MATH 0322 Intermediate Algebra Unit 4 Review: Slope and Slope-Intercept

Slope and Slope-Intercept Form 𝑚 is ______ and 𝑏 is the __________ of the line. Graph is a straight non-vertical line. Formula for slope using two points on a line: 𝑚= , where 𝑥 2 − 𝑥 1 ≠0. Slope of a line is known as . Grade of a surface is its slope in % form. Slope of a line can be 𝑦=𝑚𝑥+𝑏 slope 𝑦-intercept 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1 If 𝑥 2 − 𝑥 1 =0, then slope is undefined. 𝑚= 𝑟𝑖𝑠𝑒 𝑟𝑢𝑛 It means 𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑦 𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑥 . >0, <0, 𝑜𝑟=0.

Slope and Slope-Intercept Form Slope of a line either increases, decreases, or is constant from left-right. If 𝑚>0, the line _________ left-right. Calculate slope to verify 𝑚 is positive. If 𝑚<0, the line _________ left-right. Calculate slope to verify 𝑚 is negative. 𝑚=______ Calculate slope. 𝑚=______ If 𝑚=0, the line is ________ left-right. 𝑥 𝑦 (−3,−2) (0,−1) increases 𝒙 𝟏 , 𝒚 𝟏 𝒙 𝟐 , 𝒚 𝟐 (−1)−(−2) ( 0 )−(−3) = 𝑚= 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1 = − − 1 3 𝑥 𝑦 (0,6) (2,0) decreases −3 𝑥 𝑦 (−1,5) (4,5) constant

Slope and Slope-Intercept Form Two lines can be either: Parallel equal slopes 𝒎 𝟏 = 𝒎 𝟐 but different y-intercepts 𝒃 or Perpendicular slopes are reciprocals with opposite signs, in other words 𝒎 𝟏 =−( 𝟏 𝒎 𝟐 ) or 𝒎 𝟐 =−( 𝟏 𝒎 𝟏 ) or Neither (0,−𝟓) (0,𝟔) 𝑥 𝑦 (−3,−6) (−3,5) 𝒎 𝟏 = 𝒚=𝒎𝒙+𝒃 𝒚= 𝟏 𝟑 𝒙+𝟔 𝒎 𝟐 = 𝒚= 𝟏 𝟑 𝒙−𝟓 𝒚=𝒎𝒙+𝒃 𝑥 𝑦 (0,2) (2,0) (0,−1) (1,0) 𝒚= 𝒎𝒙+𝒃 𝒎 𝟐 = −𝟐 𝒚=−𝟐𝒙+𝟐 𝒚= 𝟏 𝟐 𝒙−𝟏 𝟏 𝟐 𝒎 𝟏 = 𝒚=𝒎𝒙+𝒃

MATH 0322 Intermediate Algebra Unit 4 Section 3.5 Point-Slope Equation

Point-Slope Form What is the formula for slope, as discussed in class? What is the Slope-Intercept Form of a line, as discussed in class? Point-Slope Form of a non-vertical line: This equation needs: slope 𝑚 of the line and a point ( 𝑥 1 , 𝑦 1 ) on the line 𝑚= 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1 ; 𝑥 2 − 𝑥 1 ≠0 𝑦=𝑚𝑥+𝑏 𝑦− 𝑦 1 =𝑚 𝑥− 𝑥 1

Point-Slope Form Forms of lines and their names are: Equation Name 𝐴𝑥+𝐵𝑦=𝐶 𝑥=𝑎 𝑦=𝑏 𝑦=𝑚𝑥+𝑏 𝑦− 𝑦 1 =𝑚(𝑥− 𝑥 1 ) Standard Form Vertical Line Form Horizontal Line Form Slope-Intercept Form Point-Slope Form

Point-Slope Form Point-Slope Form 𝑦− 𝑦 1 =𝑚(𝑥− 𝑥 1 ) is used: to write equation of a line having slope 𝑚 and point ( 𝑥 1 , 𝑦 1 ), ** (Only substitute values for 𝑥 1 , 𝑦 1 , and 𝑚.) to write equation of a line 𝐿 1 that is Parallel or Perpendicular to another line 𝐿 2 , (Parallel: 𝑚 1 = 𝑚 2 , but lines have different points.) (Perpendicular: 𝑚 1 =− 1 𝑚 2 , lines have one common point.) to graph lines using 𝑥 1 , 𝑦 1 , and 𝑚.

