1 / 88

890 likes | 1.21k Views

Precalculus – MAT 129. Instructor: Rachel Graham Location: BETTS Rm. 107 Time: 8 – 11:20 a.m. MWF. Chapter Seven. Linear Systems and Matrices. Ch. 7 Overview. Solving Systems of Equations Systems of Linear Equations in Two Variables Multivariable Linear Systems

Download Presentation
## Precalculus – MAT 129

**An Image/Link below is provided (as is) to download presentation**
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.
Content is provided to you AS IS for your information and personal use only.
Download presentation by click this link.
While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

**Precalculus – MAT 129**Instructor: Rachel Graham Location: BETTS Rm. 107 Time: 8 – 11:20 a.m. MWF**Chapter Seven**Linear Systems and Matrices**Ch. 7 Overview**• Solving Systems of Equations • Systems of Linear Equations in Two Variables • Multivariable Linear Systems • Matrices and Systems of Equations**Ch. 7 Overview (cont.)**• Operations with Matrices • The Inverse of a Square Matrix • The Determinant of a Square Matrix • Applications of Matrices and Determinants**7.1 – Solving Systems of Equations**• The Method of Substitution**7.1 – The Method of Substitution**• Systems of equations • thenumber of equations must equal the number of unknown variables • A solution of a system is an ordered pair that satisfies each equation**7.1 – The Method of Substitution**• Solve one of the equations for one variable. • Substitute into the other equation. • Solve the equation you substituted into (it will now only be in one variable) • Back-substitute into the first equation. • Check the solution(s).**Example 1.7.1**Pg. 481 #5 Solve the system by the method of substitution. 2x + y = 6 -x + y = 0**Solution Example 1.7.1**Pg. 481 #5 Following the steps: • Solving for y in the second equation gives y=x. • Substituting in the first equation gives 2x + x = 6 3x = 6 x = 2. • Back substituting gives y = 2. • Check (2,2) in both equations.**Things that can happen**• No-Solution Case • The graphs don’t touch. • Two-Solution Case • The graphs touch in two places.**Example 2.7.1**Solve the system by the method of substitution. 6x + 5y = -3 -x – (5/6)y = -7**Solution Example 2.7.1**Solving for y in the first equation yields: y = -(3/5) – (6/5)x Substituting for y in the second equation: -x – (5/6)(-3/5 – (6/5)x ) ½ = -7. This means that there is no solution.**Example 3.7.1**Pg. 482 #27 Solve the system by the method of substitution. x3 – y = 0 x – y = 0**Solution Example 3.7.1**Pg. 482 #27 Solving for y in the second equation yields: y = x Substituting for y in the first equation: x3 – x = 0 x(x2 – 1) = 0 x = 1, -1, and 0. This means that there are multiple solutions.**Example 4.7.1**Pg. 478 Example 5 Solve the system of equations graphically. y = ln(x) x + y = 1**Solution Example 4.7.1**Looking at the graph of these two equations shows us that the point of intersection is (1,0). Make sure to still check graphical solutions in the equations!**Activities (479)**1. Solve the system by substitution. 3x + 2y = 14 and x – 2y = 10. 2. Find all points of intersection. 4x – y – 5 = 0 and 4x2 – 8x + y + 5 = 0 3. Solve the system graphically. 3x + 2y = 6 and y = ln(x-1)**7.2 – Systems of Linear Equations in Two Variables**• The Method of Elimination • Graphical Interpretation of Two-Variable Systems • Application**7.2 – The Method of Elimination**• Find coefficients that when added will cancel each other out. • Remember you can multiply to make this happen! • Add the equations to eliminate one variable; solve the resulting equation. • Back-substitute into the first equation. • Check your solutions in both original equations.**Example 1.7.2**Pg. 486 Example 2 Solve the system of linear equations. 5x + 3y = 9 2x - 4y = 14**Solution Example 1.7.2**Pg. 486 Example 2 (3, -2)**Example 2.7.2**Pg. 491 #11 Solve the system of linear equations. 