N = -25p + 7800N = 5p + 6000

1) Solve either of the equations for one variable in terms of the other. The system of equations already has N isolated in both equations.

(b) Find the ticket price at which supply and demand are equal. At this price, how many tickets will be supplied and sold?


Substitution Method

CONTINUECONTINUEDD

2) Substitute the expression from step 1 into the other equation. I’ll replace the N in the second equation with -25p + 7800.

-25p + 7800 = 5p + 6000

Add 25p to both sides

3) Solve the resulting equation containing one variable.

Divide both sides by 30

Subtract 6000 from both sides

-25p + 7800 = 5p + 6000

7800 = 30p + 6000

1800 = 30p

60 = p


Substitution Method

Replace p with 60

CONTINUECONTINUEDD

Multiply

Add

N = -25p + 7800

N = -25(60) + 7800

4) Back-substitute the obtained value into one of the original equations. We back-substitute 60 for p into one of the original equations to find N. Let’s use the first equation.

N = -1500 + 7800

N = 6300


Substitution Method

CONTINUECONTINUEDD

Therefore, the solution is (60,6300). Therefore, supply and demand will be equal when the ticket price is $60 and 6300 tickets are sold.

5) Check. Now we will show that (60,6300) is a solution for both of the original equations.

? ?

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truetrue

N = -25p + 7800 N = 5p + 6000

6300 = -25(60) + 7800 6300 = 5(60) + 60006300 = -1500 + 7800 6300 = 300 + 6000

6300 = 6300 6300 = 6300

Chapter 3 Systems of Linear Equations. § 3.1 Systems of Linear Equations in Two Variables.

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