P P P Point-Slope Form Practice: 𝑚=−2 𝑚=− 1 5 𝑚= 4 3 What slope is parallel to 𝑚=−2? What slope is perpendicular to 𝑚=5? What slope is perpendicular to 𝑚=− 3 4 ? P 𝑚=−2 𝑚=− 1 5 P 𝑚= 4 3 P

P P Point-Slope Form Practice: What slope of a line is: Parallel to 7𝑥+2𝑦=−6 Write in Slope-Intercept form. Perpendicular to 7𝑥+2𝑦=−6. Need negative reciprocal of slope. 7𝑥+2𝑦=−6 P 𝑚=− 7 2 2𝑦=−7𝑥−6 𝑦=− 7 2 𝑥−3 𝑚= 2 7 P

Point-Slope Form Practice: Write the equation of a line, first in Point-Slope form, then in Slope-Intercept form, having 𝑚=−2 and passing through (3,1). Write your form. Substitute values for 𝑥 1 , 𝑦 1 , 𝑚. Simplify to write in Point-Slope Form. Slope-Intercept Form. 𝑦− 𝑦 1 =𝑚(𝑥− 𝑥 1 ) 𝑦− = (𝑥−( )) 1 −2 3 P 𝑦−1=−2(𝑥−3) 𝑦−1=−2𝑥+6 P 𝑦= −2𝑥+7

Point-Slope Form Your turn: Write the equation of a line, first in Point-Slope form, then in Slope-Intercept form, having 𝑚=5 and passing through (1,2). Write your form. Substitute values for 𝑥 1 , 𝑦 1 , 𝑚. Simplify to write in Point-Slope Form. Slope-Intercept Form. 𝑦− 𝑦 1 =𝑚(𝑥− 𝑥 1 ) 𝑦− = (𝑥−( )) 2 5 1 P 𝑦−2=5(𝑥−1) P 𝑦=5𝑥−3

Point-Slope Form Practice: Write the equation of a line, first in Point-Slope form, then in Slope-Intercept form, passing through (4,−3) and (−2,6). Need a slope. Substitute values for 𝑥 1 , 𝑦 1 , 𝑚. Simplify to write in Point-Slope Form. Slope-Intercept Form. ( 𝒙 𝟏 , 𝒚 𝟏 ) ( 𝒙 𝟐 , 𝒚 𝟐 ) 𝑚= ( 𝑦 2 − 𝑦 1 ) ( 𝑥 2 − 𝑥 1 ) = ((6)−(−3)) ((−2)−(4)) = 9 −6 =− 3 2 − 3 2 𝑦− = (𝑥−( )) −3 4 P 𝑦+3=− 3 2 (𝑥−4) 𝑦+3=− 3 2 𝑥+6 P − 3 2 𝑥+3 𝑦=

Point-Slope Form Your turn: Write the equation of a line, first in Point-Slope form, then in Slope-Intercept form, passing through (−2,−1) and (−1,−6). Need a slope. Substitute values for 𝑥 1 , 𝑦 1 , 𝑚. Simplify to write in Point-Slope Form. Slope-Intercept Form. 𝑚= ( 𝑦 2 − 𝑦 1 ) ( 𝑥 2 − 𝑥 1 ) = ((−6)−(−1)) ((−1)−(−2)) =−5 𝑦− = (𝑥−( )) −1 −5 −2 P 𝑦+1=−5(𝑥+2) P 𝑦=−5𝑥−11

