[email protected] mth55_lec-13_sec_3-3a_3var_lin_sys.ppt 1 bruce mayer, pe chabot college...

[email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt1

Bruce Mayer, PE Chabot College Mathematics

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

§3.3a 3-Var§3.3a 3-VarLinear SystemsLinear Systems



Review §Review §

Any QUESTIONS About• §3.2b → More System Applications

Any QUESTIONS About HomeWork• §3.2 → HW-09

3.2 MTH 55



Systems of 3-VariablesSystems of 3-Variables

A linear equation in three variables is an equation equivalent to form Ax + By + Cz = D• Where A, B, C, and D are real numbers

and A, B, and C are not all 0

A solution of a system of three equations in three variables is an ORDERED TRIPLE (x1, y1, z1) that makes all three equations true.



Example Example Ordered Triple Ordered Triple SolnSoln Determine whether

(2, −1, 3) is a solution of the system

4,

2 2 3,

4 2 3.

x y z

x y z

x y z

SOLUTION: In all three equations, replace x with 2, y with −1, and z with 3.

x + y + z = 4

2 + (–1) + 3 = 4

4 = 4 3 = 3

2x – 2y – z = 3

2(2) – 2(–1) – 3 = 3

– 4x + y + 2z = –3

– 4(2) + (–1) + 2(3) = –3

–3 = –3



Example Example Ordered Triple Ordered Triple SolnSoln Determine whether

(2, −1, 3) is a solution of the system

4,

2 2 3,

4 2 3.

x y z

x y z

x y z

SOLUTION: Because (2, −1, 3) satisfies all three equations in the system, this triple IS a solution for the system.



Solve 3Var Sys by EliminationSolve 3Var Sys by Elimination

1. Write each equation in the standard form of Ax + By+ Cz = D.

2. Eliminate one variable from one PAIR of equations using the elimination method.

3. If necessary, eliminate the same variable from another PAIR of equations to produce a system of TWO varialbles in TWO Equations




4. Steps 2 and 3 result in two equations with the same two variables. Solve these equations using the elimination method.

5. To find the third variable, substitute the values of the variables found in step 4 into any of the three original equations that contain the 3rd variable

6. Check the ordered triple in all three of the original equations



Example Example Solve by Elimination Solve by Elimination

Solve System of Equations

6,

2 2,

3 8.

x y z

x y z

x y z

(1)

(2)

(3)

SOLUTION: select any two of the three equations and work to get one equation in two variables. Let’s add eqs (1) & (2)

6

2 2

x y z

x y z

(1)

(2)

(4)2x + 3y = 8

Adding to eliminate z




Next, select a different pair of equations and eliminate the same variable. Use (2) & (3) to again eliminate z.

2 2

3 8

x y z

x y z

(5) x – y + 3z = 8

Multiplying equation (2) by 3

4x + 5y = 14.

3x + 6y – 3z = 6




Now solve the resulting system of eqns (4) & (5). That will give us two of the numbers in the solution of the original system

Multiply both sides of eqn (4) by −2 and then add to eqn (5):

4x + 5y = 14

2x + 3y = 8(5)

(4)

4x + 5y = 14–4x – 6y = –16,

–y = –2 y = 2




Substituting into either equation (4) or (5) we find that x = 1

Now we have x = 1 and y = 2. To find the value for z, we use any of the three original equations and substitute to find the third number z.

Let’s use eqn (1) and substitute our two numbers in it:

x + y + z = 6

1 + 2 + z = 6z = 3.




At this point we have solved for All Three variables in the System:

z = 3y = 2x = 1

Thus We have obtained the ordered triple (1, 2, 3). It should be checked in all three equations

Finally the Solution Set: (1, 2, 3)




Solve System of Equations

SOLUTION: The equations are in standard form and do not contain decimals or fractions.

Eliminate z from eqns (2) & (3).

3 9 6 3

2 2

2

x y z

x y z

x y z

(1)

(2)

(3)

2 2

2

x y z

x y z

(2)

(3)

(4)3x + 2y = 4




Eliminate z from equations (1) and (2).

Eliminate x using above and eqn (4)

Multiplying equation (2) by 6

3 9 6 3

2 2

x y z

x y z

3 9 6 3

12 6 6 12

x y z

x y z

Adding Eqns15x + 15y = 15

3x + 2y = 415x + 15y = 15

Multiplying top by 5

15x – 10y = 2015x + 15y = 15

AddingEqns

5y = 5y = 1




Using y = −1, find x from equation 4 by substituting

Substitute x = 2 and y = −1 to find z

3x + 2y = 43x + 2(1) = 4

x = 2

x + y + z = 22 – 1 + z = 2 1 + z = 2 z = 1 Thus The solution

is the ordered triple (2, −1, 1)

3 9 6 3

2 2

2

x y z

x y z

x y z




x – 4y + 3z = –9 (3)

Solve the system. 2x + 3y – z = 5 (1)

–3x + 2y – 4z = 2 (2)

x – 4y + 3z = –9 (3)

Step 1 Eliminate a variable from the sum of two equations. The choice of

the variable to eliminate is arbitrary. We will eliminate z.

