171S9.2p Systems of Equations in Three Variables

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CHAPTER 9: Systems of Equations and Matrices9.1 Systems of Equations in Two Variables 9.2 Systems of Equations in Three Variables 9.3 Matrices and Systems of Equations

MAT 171 Precalculus AlgebraDr. Claude Moore

Cape Fear Community College

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9.2 Systems of Equations in Three Variables

• Solve systems of linear equations in three variables.• Use systems of three equations to solve applied problems.• Model a situation using a quadratic function.

To model a quadratic function with the TI, see use of the TI Calculator on the Important Links webpage and click on Modeling – Statistical Modeling with TI­83 Calculator – linear, quadratic, and other regressions. (http://cfcc.edu/faculty/cmoore/TI83Modeling.htm)

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Solving Systems of Equations in Three VariablesA linear equation in three variables is an equation equivalent to one of the form Ax + By + Cz = D. A, B, C, and D are real numbers and A, B, and C are not 0.

A solution of a system of three equations in three variables is an ordered triple that makes all three equations true. Example: The triple (4, 0, ­3) is the solution of this system of equations. We can verify this by substituting 4 for x, 0 for y, and ­3 for z in each equation. x ­ 2y + 4z = ­8 2x + 2y ­ z = 11 x + y ­ 2z = 10

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Gaussian Elimination

An algebraic method used to solve systems in three variables.The original system is transformed to an equivalent

one of the form: Ax + By + Cz = D, Ey + Fz = G, Hz = K. Then the third equation is solved for z and back­substitution is used to find y and then x.

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Legal Operations

The following operations can be used to transform the original system to an equivalent system in the desired form.

1. Interchange any two equations.2. Multiply both sides of one of the equations by a nonzero constant.3. Add a nonzero multiple of one equation to another equation.

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Example

Solve the system(1) x + 3y + 2z = 9 (2) x ­ y + 3z = 16(3) 3x ­ 4y + 2z = 28

Solution: Choose 1 variable to eliminate using 2 different pairs of equations. Let’s eliminate x from equations (2) and (3).

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Example

−x − 3y − 2z = −9 Mult. (1) by −1 x − y + 3z = 16 (2) −4y + z = 7 (4)

−3x − 9y − 6z = −27 Mult. (1) by −3 3x − 4y + 2z = 28 (3) −13y − 4z = 1 (5)

Solve the system

(1) x + 3y + 2z = 9 (2) x ­ y + 3z = 16(3) 3x ­ 4y + 2z = 28

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Example continued

Now we have… x + 3y + 2z = 9 (1) ­4y + z = 7 (4) ­13y ­ 4z = 1 (5) Next, we multiply equation (4) by 4 to make the z coefficient a multiple of the z coefficient in the equation below it. x + 3y + 2z = 9 (1) ­16y + 4z = 28 (6) ­13y ­ 4z = 1 (5)

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Example continued

Now, we add equations (5) and (6). ­13y ­ 4z = 1 (5) ­16y + 4z = 28 (6) ­29y = 29

Now, we have the system of equations: x + 3y + 2z = 9 (1) ­13y ­ 4z = 1 (5) ­29y = 29 (7)

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Example continued

Next, we solve equation (7) for y: ­29y = 29 y = ­1

Then, we back­substitute ­1 in equation (5) and solve for z. ­13(­1) ­ 4z = 1 13 ­ 4z = 1 ­4z = ­12 z = 3

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Example continued

Finally, we substitute −1 for y and 3 for z in equation (1) and solve for x: x + 3(−1) + 2(3) = 9 x − 3 + 6 = 9 x = 6

The triple (6, −1, 3) is the solution of this system.

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GraphsThe graph of a linear equation in three variables is a plane. Thus the solution set of such a system is the intersection of three planes.

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Application

A food service distributor conducted a study to predict fuel usage for new delivery routes, for a particular truck. Use the chart to find the rates of fuel in rush hour traffic, city traffic, and on the highway.

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34186Week 32487Week 21592Week 1

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Solution1. Familiarize. We let x, y, and z represent the hours in rush hour traffic, city traffic, and highway, respectively. 2. Translate. We have three equations: 2x + 9y + 3z = 15 (1) 7x + 8y + 3z = 24 (2) 6x + 18y + 6z = 34 (3)3. Carry Out. We will solve this equation by eliminating z from equations (2) and (3). ­2x ­ 9y ­ 3z = ­15 Mult. (1) by ­1 7x + 8y + 3z = 24 (2) 5x ­ y = 9 (4)

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Solution continued Next, we can solve for x: ­4x ­ 18y ­ 6z = ­30 Mult. (1) by ­2

6x + 18y + 6z = 34 (3) 2x = 4 x = 2 Next, we can solve for y by substituting 2 for x in equation (4): 5(2) ­ y = 9 y = 1 Finally, we can substitute 2 for x and 1 for y in equation (1) to solve for z: 2(2) + 9(1) + 3z = 15 4 + 9 + 3z = 15 3z = 2

z = Solving the system we get (2, 1, ).