Point-Slope Form Practice: Find the equation of a line in Slope-Intercept Form that passes through the point (−3,4) and is parallel to 𝑥−3𝑦=7. 1) Use Point-Slope Form to begin. Need 𝑚 parallel to equation. and point ( 𝑥 1 , 𝑦 1 ). 2) Substitute 𝑚, 𝑥 1 , 𝑦 1 into Point-Slope Form, then simplify to Slope-Intercept Form. 𝑦− 𝑦 1 =𝑚(𝑥− 𝑥 1 ) 𝑥−3𝑦=7 (Solve for 𝑦.) −3𝑦=−𝑥+7 (−3,4) 𝑦= 1 3 𝑥− 7 3 𝑚= 1 3 𝑦−4 = 1 3 𝑥+1 𝑦− 𝑦 1 =𝑚(𝑥− 𝑥 1 ) 𝑦−(4) = 1 3 (𝑥−(−3)) P 𝑦= 1 3 𝑥+5 𝑦−4 = 1 3 (𝑥+3)

Point-Slope Form Your turn: Find the equation of a line in Slope-Intercept Form that passes through the point (−2,5) and is parallel to −3𝑥+𝑦=1. 1) Use Point-Slope Form to begin. Need slope 𝑚 parallel to −3𝑥+𝑦=1. and a point ( 𝑥 1 , 𝑦 1 ). 2) Substitute 𝑚, 𝑥 1 , 𝑦 1 into Point-Slope Form, then simplify to Slope-Intercept Form. 𝑦− 𝑦 1 =𝑚(𝑥− 𝑥 1 ) +3𝑥 +3𝑥 (Solve for 𝑦.) 𝑦=3𝑥+1 𝑚=3 (−2,5) 𝑦− 𝑦 1 =𝑚(𝑥− 𝑥 1 ) 𝑦−5 =3𝑥+6 P 𝑦−(5) =3(𝑥−(−2)) 𝑦=3𝑥+11 𝑦−5 =3(𝑥+2)

Complete and Practice HW 3.5

MATH 0322 Intermediate Algebra Unit 4 Solving Systems of Linear Equations (Graphing Method) Section: 4.1

Solving Systems of Linear Equations (Graphing Method) Chapter 4 studies techniques used to solve a System of Linear Equations. What is a System of Linear Equations? A set of 2 or more Linear Equations in 2 or more variables. Example: 𝑥+𝑦=−5 −2𝑥+𝑦=1 Your goal: Solve a Linear System & find its solution. Straight lines do 3 things, so 3 types of solutions. Intersect once Intersect many times Never intersect 𝑥 𝑦 Eqt#1 Eqt#2 𝑥 𝑦 Eqt#1 Eqt#2 𝑥 𝑦 Eqt#1 Eqt#2 One solution Infinitely many solutions No solution

Solving Systems of Linear Equations (Graphing Method) A solution to a Linear System must satisfy all equations in the system. A Linear System having infinitely many solutions is said to have dependent equations. The solution set for dependent equations is written in Set Notation. Example: (𝑥,𝑦) A Linear System having no solution is called an Inconsistent System. Its solution set is the called the empty set, ∅. Equation #1 or Equation #2 goes here.

Solving Systems of Linear Equations (Graphing Method) Practice: Determine whether (−3,1) is a solution to the given Linear System. −𝑥+4𝑦=7 2𝑥+𝑦=−1 Substitute to check Equation #1. Substitute to check Equation #2. Is (−3,1) a solution? P − −𝟑 +4 𝟏 = 7 O 2 −𝟑 + 𝟏 = −5 P Not a solution.

