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10.2 Systems of Linear Equations in Several Variables

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Objectives

► Solving a Linear System

► The Number of Solutions of a Linear System

► Modeling Using Linear Systems

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Systems of Linear Equations in Several Variables

A linear equation in n variables is an equation that can be put in the form

a1x1 + a2x2 + · · · + anxn = c

where a1, a2, . . . , an and c are real numbers, and x1, x2, . . . , xn are the variables.

If we have only three or four variables, we generally use x, y, z, and ω instead of x1, x2, x3, and x4.

Such equations are called linear because if we have just two variables, the equation is a1x + a2y = c, which is the equation of a line.

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Systems of Linear Equations in Several Variables

Here are some examples of equations in three variables that illustrate the difference between linear and nonlinear equations.

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Solving a Linear System

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Solving a Linear SystemThe following are two examples of systems of linear equations in three variables.

The second system is in triangular form; that is, the variable x doesn’t appear in the second equation, and the variables x and y do not appear in the third equation.

A system of linear A system in triangular equations form

x – 2y – z = 1 x – 2y – z = 1 –x + 3y + 3z = 4 y + 2z = 52x – 3y + z = 10 z = 3

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Solving a Linear SystemIt’s easy to solve a system that is in triangular form by using back-substitution.

So our goal in this section is to start with a system of linear equations and change it to a system in triangular form that has the same solutions as the original system.

We begin by showing how to use back-substitution to solve a system that is already in triangular form.

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Example 1 – Solving a Triangular System Using Back-Substitution

Solve the system using back-substitution:

x – 2y – z = 1 y + 2z = 5

z = 3Solution:From the last equation we know that z = 3.

We back-substitute this into the second equation and solve for y.

y + 2(3) = 5

Equation 1

Equation 2

Equation 3

Back-substitute z = 3 into Equation 2

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Example 1 – Solutiony = –1

Then we back-substitute y = –1 and z = 3 into the first equation and solve for x.

x – 2(–1) – (3) = 1

x = 2

The solution of the system is x = 2, y = –1, z = 3. We can also write the solution as the ordered triple (2, –1, 3).

cont’d

Solve for y

Back-substitute y = –1 and z = 3 into Equation 1

Solve for x

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Solving a Linear SystemTo change a system of linear equations to an equivalent system (that is, a system with the same solutions as the original system), we use the elimination method.

This means that we can use the following operations.

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Solving a Linear SystemTo solve a linear system, we use these operations to change the system to an equivalent triangular system. Then we use back-substitution as in Example 1.

This process is called Gaussian elimination.

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Example 2 – Solving a System of Three Equations in Three Variables

Solve the system using Gaussian elimination.

x – 2y + 3z = 1 x + 2y – z = 13

3x + 2y – 5z = 3

Solution:We need to change this to a triangular system, so we begin by eliminating the x-term from the second equation.

Equation 1

Equation 2

Equation 3

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Example 2 – Solutionx + 2y – z = 13

x – 2y + 3z = 1

4y – 4z = 12

This gives us a new, equivalent system that is one step closer to triangular form.

x – 2y + 3z = 1 4y – 4z = 12

3x + 2y – 5z = 3

cont’d

Equation 1

Equation 2

Equation 2 + (–1) Equation 1 = new Equation 2

Equation 1

Equation 2

Equation 3

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Example 2 – SolutionNow we eliminate the x-term from the third equation.

x – 2y + 3z = 14y – 4z = 12

8y – 14z = 0

Then we eliminate the y-term from the third equation.

x – 2y + 3z = 14y – 4z = 12

–6z = –24

cont’d

Equation 3 + (–3) Equation 1 = new Equation 3

Equation 3 + (–2) Equation 2 = new Equation 3

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Example 2 – SolutionThe system is now in triangular form, but it will be easier to work with if we divide the second and third equations by the common factors of each term.

x – 2y + 3z = 1y – z = 3

z = 4

Now we use back-substitution to solve the system. From the third equation we get z = 4. We back-substitute this into the second equation and solve for y.

y – (4) = 3

cont’d

Equation 2 = new Equation 2

Equation 3 = new Equation 3

Back-substitute z = 4 into Equation 2

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Example 2 – Solutiony = 7

Now we back-substitute y = 7 and z = 4 into the first equation and solve for x.

x – 2(7) + 3(4) = 1

x = 3

The solution of the system is x = 3, y = 7, z = 4, which we can write as the ordered triple (3, 7, 4).

cont’d

Solve for y

Back-substitute y = 7 and z = 4 into Equation 1

Solve for x

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Example 2 – SolutionCheck Your Answer:x = 3, y = 7, z = 4:

(3) – 2(7) + 3(4) = 1

(3) + 2(7) – (4) = 13

3(3) + 2(7) – 5(4) = 3

cont’d

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The Number of Solutions of a Linear System

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The Number of Solutions of a Linear System

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The Number of Solutions of a Linear System

A system with no solutions is said to be inconsistent, and a system with infinitely many solutions is said to be dependent.

As we see in the next example, a linear system has no solution if we end up with a false equation after applying Gaussian elimination to the system.

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Example 3 – A System with No Solution

Solve the following system.

x + 2y – 2z = 1 2x + 2y – z = 6

3x + 4y – 3z = 5

Solution:To put this in triangular form, we begin by eliminating the x-terms from the second equation and the third equation.