3r + 2s = 10 2r + 5s = 3**Solution Example 2.7.2**Pg. 491 #11 Multiplying: -2*(3r + 2s = 10) and 3*(2r + 5s = 3) Adding: 11s = -11 s = -1**Solution Example 2.7.2 (cont.)**Substituting: 3r + 2(-1) = 10 3r = 12 r = 4 Checking r = 4 and s = -1 into both original solutions: 12 – 2 = 10 √ 8 – 5 = 3 √**7.2 – Graphical Interpretation of Two-Variable Systems**There are three kinds of systems that you will encounter: • One solution • Infinitely many solutions (identical lines) • Equations with at least one solution are called consistent. • No solution (parallel lines) • These equations are inconsistent.**Example 3.7.2**Pg. 487 Example 3 The graphs of the three different types of systems.**Example 4.7.2**Pg. 491 #21 Solve the system of linear equations by elimination. 4x + 3y = 3 3x + 11y = 13**Solution Example 4.7.2**Pg. 491 #21 Multiplying: -11*(4x + 3y = 3) and 3*(3x + 11y = 13) Adding: -35x = 6 x = -(6/35)**Solution Example 4.7.2 (cont.)**Substituting: y = 43/35 Checking the solution shows that they work in both equations.**7.3 – Multivariable Linear Systems**• Row-Echelon Form and Back Substitution • Gaussian Elimination • Nonsquare Systems • Graphical Interpretation of Three-Variable Systems • Partial Fraction Decomposition and Other Applications**7.3 – Row-Echelon Form and Back Substitution**Row-echelon form is a “stair-step” pattern and all leading coefficients are 1. See Example 1 on pg. 495**Example 1.7.3**Pg. 505 #5 Use back-substitution to solve the system of linear equations. 2x – y + 5z = 16 y + 2z = 2 z = 2**Solution Example 1.7.3**Pg. 505 #5 Substitute z=6 into the second equation: y + 2(6) = 4 y = -8 Then substitute both z and y into the first: 2x + 8 + 30 = 24 2x = -14 x = -7**Solution Example 1.7.3 (cont.)**Pg. 505 #5 So the ordered triple that is the solution of this system of equations is: (-7, -8, 6) Check these solutions!**7.3 – Gaussian Elimination**Elementary Row Operations • Interchange two equations. • Multiply one of the equations by a nonzero constant. • Add a multiple of one equation to another equation.**7.3 – Gaussian Elimination**Use elementary row operations to get the equations in row-echelon form.**Example 2.7.3**Pg. 496 Example 2 Read the row operations that were used in the red boxes to the right of the equations.**Example 3.7.3**Pg. 506 #15 Solve the system of equations and check any solution. x + y + z = 6 2x – y + z = 3 3x - z = 0 Do on the board.**Solution Example 3.7.3**Pg. 506 #15 So the ordered triple that is the solution of this system of equations is: (1, 2, 3) Don’t forget to check these solutions.**Example 4.7.3**Pg. 497-498 Examples 3 and 4 Recall these special cases. The same rules apply as when we are in two variables.**7.3 – Nonsquare Systems**In a non-square system of equations, the number of equations differs from the number of variables.**Example 5.7.3**Pg. 499 Example 5 Pay close attention these are pretty hard.**Example 6.7.3**Pg. 506 #37 Solve the system of linear equations and check your solutions! x – 2y + 5z = 2 4x – z = 0**Solution Example 6.7.3**Pg. 506 #37 So the ordered triple that is the solution of this system of equations is: (2a, 21a – 1, 8a) Don’t forget to check these solutions!**7.3 – Graphs in Three Dimensions**Note the pictures of what you are finding for solutions on the bottom of pg. 500.**7.3 – Partial Fractions**• Distinct Linear Factors • Repeated Linear Factors**Example 7.7.3**Pg. 502 Example 6**Example 8.7.3**Now look at Example 7 on pg. 503 Note: When you get double roots you have to include the single factor and the squared factor.**Activities (499 & 502)**1. Solve the system: x – y + z = 4 x +3y – 2z = -3 3x + 2y + z = 5 2. Solve the system: x – 2y – z = -5 2x + y + z = 5 3 and 4 are partial fractions that I will put on the board.**7.4 – Matrices and Systems of Equations**• Matrices • Elementary Row Operations • Gaussian Elimination with Back Substitution • Gauss-Jordan Elimination

More Related