Solving Systems of Linear Equations (Graphing Method) Your turn: Determine whether (2,−5) is a solution to the given Linear System. 3𝑥+2𝑦=−4 −2𝑥−𝑦=1 Substitute to check Equation #1. Substitute to check Equation #2. Is (2,−5) a solution? P 3 𝟐 +2 −𝟓 = −4 P −2 𝟐 − −𝟓 = 1 P Yes

Solving Systems of Linear Equations (Graphing Method) Practice: Solve by the Graphing Method. −2𝑥+𝑦=7 2𝑥+𝑦=−1 Rewrite in Slope-Intercept form. Graph. Determine solution, then Check. (−2,3) P −2 −𝟐 + 𝟑 =7 ? 𝑦=2𝑥+7 P 2 −𝟐 + 𝟑 =−1 ? 𝑦=−2𝑥−1 𝑥 𝑦 10 −10 (−𝟐,𝟑) P

Solving Systems of Linear Equations (Graphing Method) Your turn: Solve by the Graphing Method. 3𝑥−𝑦=−2 −3𝑥+𝑦=−5 Rewrite in Slope-Intercept form. Graph. Determine solution. No Solution. 𝑦=3𝑥+2 𝑦=3𝑥−5 𝑥 𝑦 10 −10 Parallel lines P

Solving Systems of Linear Equations (Graphing Method) Without graphing, how can you tell if a Linear System in Slope-Intercept form has: One solution Infinitely many solutions No solution *Lines intersect once. 𝑚 1 ≠ 𝑚 2 *Lines are the same. 𝑚 1 = 𝑚 2 and 𝑏 1 = 𝑏 2 *Lines are parallel. 𝑚 1 = 𝑚 2 , but 𝑏 1 ≠ 𝑏 2

Solving Systems of Linear Equations (Graphing Method) Your turn: Without a graphing calculator, what can you determine from the Linear System below? −130𝑥−2𝑦=−1780 −357𝑥+3𝑦=546 A) It has only one solution? B) It has infinitely many solutions? C) It has no solution? Pros/Cons of Graphing Method Pros: works well if solution is an integer, it allows you to “see” if a solution exists. Cons: difficult to determine exact value of solution if it is very large or is a fraction or decimal. 𝑦=−65𝑥+890 𝒎 𝟏 ≠ 𝒎 𝟐 𝑦=119𝑥+182 P

Complete and Practice HW 4.1 Graphing Method Complete and Practice HW 4.1

MATH 0322 Intermediate Algebra Unit 4 Solving Systems of Linear Equations (Substitution Method) Section: 4.2

Solving Systems of Linear Equations (Substitution Method) This section introduces the Substitution Method: an algebraic method that substitutes the value of a variable from one equation into the other equation to solve a Linear System. To solve a Linear System using this method: Isolate a variable.(simplest one) Substitute into other equation, solve 1st solution. Back-substitute into an original equation, solve 2nd solution, then Check.

Solving Systems of Linear Equations (Substitution Method) 3 types of results: One solution Infinitely many solutions No solution 𝑥 𝑦 𝑥 𝑦 𝑥 𝑦 Graphing Method Step 2 & 3 produce single value for 𝑥, 𝑦. Step 2 produces 𝑐=𝑐. (𝑐 is a constant) Step 2 produces 𝑐≠𝑐. (𝑐 is a constant) SubstitutionMethod

Solving Systems of Linear Equations (Substitution Method) Answer the following questions for the given Linear System. 𝑥=2𝑦−7 −3𝑥=𝑦+1 Which equation has the simplest variable to isolate in Step 1 of our strategy? Which variable is that? P Equation #1 P 𝑥-variable

Solving Systems of Linear Equations (Substitution Method) Answer the following questions for the given Linear System. 4𝑥−𝑦=1 −𝑥+𝑦=−6 Which equation has the simplest variable to isolate in Step 1 of our strategy? Which variable is that? P Equation #2 P 𝑦-variable