Equation 1

Equation 2

Equation 3

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Example 3 – Solutionx + 2y – 2z = 1

–2y + 3z = 4 3x + 4y + 3z = 5

x + 2y – 2z = 1 –2y + 3z = 4–2y + 3z = 2

Now we eliminate the y-term from the third equation.

x + 2y – 2z = 1 –2y + 3z = 4

0 = 2

cont’d

Equation 2 + (–2) Equation 1 = new Equation 2

Equation 3 + (–3) Equation 1 = new Equation 3

Equation 3 + (–1) Equation 2 = new Equation 3

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Example 3 – SolutionThe system is now in triangular form, but the third equation says 0 = 2, which is false.

No matter what values we assign to x, y, and z, the third equation will never be true. This means that the system has no solution.

cont’d

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Example 4 – A System with Infinitely Many Solutions

Solve the following system.

x – y + 5z = –2 2x + y + 4z = 2

2x + 4y – 2z = 8

Solution:To put this in triangular form, we begin by eliminating the x-terms from the second equation and the third equation.

Equation 1Equation 2Equation 3

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Example 4 – Solutionx – y + 5z = –2

3y – 6z = 6 2x + 4y – 2z = 8

x – y + 5z = –2 3y – 6z = 6

6y – 12z = 12

Now we eliminate the y-term from the third equation.

x – y + 5z = –2 3y – 6z = 6

0 = 0

cont’d

Equation 2 + (–2) Equation 1 = new Equation 2

Equation 3 + (–2) Equation 1 = new Equation 3

Equation 3 + (–2) Equation 2 = new Equation 3

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Example 4 – SolutionThe new third equation is true, but it gives us no new information, so we can drop it from the system.

Only two equations are left. We can use them to solve for x and y in terms of z, but z can take on any value, so there are infinitely many solutions.

To find the complete solution of the system, we begin by solving for y in terms of z, using the new second equation.

3y – 6z = 6

y – 2z = 2

cont’d

Equation 2

Multiply by

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Example 4 – Solutiony = 2z + 2

Then we solve for x in terms of z, using the first equation.

x – (2z + 2) + 5z = –2

x + 3z – 2 = –2

x = –3z

cont’d

Solve for y

Substitute y = 2z + 2 into Equation 1

Simplify

Solve for x

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Example 4 – SolutionTo describe the complete solution, we let t represent any real number. The solution is

x = –3t

y = 2t + 2

z = t

We can also write this as the ordered triple (–3t, 2t + 2, t).

cont’d

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The Number of Solutions of a Linear System

In the solution of Example 4 the variable t is called a parameter. To get a specific (so-called particular) solution, we give a specific (particular) value to the parameter t.

For instance, if we set t = 2, we get

x = –3(2) = –6

y = 2(2) + 2 = 6

z = 2

Thus (–6, 6, 2) is a solution of the system.

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The Number of Solutions of a Linear System

Here are some other solutions of the system obtained by substituting other values for the parameter t.

You should check that these points satisfy the original equations.

Since there are infinitely many choices for the parameter t, the system has infinitely many solutions.

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Modeling Using Linear Systems

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Modeling Using Linear SystemsLinear systems are used to model situations that involve several varying quantities.

In the next example we consider an application of linear systems to finance.

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Example 5 – Modeling a Financial Problem Using a Linear System

Jason receives an inheritance of $50,000. His financial advisor suggests that he invest this in three mutual funds: a money-market fund, a blue-chip stock fund, and a high-techstock fund.

The advisor estimates that the money-market fund will return 5% over the next year, the blue-chip fund 9%, and the high-tech fund 16%. Jason wants a total first-year return of $4000.

To avoid excessive risk, he decides to invest three times as much in the money-market fund as in the high-tech stock fund. How much should he invest in each fund?

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Example 5 – SolutionLet x = amount invested in the money-market fund

y = amount invested in the blue-chip stock fund

z = amount invested in the high-tech stock fund

We convert each fact given in the problem into an equation.

x + y + z = 50,000

0.05x + 0.09y + 0.16z = 4000

x = 3z

Total amount invested is $50,000

Total investment return is $4000

Money-market amount is 3 high-tech amount

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Example 5 – SolutionMultiplying the second equation by 100 and rewriting the third gives the following system, which we solve using Gaussian elimination.

x + y + z = 50,0005x + 9y + 16z = 400,000

x – 3z = 0

x + y + z = 50,0004y + 11z = 150,000–y – 4z = –50,000

cont’d

100 Equation 2

Subtract 3z

Equation 2 + (–5) Equation 1 = new Equation 2

Equation 3 + (–1) Equation 1 = new Equation 3

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Example 5 – Solutionx + y + z = 50,000

– 5z = –50,000–y – 4z = –50,000

x + y + z = 50,000z = 10,000

y + 4z = 50,000

x + y + z = 50,000y + 4z = 50,000

z = 10,000

cont’d

Equation 2

Equation 2 + 4 Equation 3 = new Equation 3

Interchange Equations 2 and 3

(–1) Equation 3

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Example 5 – SolutionNow that the system is in triangular form, we use back-substitution to find that x = 30,000, y = 10,000, and z = 10,000.

This means that Jason should invest

$30,000 in the money-market fund

$10,000 in the blue-chip stock fund

$10,000 in the high-tech stock fund

cont’d