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Chabot Mathematics §3.3a 3-Var Linear Systems Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected] Chabot College Mathematics 1 Bruce Mayer, PE [email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt Review § 3.2 MTH 55 Any QUESTIONS About • §3.2b → More System Applications Any QUESTIONS About HomeWork • §3.2 → HW-09 Chabot College Mathematics 2 Bruce Mayer, PE [email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt Systems 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. Chabot College Mathematics 3 Bruce Mayer, PE [email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt Example Ordered Triple Soln Determine whether (2, −1, 3) is a solution of the system x y z 4, 2 x 2 y z 3, 4 x y 2 z 3. 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 Chabot College Mathematics 4 2x – 2y – z = 3 – 4x + y + 2z = –3 2(2) – 2(–1) – 3 = 3 – 4(2) + (–1) + 2(3) = –3 3=3 –3 = –3 Bruce Mayer, PE [email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt Example Ordered Triple Soln Determine whether (2, −1, 3) is a solution of the system x y z 4, 2 x 2 y z 3, 4 x y 2 z 3. SOLUTION: Because (2, −1, 3) satisfies all three equations in the system, this triple IS a solution for the system. Chabot College Mathematics 5 Bruce Mayer, PE [email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt Solve 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 Chabot College Mathematics 6 Bruce Mayer, PE [email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt Solve 3Var Sys by Elimination 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 Chabot College Mathematics 7 Bruce Mayer, PE [email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt Example Solve by Elimination Solve System of Equations x y z 6, (1) x 2 y z 2, (2) x y 3 z 8. (3) SOLUTION: select any two of the three equations and work to get one equation in two variables. Let’s add eqs (1) & (2) x yz6 x 2y z 2 2x + 3y Chabot College Mathematics 8 =8 (1) (2) (4) Adding to eliminate z Bruce Mayer, PE [email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt Example Solve by Elimination Next, select a different pair of equations and eliminate the same variable. Use (2) & (3) to again eliminate z. x 2y z 2 x y 3z 8 Multiplying equation (2) by 3 3x + 6y – 3z = 6 x – y + 3z = 8 4x + 5y Chabot College Mathematics 9 = 14. Bruce Mayer, PE [email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt (5) Example Solve by Elimination Now solve the resulting system of eqns (4) & (5). 2x + 3y = 8 That will give us two of 4x + 5y = 14 the numbers in the solution of the original system Multiply both sides of eqn (4) by −2 and then add to eqn (5): Chabot College Mathematics 10 –4x – 6y = –16, 4x + 5y = 14 –y = –2 y=2 Bruce Mayer, PE [email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt (4) (5) Example Solve by Elimination 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) x+y+z=6 and substitute our 1 + 2 + z = 6 two numbers in it: z = 3. Chabot College Mathematics 11 Bruce Mayer, PE [email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt Example Solve by Elimination At this point we have solved for All Three variables in the System: x=1 y=2 z=3 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) Chabot College Mathematics 12 Bruce Mayer, PE [email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt Example Solve by Elimination Solve System 3x 9 y 6 z 3 (1) of Equations 2 x y z 2 (2) x yz2 (3) SOLUTION: The equations are in standard form and do not contain decimals or fractions. 2 x y z 2 (2) Eliminate z from (3) x y z 2 eqns (2) & (3). 3x + 2y = 4 Chabot College Mathematics 13 (4) Bruce Mayer, PE [email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt Example Solve by Elimination Eliminate z from equations (1) and (2). 3x 9 y 6 z 3 2x y z 2 Multiplying equation (2) by 6 3x 9 y 6 z 3 12 x 6 y 6 z 12 15x + 15y = 15 Adding Eqns Eliminate x using above and eqn (4) 3x + 2y = 4 15x + 15y = 15 Chabot College Mathematics 14 Multiplying top by 5 15x – 10y = 20 15x + 15y = 15 5y = 5 y = 1 Bruce Mayer, PE [email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt Adding Eqns Example Solve by Elimination Using y = −1, find x from equation 4 by substituting 3x + 2y = 4 3x + 2(1) = 4 Substitute x = 2 and y = −1 to find z Thus The solution is the ordered triple (2, −1, 1) Chabot College Mathematics 15 x=2 x+y+z=2 2–1+z=2 1+z=2 z=1 3x 9 y 6 z 3 2x y z 2 x yz2 Bruce Mayer, PE [email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt Example Solve by Elimination Solve the system. 