Solving Systems of Linear Equations (Substitution Method) Answer the following questions for the given Linear System. 4𝑥+2𝑦=−10 3𝑥−5𝑦=12 Which equation has the simplest variable to isolate in Step 1 of our strategy? Which variable is that? Why? P Equation #1 P 𝑦-variable Solving for 𝑦 does not produce fractions. Eqt#1 for 𝒚 4𝑥+2𝑦=−10 2𝑦=−4𝑥−10 𝑦=−2𝑥−5 𝑦=− 1 2 𝑥− 5 2 4𝑥+2𝑦=−10 4𝑥=−2𝑦−10 Eqt#1 for 𝒙

Solving Systems of Linear Equations (Substitution Method) Practice: Solve by the Substitution Method. 𝑥=2𝑦−7 −2𝑥=−𝑦+5 Isolate a variable. Substitute into other equation, then solve. Back-substitute to solve, then check. 𝑥=2𝑦−7 Eqt#𝟏 −2 =−𝑦+5 −2𝑥=−𝑦+5 Eqt#𝟐 2𝑦−7 −4𝑦+14=−𝑦+5 −3𝑦+14=5 −3𝑦=−9 ? 𝑦= 3 𝑥=2( )−7 Eqt#𝟏 3 𝑥=6−7 ? 𝑥=−1

Solving Systems of Linear Equations (Substitution Method) Your turn: Solve by Substitution Method. −5𝑦=4𝑥−1 𝑦=−𝑥+3 Isolate a variable. Substitute into other equation, then solve. Back-substitute to solve, then check. 𝑦=−𝑥+3 Eqt#𝟐 −5 =4𝑥−1 −5𝑦=4𝑥−1 Eqt#𝟏 −𝑥+3 5𝑥−15=4𝑥−1 𝑥−15=−1 ? 𝑥= 14 𝑦=− +3 Eqt#𝟐 14 ? 𝑦=−11

Solving Systems of Linear Equations (Substitution Method) Your turn: Solve by Substitution Method. −4𝑥+𝑦=−11 2𝑥−3𝑦=5 Isolate a variable. Substitute into other equation, then solve. −4𝑥+𝑦=−11 Eqt#𝟏 𝑦=4𝑥−11 2𝑥−3( )=5 Eqt#𝟏 4𝑥−11 2𝑥−12𝑥+33=5 −10𝑥+33=5 −10𝑥=−28 14 5 𝑥=

Solving Systems of Linear Equations (Substitution Method) Your turn: Solve by Substitution Method. −4𝑥+𝑦=−11 2𝑥−3𝑦=5 Back-substitute to solve, then check. 14 5 −4( )+𝑦=−11 Eqt#𝟏 − 56 5 + 𝑦 =−11 𝟏 (𝟓) (𝟓) (𝟓) 𝟏 −56+5𝑦=−55 5𝑦=1 ? ? 𝑥= 14 5 1 5 𝑦=

Solving Systems of Linear Equations (Substitution Method) Your turn: Solve by Substitution Method. −4𝑥+𝑦=−11 2𝑥−3𝑦=5 Back-substitute to solve, then check. P P 𝑥= 14 5 𝑦= 1 5 14 5 1 5 14 5 1 5 −4( )+( )=−11 Eqt#𝟏 2 −3( )=5 Eqt#𝟐 − 56 5 + 1 5 =−11 28 5 − 3 5 =5 − 55 5 =−11 25 5 =5 −11=−11 5=5

Solving Systems of Linear Equations (Substitution Method) Your turn: Solve by Substitution Method. 6𝑥+2𝑦=7 𝑦−1=−3𝑥 Isolate a variable. Substitute into other equation, then solve. 𝑦−1=−3𝑥 Eqt#𝟐 𝑦=−3𝑥+1 6𝑥+2( )=7 Eqt#𝟏 −3𝑥+1 6𝑥−6𝑥+2=7 2=7 𝑐≠𝑐 P System has no solution.

