MATH 0322 Intermediate Algebra Unit 4

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 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

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

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

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

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.

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.

Your turn: Determine whether (2,−5) is a solution to the given Linear System 𝑥+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

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

Your turn: Solve by the Graphing Method 𝑥−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

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

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

Graphing Method Complete and Practice HW 4.1

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.

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

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

Answer the following questions for the given Linear System 𝑥−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

Answer the following questions for the given Linear System 𝑥+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

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 − =−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

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 − =4𝑥−1 −5𝑦=4𝑥−1 Eqt#𝟏 −𝑥+3 5𝑥−15=4𝑥−1 𝑥−15=−1 ? 𝑥= 14 𝑦=− Eqt#𝟐 14 ? 𝑦=−11

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=

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 − 𝑦 =−11 𝟏 (𝟓) (𝟓) (𝟓) 𝟏 −56+5𝑦=−55 5𝑦=1 ? ? 𝑥= 14 5 1 5 𝑦=

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 −3( )=5 Eqt#𝟐 − =−11 28 5 − 3 5 =5 − 55 5 =−11 25 5 =5 −11=−11 5=5

Your turn: Solve by Substitution Method 𝑥+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.

Substitution Method Complete and Practice HW 4.2

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.)

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.

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

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

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.

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

Practice: Solve by the Addition Method 𝑥=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

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) −3𝑦=5 Eqt#𝟐 𝟏 28 5 − 3𝑦 =5 (𝟓) (𝟓) (𝟓) 𝟏 28−15𝑦=25 −15𝑦=−3 P 𝑦= 1 5

Addition Method Complete and Practice HW 4.3

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= → −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= → − 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:

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.

Matrices Practice: Is this matrix in Row-Echelon form as discussed in class? Yes No 1 − −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. − − 𝑅 1 ↔ 𝑅 2 1 2 𝑅 1 Multiply 𝑅 1 by Interchange 𝑅 1 & 𝑅 2 . − −6 P 𝟏 𝟐 ( ) 𝟏 𝟐 𝟏 𝟐 − − 1 2 −1 5 −3 3 P

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

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

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

Matrices Apply Row-Operations to get Row-Echelon form. 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 −𝟐 −𝟐 + 𝟐 𝟑 𝟕 −4 7 −13 1 − −13 7

Matrices Apply Row-Operations to get Row-Echelon form. 1 −4 2 3 −13 7
1 − −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.) − − −𝟐 −𝟐 −𝟐 𝟏 −𝟒 −𝟏𝟑 =𝟎 =𝟏𝟏 =𝟑𝟑 1 − −13 33

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

Matrices Rewrite rows as equations, then back-substitute to determine solution set. 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 − −

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 𝑥 2 − 𝑥 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

MATH 0322 Intermediate Algebra Unit 4

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MATH 0322 Intermediate Algebra Unit 4

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