6x + 9y – 3z = 15

7x + 5y = 6 Add. (4)

Multiply each side of (1) by 3.

7x + 5y = 6 (4)




4x – 16y + 12z = –36


–3x + 2y – 4z = 2 (2)

x – 4y + 3z = –9 (3)

Step 2 Eliminate the same variable, z, from any other equations.

–9x + 6y – 12z = 6

–5x – 10y = –30 Add. (5)


7x + 5y = 6 (4)


–5x – 10y = –30 (5)




–5x – 10y = –30


–3x + 2y – 4z = 2 (2)

x – 4y + 3z = –9 (3)

Step 3 Eliminate a different variable and solve.

14x + 10y = 12

9x = –18 Add. (6)


7x + 5y = 6 (4)

(5)

–5x – 10y = –30 (5)

x = –2



Example Example Solve by Elimination Solve by EliminationSolve the system. 2x + 3y – z = 5 (1)

–3x + 2y – 4z = 2 (2)

x – 4y + 3z = –9 (3)

Step 4 Find a second value by substituting –2 for x in (4) or (5).

7x + 5y = 6 (4)

7x + 5y = 6 (4)

Recall x = –2.

–5x – 10y = –30 (5)

y = 4

7(–2) + 5y = 6

–14 + 5y = 6

5y = 20




–3x + 2y – 4z = 2 (2)

x – 4y + 3z = –9 (3)

Step 5 Find a third value by substituting –2 for x and 4 for y into any of the

three original equations.

2x + 3y – z = 5 (1)

7x + 5y = 6 (4)

From Before x = –2 and y = 4

–5x – 10y = –30 (5)

2(–2) + 3(4) – z = 5

8 – z = 5

z = 3




–3x + 2y – 4z = 2 (2)

x – 4y + 3z = –9 (3)

Step 6 Check that the ordered triple (–2, 4, 3) is the solution of the system.

The ordered triple must satisfy all three original equations.

2x + 3y – z = 5

2(–2) + 3(4) – 3 = 5

–4 + 12 – 3 = 5

5 = 5

(1)

–3x + 2y – 4z = 2

–3(–2) + 2(4) – 4(3) = 2

6 + 8 – 12 = 2

2 = 2

(2)

x – 4y + 3z = –9

(–2) – 4(4) + 3(3) = –9

–2 – 16 + 9 = –9

–9 = –9

(3)



Example Example Missing Term(s) Missing Term(s)

Multiply each side of (2) by 4.–20y – 8z = 12

Solve the system. 5x + 4y = 23 (1)

–5y – 2z = 3 (2)

–8x + 9z = –2 (3)

Since equation (3) is missing the variable y, a good way to begin the

solution is to eliminate y again by using equations (1) and (2).

25x + 20y = 115

25x – 8z = 127 Add. (4)

25x – 8z = 127 (4)





Multiply each side of (4) by 9. 225x – 72z = 1143


–5y – 2z = 3 (2)

–8x + 9z = –2 (3)

Now use equations (3) and (4) to eliminate z and solve.

–64x + 72z = –16

161x = 1127 Add. (5)

25x – 8z = 127 (4)


x = 7




5(7) + 4y = 23


–5y – 2z = 3 (2)

–8x + 9z = –2 (3)

Substituting into equation (1) gives

5x + 4y = 23

35 + 4y = 23

25x – 8z = 127 (4)

4y = –12

y = –3.

–5(–3) – 2z = 3

Substituting into equation (2) gives

–5y – 2z = 3

15 – 2z = 3

–2z = –12

z = 6.





–5y – 2z = 3 (2)

–8x + 9z = –2 (3)

Thus, x = 7, y = −3, and z = 6. Check these values in each of the original equations of the system to verify that the solution set of the system is the triple (7, −3, 6)



Inconsistent SystemInconsistent System

If, in the process of performing elimination on a linear system, an equation of the form 0 = a occurs, where a ≠ 0, then the system has no solution and is inconsistent.



Example Example Inconsistent Sys Inconsistent Sys

Solve the System of Equations

SOLUTION: To eliminate x from Eqn (2), add −2 times Eqn (1) to Eqn (2)

2x 2y 4z 10

2x y z 7 (3)

3y 3z 3 (4)




Next add −3 times Eqn (1) to Eqn (3) to eliminate x from Eqn (3)

We now have the following system

3x 3y 6z 15

3x 2y 5z 20 (3)

y z 5 (5)




Now Multiply Eqn (4) by 1/3 to obtain

To eliminate y from Eqn (5), add −1 times Eqn (6) to Eqn (5)

y z 1 (6)

y z 1

y z 5 (5)

0 6 (7)




We now have the system in simplified elimination form:

This system is equivalent to the original system. Since equation (7) is false (a contradiction), we conclude that the solution set of the system is Ø (the NULL Set), and the system is INconsistent.