Complete and Practice HW 4.2 Substitution Method Complete and Practice HW 4.2

MATH 0322 Intermediate Algebra Unit 4 Solving Systems of Linear Equations (Addition Method) Section: 4.3

Solving Systems of Linear Equations (Addition Method) This section introduces the Addition Method: an algebraic method that eliminates a variable by adding opposites to solve a Linear System. The goal of this method is to reduce the Linear System to only 1 equation in 1 variable so that a solution may be reached. To do this: the system must be in Standard Form, and have opposite 𝑥 or 𝑦 terms, so equations can be added to combine like terms into 1 equation in 1 variable. (opposite 𝑥 or 𝑦 terms are eliminated to zero.)

Solving Systems of Linear Equations (Addition Method) How to solve a Linear System using this method: (*Make sure system is in Standard Form.*) Reduce the system: add equations to eliminate opposite 𝑥 or 𝑦 terms. *Multiply equation(s) by a constant to form opposites if needed. Solve remaining equation for 1st solution. Back-substitute into an original equation, solve for 2nd solution, then Check.

Solving Systems of Linear Equations (Addition Method) 3 types of results: One solution Infinitely many solutions No solution 𝑥 𝑦 𝑥 𝑦 𝑥 𝑦 Graphing Method Step 2 & 3 produce single value for 𝑥, 𝑦. Step 2 produces 𝑐=𝑐. (𝑐 is a constant) Step 2 produces 𝑐≠𝑐. (𝑐 is a constant) SubstitutionMethod Step 2 & 3 produce single value for 𝑥, 𝑦. Step 1 produces 0=0. Step 1 produces 0≠0. Addition Method

Solving Systems of Linear Equations (Addition Method) Answer the following questions for the given Linear System. −3𝑥−2𝑦=−7 −3𝑥+2𝑦=1 Is the Linear System in the proper form to begin Step 1 of our strategy? What form is that? For Step 1 of our strategy, which variables, 𝑥 or 𝑦, are opposites? P Yes P Standard Form 𝐴𝑥+𝐵𝑦=𝐶 P 𝑦-variable

Solving Systems of Linear Equations (Addition Method) Answer the following questions for the given Linear System. Is the Linear System in the proper form to begin Step 1 of our strategy? Which equation must be rewritten? For Step 1 of our strategy, which variable can be made to be opposites easier, 𝑥 or 𝑦? How? 3𝑦=−𝑥+1 2𝑥+5𝑦=−4 (−𝟐) 𝑥+3𝑦=1 2𝑥+5𝑦=−4 −2𝑥−6𝑦=−2 2𝑥+5𝑦=−4 P No P Equation #1 P 𝑥-variable P Multiply Equation#1 by −2.

Solving Systems of Linear Equations (Addition Method) Practice: Solve by the Addition Method. 𝑥−𝑦=−7 −2𝑥=−𝑦+5 Standard Form? Reduce system: opposites 𝑦-terms, add equations. Solve for 1st solution. Back-substitute for 2nd solution, then check. 𝑥−𝑦=−7 −2𝑥+𝑦=5 Step 1) + −𝑥=−2 P Step 2) 𝑥=2 Step 3) −𝑦=−7 Eqt#𝟏 2 −𝑦=−9 P 𝑦=9 ( )−( )=−7 −2 =− +5 𝟐 𝟗 𝟐 𝟗

Solving Systems of Linear Equations (Addition Method) Practice: Solve by the Addition Method. 4𝑥=36+8𝑦 𝑥−2𝑦=9 Standard Form? Reduce system: form opposite 𝑥-terms, add equations. 4𝑥−8𝑦=36 𝑥−2𝑦=9 (−𝟒) 4𝑥−8𝑦=36 −4𝑥+8𝑦=−36 Step 1) + 0=0 P Infinitely many solutions