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Solution continued

4. Check: Substituting 2 for x, 1 for y, and for z, we see that the solution makes each of the three equations true.

5. State: In rush hour traffic the distribution truck uses fuel at a rate of 2 gallons per hour. In city traffic, the same truck uses 1 gallon of fuel per hour. In highway traffic, the same truck used gallon of fuel per hour.

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764/2. Solve the system of equations. x + 6y + 3z = 42x + y + 2z = 33x ­ 2y + z = 0

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764/8. Solve the system of equations. x + 2y ­ z = 44x ­ 3y + z = 85x ­ y = 12

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764/10. Solve the system of equations. x + 3y + 4z = 13x + 4y + 5z = 3 x + 8y + 11z = 2

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764/16. Solve the system of equations. w + x ­ y + z = 0 ­w + 2x + 2y + z = 5 ­w + 3x + y ­ z = ­4­2w + x + y ­ 3z = ­7

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765/20. Adopting Abroad. The three foreign countries from which the largest number of children were adopted in 2009 were China, Ethiopia, and Russia. A total of 6864 children were adopted from these countries. The number of children adopted from China was 862 fewer than the total number adopted from Ethiopia and Russia. Twice the number adopted from Russia is 171 more than the number adopted from China. (Source: U.S. Department of State) Find the number of children adopted from each country.

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765/22. Mother’s Day Spending. The top three Mother’s Day gifts are flowers, jewelry, and gift certificates. The total of the average amounts spent on these gifts is $53.42. The average amount spent on jewelry is $4.40 more than the average amount spent on gift certificates. Together, the average amounts spent on flowers and gift certificates is $15.58 more than the average amount spent on jewelry. (Source: BIGresearch) What is the average amount spent on each type of gift?

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766/29. Investment. Walter earns a year­end bonus of $5000 and puts it in 3 one­year investments that pay $243 in simple interest. Part is invested at 3%, part at 4%, and part at 6%. There is $1500 more invested at 6% than at 3%. Find the amount invested at each rate.

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766/32. Cost of Snack Food. Martin and Eva pool their loose change to buy snacks on their coffee break. One day, they spent $6.75 on 1 carton of milk, 2 donuts, and 1 cup of coffee. The next day, they spent $8.50 on 3 donuts and 2 cups of coffee. The third day, they bought 1 carton of milk, 1 donut, and 2 cups of coffee and spent $7.25. On the fourth day, they have a total of $6.45 left. Is this enough to buy 2 cartons of milk and 2 donuts?

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766/34. Student Loans. The table below lists the volume of nonfederal student loans, in billions of dollars, represented in terms of the number of years since 2004.

See Modeling tutorial for calculator at http://cfcc.edu/faculty/cmoore/TI­STAT.htm#algebra

a) Fit a quadratic function f (x) = ax2 + bx + c to the data, where x is the number of years since 2004. b) Use the function to estimate the volume of nonfederal student loans in 2007.

f(x) = ­1.8625x2 + 6.425x + 15.1; 2007, x = 3 & y = $17,613 billions in nonfederal student loans in 2007.

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767/37. Unemployment Rate. The table below lists the U.S. unemployment rate in October for selected years.

See Modeling tutorial for calculator at http://cfcc.edu/faculty/cmoore/TI­STAT.htm#algebra

a) Use a graphing calculator to fit a quadratic function f (x) = ax2 + bx + c to the data, where x is the number of years since 2002. b) Use the function found in part (a) to estimate the unemployment rate in 2003, in 2007, and in 2009.

f(x) = 0.1437x2 ­ 0.6922x + 5.8825; US unemployment rates for years 2003: x = 1 & y = 5.3%; 2007: x = 5 & y = 6.0%; 2009: x = 7 & y = 8.1%

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766/33. Job Loss. The table below lists the percent of American workers who responded that they were likely to be laid off from their jobs in the coming year, x is the number of years since 1990.

a) Fit a quadratic function f (x) = ax2 + bx + c

to the data, where x is the number of years since 1990.

b) Use the function to estimate the percent of workers who responded that they were likely to be laid off in the coming year in 2003.

Do NOT use calculator.

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766/33. Job Loss. The table below lists the percent of American workers who responded that they were likely to be laid off from their jobs in the coming year, x represents the number of years since 1990.

a) Fit a quadratic function f (x) = ax2 + bx + c

to the data, where x is the number of years since 1990.

b) Use the function to estimate the percent of workers who responded that they were likely to be laid off in the coming year in 2003.

Do NOT use calculator.

Sep 12­1:44 PM

766/33. Job Loss. The table below lists the percent of American workers who responded that they were likely to be laid off from their jobs in the coming year, x represents the number of years since 1990.

a) Fit a quadratic function f (x) = ax2 + bx + c

to the data, where x is the number of years since 1990.

b) Use the function to estimate the percent of workers who responded that they were likely to be laid off in the coming year in 2003.

Do NOT use calculator.

Sep 12­1:44 PM

766/33. Job Loss. The table below lists the percent of American workers who responded that they were likely to be laid off from their jobs in the coming year, x represents the number of years since 1990.

a) Fit a quadratic function f (x) = ax2 + bx + c

to the data, where x is the number of years since 1990.

b) Use the function to estimate the percent of workers who responded that they were likely to be laid off in the coming year in 2003.

Do NOT use calculator.