2x + 3y – z = 5 (1) –3x + 2y – 4z = 2 (2) x – 4y + 3z = –9 7x + 5y (3) = 6 (4) 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 x – 4y + 3z = –9 7x + 5y Chabot College Mathematics 16 = 6 Multiply each side of (1) by 3. (3) Add. (4) Bruce Mayer, PE [email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt Example Solve by Elimination Solve the system. 2x + 3y – z = 5 (1) –3x + 2y – 4z = 2 (2) x – 4y + 3z = –9 7x + 5y –5x – 10y (3) = 6 (4) = –30 (5) Step 2 Eliminate the same variable, z, from any other equations. –9x + 6y – 12z = 6 Multiply each side of (2) by 3. 4x – 16y + 12z = –36 Multiply each side of (3) by 4. –5x – 10y Chabot College Mathematics 17 = –30 Add. (5) Bruce Mayer, PE [email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt Example Solve by Elimination Solve the system. 2x + 3y – z = 5 (1) –3x + 2y – 4z = 2 (2) x – 4y + 3z = –9 7x + 5y –5x – 10y (3) = 6 (4) = –30 (5) Step 3 Eliminate a different variable and solve. 14x + 10y = 12 Multiply each side of (4) by 2. –5x – 10y = –30 (5) = –18 9x Add. (6) x = –2 Chabot College Mathematics 18 Bruce Mayer, PE [email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt Example Solve by Elimination Solve the system. 2x + 3y – z = 5 (1) –3x + 2y – 4z = 2 (2) x – 4y + 3z = –9 7x + 5y –5x – 10y (3) = 6 (4) = –30 (5) Step 4 Find a second value by substituting –2 for x in (4) or (5). 7x + 5y = 6 (4) 7(–2) + 5y = 6 –14 + 5y = 6 Recall x = –2. 5y = 20 y = 4 Chabot College Mathematics 19 Bruce Mayer, PE [email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt Example Solve by Elimination Solve the system. 2x + 3y – z = 5 (1) –3x + 2y – 4z = 2 (2) x – 4y + 3z = –9 7x + 5y –5x – 10y (3) = 6 (4) = –30 (5) 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 2(–2) + 3(4) – z = 5 8– z = 5 (1) From Before x = –2 and y = 4 z = 3 Chabot College Mathematics 20 Bruce Mayer, PE [email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt Example Solve by Elimination Solve the system. 2x + 3y – z = 5 (1) –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. (1) (2) (3) 2x + 3y – z = 5 –3x + 2y – 4z = 2 x – 4y + 3z = –9 2(–2) + 3(4) – 3 = 5 –3(–2) + 2(4) – 4(3) = 2 (–2) – 4(4) + 3(3) = –9 –4 + 12 – 3 = 5 6 + 8 – 12 = 2 –2 – 16 + 9 = –9 5=5 2=2 –9 = –9 Chabot College Mathematics 21 Bruce Mayer, PE [email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt Example Missing Term(s) Solve the system. 5x + 4y = 23 (1) –5y – 2z = 3 (2) –8x + 9z = –2 (3) 25x – 8z = 127 (4) 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 –20y – 8z = 12 25x – 8z = 127 Chabot College Mathematics 22 Multiply each side of (1) by 5. Multiply each side of (2) by 4. Add. (4) Bruce Mayer, PE [email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt Example Missing Term(s) 5x + 4y = 23 Solve the system. (1) –5y – 2z = 3 (2) –8x + 9z = –2 (3) 25x – 8z = 127 (4) Now use equations (3) and (4) to eliminate z and solve. –64x + 72z = –16 Multiply each side of (3) by 8. 225x – 72z = 1143 Multiply each side of (4) by 9. 161x Add. = 1127 (5) x = 7 Chabot College Mathematics 23 Bruce Mayer, PE [email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt Example Missing Term(s) Solve the system. 5x + 4y = 23 (1) –5y – 2z = 3 (2) –8x + 9z = –2 (3) 25x – 8z = 127 (4) Substituting into equation (1) gives Substituting into equation (2) gives 5x + 4y = 23 –5y – 2z = 3 5(7) + 4y = 23 –5(–3) – 2z = 3 35 + 4y = 23 15 – 2z = 3 4y = –12 –2z = –12 y = –3. z = 6. Chabot College Mathematics 24 Bruce Mayer, PE [email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt Example Missing Term(s) Solve the system. 5x + 4y = 23 (1) –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) Chabot College Mathematics 25 Bruce Mayer, PE [email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt Inconsistent 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. Chabot College Mathematics 26 Bruce Mayer, PE [email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt Example Inconsistent Sys Solve the System x y 2z 5 (1) of Equations 2x y z 7 (2) 3x 2y 5z 20 (3) 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 Chabot College Mathematics 27 (4) Bruce Mayer, PE [email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt Example Inconsistent Sys Next add −3 times 3x 3y 6z 15 Eqn (1) to Eqn (3) 3x 2y 5z 20 (3) to eliminate x y z 5 (5) from Eqn (3) We now have the following system (1) x y 2z 5 (4) 3y 3z 3 y z 5 (5) Chabot College Mathematics 28 Bruce Mayer, PE [email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt Example Inconsistent Sys Now Multiply Eqn (4) by 1/3 to obtain y z 1 To eliminate y from Eqn (5), add −1 times Eqn (6) to Eqn (5) Chabot College Mathematics 29 y z 1 yz 5 (5) 06 (7) Bruce Mayer, PE [email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt (6) Example Inconsistent Sys We now have the system in simplified elimination form: x y 2z 5 (1) y z 1 (5) 0 6 (7) 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. Chabot