Solving Systems of Linear Equations (Addition Method) Practice: Solve by the Addition Method. −4𝑥+𝑦=−11 2𝑥−3𝑦=5 Standard Form? Reduce system: make opposite 𝑦-terms, add equations. Solve for 1st solution. Back-substitute for 2nd solution, then check. (𝟑) −12𝑥+3𝑦=−33 2𝑥−3𝑦=5 Step 1) + Yes −10𝑥=−28 P Step 2) 𝑥= 14 5 14 5 Step 3) 2 −3𝑦=5 Eqt#𝟐 𝟏 28 5 − 3𝑦 =5 (𝟓) (𝟓) (𝟓) 𝟏 28−15𝑦=25 −15𝑦=−3 P 𝑦= 1 5

Complete and Practice HW 4.3 Addition Method Complete and Practice HW 4.3

MATH 0322 Intermediate Algebra Unit 4 Appendix B: Matrices

Matrices A matrix is a rectangular array of numbers. It is a short way of writing a System of Equations that uses only coefficient numbers in each equation. 𝑥+3𝑦=8 𝑥−2𝑦=7 → 1 3 8 1 −2 7 The numbers in a matrix are called elements or entries. An augmented matrix is a matrix that uses a vertical bar to separate variable coefficients from constants. 𝑥+3𝑦=8 𝑥−2𝑦=7 → 1 3 1 −2 8 7 System of Equations Matrix

Matrices The Gaussian Method is used to find the solution to a system of equations using its augmented matrix. It uses 3 Row Operations to eliminate entries to zeros and obtain Row-Echelon form. Row-Echelon form has 1’s along the diagonal from upper-left to lower-right, with zeros below the diagonal entries. Example of a matrix in Row-Echelon form: 1 3 0 1 8 3 1 1 2 0 1 2 0 0 1 19 13 5

Matrices Row-Echelon form: 1 3 0 1 8 3 1 3 0 0 8 0 1 3 0 0 8 3 If last row looks like this, System has One Solution. System is Consistent. Equations are Independent. System has Infinitely Many Solutions. Equations are Dependent. System has No Solution. System is Inconsistent. 1 3 0 1 8 3 1 3 0 0 8 0 1 3 0 0 8 3

Matrices Practice: Is this matrix in Row-Echelon form as discussed in class? Yes No 1 1 0 2 6 1 1 −2 0 0 1 8 0 0 1 4 −1 0 P P

Matrices The 3 Row Operations used are: Interchange any two rows. 𝑅 𝑖 ↔ 𝑅 𝑗 (This is like switching two equations in the system.) Multiply row entries by a nonzero number 𝑘. 𝑘 𝑅 𝑖 (This is like multiplying an equation by a nonzero number.) Add a nonzero multiple of a row 𝑅 𝑖 to another row 𝑅 𝑗 . 𝑘𝑅 𝑖 + 𝑅 𝑗 (This is like multiplying an equation by a nonzero number and adding it to another equation.)

Matrices To solve a System of Equations using matrices: Write augmented matrix. Apply Row Operations to get Row-Echelon form. Rewrite rows as equations, then Back-Substitute to determine solution set.

Matrices Practice: Perform the given Row Operations, then rewrite the new matrix. 2 4 −1 5 −6 3 𝑅 1 ↔ 𝑅 2 1 2 𝑅 1 Multiply 𝑅 1 by 1 2 . Interchange 𝑅 1 & 𝑅 2 . −1 5 2 4 3 −6 P 𝟏 𝟐 ( ) 𝟏 𝟐 𝟏 𝟐 2 4 −1 5 −6 3 1 2 −1 5 −3 3 P

Matrices Practice: Perform the given Row Operation, then rewrite the new matrix. 1 2 −2 3 −4 1 2 𝑅 1 + 𝑅 2 Multiply 𝑅 1 by 2, then add to 𝑅 2 . 1 2 −2 3 −4 1 𝟐 + 𝟐 + 𝟐 + 𝟏 𝟐 −𝟒 𝟎 𝟕 −𝟕 P 1 2 0 7 −4 −7

Matrices Your turn: Perform the given Row Operations, then rewrite the new matrix. 1 −2 5 −1 2 −8 −5 𝑅 1 + 𝑅 2 1 −2 0 9 8 −18 P

P Matrices Practice: Solve the System of Equations using matrices. 2𝑥+3𝑦=7 5𝑥+2𝑦=1 Write augmented matrix. 𝑥 𝑦 2 3 5 2 7 1 2 3 5 2 7 1 2 3 7 P 5 2 1

Matrices Apply Row-Operations to get Row-Echelon form. 2 3 5 2 7 1 2 3 5 2 7 1 You need a 1 in 𝐶 1 to start the diagonal. Do −2 𝑅 1 + 𝑅 2 . Now do 𝑅 1 ↔ 𝑅 2 . 2 3 5 2 7 1 −𝟐 + −𝟐 + −𝟐 + 𝟐 𝟑 𝟕 2 3 1 −4 7 −13 1 −4 2 3 −13 7

Matrices Apply Row-Operations to get Row-Echelon form. 1 −4 2 3 −13 7 1 −4 2 3 −13 7 To get a 0 in place of 2 in 𝐶 1 , need to add −2 to it. Do −2 𝑅 1 + 𝑅 2 . What’s next? (Now you need a 1 where entry 11 is.) 1 −4 2 3 −13 7 −𝟐 + −𝟐 + −𝟐 + 𝟏 −𝟒 −𝟏𝟑 =𝟎 =𝟏𝟏 =𝟑𝟑 1 −4 0 11 −13 33

Matrices ( ) …continued: Row-Operations. 1 −4 0 11 −13 33 1 −4 0 11 −13 33 To get a 1 in place of 11 in 𝑅 2 , do 1 11 𝑅 2 . Row-Echelon form. 1 −4 0 11 −13 33 1 11 1 11 1 11 ( ) =𝟎 =𝟏 =𝟑 1 −4 0 1 −13 3

Matrices Rewrite rows as equations, then back-substitute to determine solution set. 1 −4 0 1 −13 3 Solve 𝑅 2 equation for 𝑦-value. 𝑦=3 Back-substitute 𝑦-value into 𝑅 1 equation, solve for 𝑥-value. 𝑥−4𝑦=−13 Type solution as an ordered pair: 𝑥−4𝑦=−13 𝑦=3 𝑥−4( )=−13 3 𝑥−12=−13 𝑥=−1 P (−1,3)

Matrices Same process followed to solve a system that contains 3 equations and 3 variables like below: 𝑥+𝑦−𝑧=−2 4𝑥−𝑦+𝑧=17 −𝑥+3𝑦−2𝑧=1 Write augmented matrix. 1 1 −1 4 −1 1 −1 3 −2 −2 17 1 Rewrite rows as equations, back-substitute to determine solution. Solution set: (3,4,19) 2) Apply Row Operations to get Row-Echelon form. 1 − 1 4 1 4 0 1 − 7 11 0 0 1 17 4 21 11 19

Appendix B: Matrices Complete and Practice

Point-Slope Form 𝑦− 𝑦 1 =𝑚 𝑥− 𝑥 1 What is the formula for slope, as discussed in class? 𝑚= 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1 , Multiply both sides by the denominator and simplify. Point-Slope Form of equation of a non-vertical line: with 𝑥 2 − 𝑥 1 ≠ 𝑚 𝑥 2 − 𝑥 1 = 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1 𝑥 2 − 𝑥 1 1 1 𝑚 𝑥 2 − 𝑥 1 = 𝑦 2 − 𝑦 1 With 𝑥 2 , 𝑦 2 fixed, this results in….. 𝑦 2 − 𝑦 1 =𝑚 𝑥 2 − 𝑥 1 𝑦− 𝑦 1 =𝑚 𝑥− 𝑥 1