Dependent SystemsDependent Systems

If, in the process of performing an elimination solution for a linear system

i. an equation of the form 0 = a (a ≠ 0) does not occur, but

ii. an equation of the form 0 = 0 does occur, then the system of equations has infinitely many solutions and the equations are dependent.



Example Example Dependent System Dependent System

Solve the System of Equations

SOLUTION: Eliminate x from Eqn (2) by adding −3 times Eqn (1) to Eqn (2

3x 3y 3z 21

3x 2y 12z 11 (2)

5y 15z 10




Eliminate x from Eqn (3) by adding −4 times Eqn (1) to Eqn (3)

We now have the equivalent system

4x 4y 4z 28

4x y 11z 18 (2)

5y 15z 10




To eliminate y from Eq (5), add −1 times Eqn (4) to Eqn (5)

We now have the equivalent systemin Final Form

5y 15z 10

5y 15z 10 (5)

0 0 (6)




The equation 0 = 0 may be interpreted as 0·z = 0, which is true for every value of z. Solving eqn (4) for y, we have y = 3z – 2. Substituting into eqn (1) and solving for x.

Thus every triple (x, y, z) = (2z + 5, 3z – 2, z) is a soln of the system for each value of z. E.g, for z = 1, the triple is (7, 1, 1).

x y z 7

x 3z 2 z 7

x 2z 5



Geometry of 3Var SystemsGeometry of 3Var Systems

The graph of a linear equation in three variables, such as Ax + By + Cz = C (where A, B, and C are not all zero), is a plane in three-Dimensional (3D) space.

Following are the possible situations for a system of three linear equations in three variables.




a) Three planes intersect in a single point. The system has only one solution

b) Three planes intersect in one line. The system has infinitely many solutions




c) Three planes coincide with each other. The system has only one solution.

d) There are three parallel planes. The system has no solution.




e) Two parallel planes are intersected by a third plane. The system has no solution.

f) Three planes have no point in common. The system has no solution.



Example Quadratic Modeling The following table shows the higher-order

multiple birth rates in the USA since 1971. At the right is a scatter plot of this data.



Example Quadratic Modeling We will SELECT three

arbitrary ordered pairs to construct the model

We Pick the Points (1,29), (11,40) and (21,100) and substitute the x & y values from these ordered pairs into the Quadratic Function

02 acbxaxy



Example Quadratic Modeling Selecting three representative ordered pairs,

we can write a system of three equations.

Solving this 3-Variable System by Elimination:

710530856124550 2 ... xxy

This produces our Model



Example Quadratic Modeling Use to the Model to

Estimate the Hi-Order Births in 1993

7130856124550 2 ...

xx

xfy

In 1993 x = 23 Sub 23 into Model

9117

7130694286129

713023856152924550

71302385612324550 2

.

...

...

...

y

y

y

y

Thus we estimate that in 1993 the Hi-Order BirthRate was about 118



WhiteBoard WorkWhiteBoard Work

Problems From §3.3 Exercise Set• 28

QuadraticFunction



All Done for TodayAll Done for Today

ACONSISTENT

System

z = -1.553x - 2.642y - 10.272 (darker green)z = 1.416x - 1.92y - 10.979 (medium green)

z = -.761x - .236y - 7.184 (lighter green)

THE THREE PLANES SHARE ONE POINT



Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

AppendiAppendixx

–

srsrsr 22



Equivalent SystemsEquivalent Systems

Operations That Produce Equivalent Systems

1. Interchange the position of any two eqns

2. Multiply any eqn by a nonzero constant

3. Add a nonzero multiple of one eqn to another

A special type of Elimination called Gaussian Elimination uses these steps to solve multivariable systems




1. Rearrange the equations, if necessary, to obtain an x-term with a nonzero coefficient in the first equation. Then multiply the first equation by the reciprocal of the coefficient of the x-term to get 1 as a leading coefficient.

2. By adding appropriate multiples of the first equation, eliminate any x-terms from the second and third equations




2. (cont.) Multiply the resulting second equation by the reciprocal of the coefficient of the y-term to get 1 as the leading coefficient.

3. If necessary by adding appropriate multiple of the second equation from Step 2, eliminate any y-term from the third equation. Solve the resulting equation for z.




4. Back-substitute the values of z from Steps 3 into one of the equations in Step 3 that contain only y and z, and solve for y.

5. Back-substitute the values of y and z from Steps 3 and 4 in any equation containing x, y, and z, and solve for x

6. Write the solution set (Soln Triple)

7. Check soln in the original equations

[email protected] mth55_lec-13_sec_3-3a_3var_lin_sys.ppt 1 bruce mayer, pe chabot college...

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