College Mathematics 30 Bruce Mayer, PE [email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt Dependent 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. Chabot College Mathematics 31 Bruce Mayer, PE [email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt Example Dependent System Solve the System x y z 7 of Equations 3x 2y 12z 11 4 x y 11z 18 (1) (2) (3) 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 Chabot College Mathematics 32 Bruce Mayer, PE [email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt Example Dependent System Eliminate x from 4x 4y 4z 28 Eqn (3) by adding 4x y 11z 18 −4 times Eqn (1) 5y 15z 10 to Eqn (3) We now 7 x y z have the 5y 15z 10 equivalent system 5y 15z 10 Chabot College Mathematics 33 Bruce Mayer, PE [email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt (2) (1) (4) (5) Example Dependent System To eliminate y from Eq (5), add −1 times Eqn (4) to Eqn (5) 5y 15z 10 5y 15z 10 00 (5) (6) We now x y z 7 (1) have the 5y 15z 10 (4) equivalent system 0 0 (6) in Final Form Chabot College Mathematics 34 Bruce Mayer, PE [email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt Example Dependent System The equation 0 = 0 may be interpreted as 0·z = 0, which is true for every value of z. Solving eqn (4) x yz7 for y, we have y = 3z – 2. x 3z 2 z 7 Substituting into eqn (1) x 2z 5 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). Chabot College Mathematics 35 Bruce Mayer, PE [email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt Geometry 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. Chabot College Mathematics 36 Bruce Mayer, PE [email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt Geometry of 3Var Systems 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 Chabot College Mathematics 37 Bruce Mayer, PE [email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt Geometry of 3Var Systems c) Three planes coincide with each other. The system has only one solution. d) There are three parallel planes. The system has no solution. Chabot College Mathematics 38 Bruce Mayer, PE [email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt Geometry of 3Var Systems 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. Chabot College Mathematics 39 Bruce Mayer, PE [email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 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. Chabot College Mathematics 40 Bruce Mayer, PE [email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 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 2 y ax bx c Chabot College Mathematics 41 a0 Bruce Mayer, PE [email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt Example Quadratic Modeling Selecting three representative ordered pairs, we can write a system of three equations. Solving this 3-Variable System by Elimination: This produces our Model y 0.2455x 1.856 x 30.7105 2 Chabot College Mathematics 42 Bruce Mayer, PE [email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt Example Quadratic Modeling Use to the Model to Estimate the HiOrder Births in 1993 y f x 0.2455 x 2 1.856 x 30.71 In 1993 x = 23 Sub 23 into Model y 0.245523 1.85623 30.71 y 0.2455529 1.85623 30.71 y 129.86 42.69 30.71 y 117.9 2 Thus we estimate that in 1993 the Hi-Order BirthRate was about 118 Chabot College Mathematics 43 Bruce Mayer, PE [email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt WhiteBoard Work Problems From §3.3 Exercise Set • 28 Quadratic Function Chabot College Mathematics 44 Bruce Mayer, PE [email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt All Done for Today A CONSISTENT 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 POINTBruce Mayer, PE Chabot College Mathematics 45 [email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt Chabot Mathematics Appendix r s r s r s 2 2 Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected] – Chabot College Mathematics 46 Bruce Mayer, PE [email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt Equivalent 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 Chabot College Mathematics 47 Bruce Mayer, PE [email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt Solve 3Var Sys by Elimination 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 Chabot College Mathematics 48 Bruce Mayer, PE [email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt Solve 3Var Sys by Elimination 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. Chabot College Mathematics 49 Bruce Mayer, PE [email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt Solve 3Var Sys by Elimination 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 Chabot College Mathematics 50 Bruce Mayer, PE [email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt