7.1 Solving Linear Systems by Graphing

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1 7.1 Solving Linear Sstems b Graphing Objectives: Learn how to solve a sstem of linear equations b graphing Learn how to model a real-life situation using a sstem of linear equations With an equation, an point on the line (x, ) is called a solution to the equation. With or more equations, an point that is true for both equations is also a solution to the sstem. Is (, 1) the solution to the sstem? 3x + = x + 3 = 5 Plug in the (x, ) values 3() + ( 1) = () + 3( 1) = 5 6 = 3 = 5 = 5 = 5 Yes Yes Check b graphing: write in slope-intercept form first 8 6 3x + = x + 3 = 5 = 3x + 3 = x 5 = 3 x + = 1 3 x x (, 1) Mr. Noes, Akimel A-al Middle School 1 Heath Algebra 1 - An Integrated Approach 6 8

2 7.1 Solving Linear Sstems b Graphing Graph and check: = x + 1 = x + 5 Check: Substitute solution into each equation (, 3) Solution 3 = = () = 3 3 = 3 Yes Yes x Solve: x+ = x = Rewrite x + = in slope-intercept form: = x + Rewrite x = in slope-intercept form: = x check: Substitute (, 0) into each equation = x + 0 = () + 0 = 0 = x 0 = 0 = (, 0) x Solution (, 0) Mr. Noes, Akimel A-al Middle School Heath Algebra 1 - An Integrated Approach

3 7.1 Solving Linear Sstems b Graphing Examples: You invest $9,000 in two different accounts. One account pas 5% interest and the other pas 6% interest. If ou earn $510 in total interest, how much did ou invest in each account? Equation #1:.05x +.06 = 510 Equation #: x + = 9,000 Solve b graphing (find the x-, -intercepts) When x = 0 When = 0 x + = 9,000 x + = 9,000 = 9,000 x = 9, = x = 510 = 8,500 x = 10, x Graph is upper right quadrant, crossing at (3000, 6000) Answer: $3,000 is invested at 5% and $6,000 is invested at 6%. Check answer to make sure! Mr. Noes, Akimel A-al Middle School 3 Heath Algebra 1 - An Integrated Approach

4 7. Solving Linear Sstems b Substitution Objectives: Learn how to use substitution to solve a linear equation Learn how to model real-life situations using a sstem of linear equations Basic Steps: 1) Solve one of the equations for one of its variables ) Substitute that expression into the other equation and solve for the other variable 3) Substitute this value into the revised first equation and solve ) Check the solution pair in each of the original equations Example: x + = 1 x + = 1) Solve Equation #1 for : Equation #1: x + = 1 = x + 1 (call this Revised Equation #1) ) Substitute the expression found for from equation #1 into equation #: Equation #: x + = x + (x + 1) = x + x + 1 = 3x + 1 = 1 1 3x = x = 1 Mr. Noes, Akimel A-al Middle School 1 Heath Algebra 1 - An Integrated Approach

5 7. Solving Linear Sstems b Substitution 3) Substitute the solution found for x back into Revised Equation #1 and solve for : Revised Equation #1: = x + 1 = ( 1) + 1 = 0 solution: ( 1, 0) ) Check b substituting the solution into both equations: Equation #1 Equation # x + = 1 x + = ( 1) + (0) = 1 ( 1) + (0) = = = 1 = 1 = Since the statements both equations make are true when substituting the values ( 1, 0) in each, we know point ( 1, 0) is a solution Graph to check: ( 1, 0) x Mr. Noes, Akimel A-al Middle School Heath Algebra 1 - An Integrated Approach

6 7. Solving Linear Sstems b Substitution Real-Life Example: Dinner at China Buffet Adult Dinner price: $11.95 Child Dinner price: $6.95 If our bill for 6 people is $61.70, how man adults and children were in our dinner part? Verbal Model: Labels: EQ #1: # of Adults + # of Children = Total # People in Dinner Part EQ #: Price per Adult # of Adults + Price per Child # of Children = Total Bill Let A = # of adults C = # of children Adult Dinner price = $11.95 Child Dinner price = $6.95 Total Bill = $61.70 Total # People in Dinner Part = 6 Algebraic Model: EQ #1: A + C = 6 EQ #: A C = Step 1: Solve Equation #1 for A EQ #1: A + C = 6 A = C + 6 (Revised Equation #1) Mr. Noes, Akimel A-al Middle School 3 Heath Algebra 1 - An Integrated Approach

7 7. Solving Linear Sstems b Substitution Step : Substitute the expression found for A into Equation # and solve for C EQ #: A C = ( C + 6) C = C C = C = C = 10 C = Step 3: Substitute the solution found for C into Revised Equation #1 and solve for A Revised EQ #1: A = C + 6 A = () + 6 A = Step : Check b substituting the solution into both equations: Equation #1 Equation # A + C = A C = () + () = () () = = = = Mr. Noes, Akimel A-al Middle School Heath Algebra 1 - An Integrated Approach

8 7.3 Solving Linear Sstems b Linear Combinations Objectives: Learn to use linear combinations to solve a linear sstem Learn how to model a real-life situation using a sstem of linear equations Linear Combinations: A Third wa of Solving a Linear Sstem of Equations Basic Steps: 1) Arrange equations with like terms in columns. ) Stud the x or coefficients. Multipl one or both equations b an appropriate number to obtain new coefficients for x or that are opposites. 3) Add the two equations together. The sum should have eliminated one of the variables. Solve for the remaining variable ) Substitute the value in Step 3 into either of the original equations and solve for the other variable. You now have an ordered pair 5) Check the solution in each original equation. If both are true, ou have a solution Mr. Noes, Akimel A-al Middle School 1 Heath Algebra 1 - An Integrated Approach

9 7.3 Solving Linear Sstems b Linear Combinations Example #1: Equation #1: x + 3 = 1 Equation #: x 3 = 1 (since 3 and 3 are opposites, add equations) 6x = 6 6 x = 1 3 x = 1 3 (substitute 1 3 back into one of the equations and solve for ) Equation #1: x + 3 = 1 (Substituting into Equation #1) 1 ( 3 ) + 3 = = 1 3 = 1 3 = 1 9 Solution: 1 (, 1) 3 9 Check: Substitute the solution into the other Equation Equation #: x 3 = 1 1 ( 3 ) 3 1 ( 9) = = 1 1 = 1 Mr. Noes, Akimel A-al Middle School Heath Algebra 1 - An Integrated Approach

10 7.3 Solving Linear Sstems b Linear Combinations Example #: Equation #1: 3x + 5 = 6 Equation #: x + = 5 (won t combine as currentl written) Tr to get rid of either x or Multipl top equation b and bottom equation b 3 to end up with 1x and 1x. When added together, the x s will cancel each other out. (3x + 5) = (6) 1x + 0 = 3( x + ) = 3(5) 1x + 6 = 15 6 = 39 = 3 Solve for x: Substitute the solution found for into either one of the equations and solve for x: Equation #1: 3x + 5 = 6 3x + 3 5( ) = 6 3x + 15 = 6 3x = 3 x = 1 Solution: ( 1, 3) Mr. Noes, Akimel A-al Middle School 3 Heath Algebra 1 - An Integrated Approach

11 7.3 Solving Linear Sstems b Linear Combinations Check: Substitute the solution into the other Equation Equation #: x + = 5 ( 1 ) + 3 ( ) + 3 = 5 5 = 5 = 5 Helpful Hint: When there are fractions in the equations, clear the fractions first. Choosing when to use each method: Graph: to approximate a solution, check a solution or show visual. Substitution or combination: for exact solution Substitution: when one variable has 1 or 1 for the coefficient Combination: when coefficients are larger Mr. Noes, Akimel A-al Middle School Heath Algebra 1 - An Integrated Approach

12 7. Problem Solving Using Linear Sstems Objective: Learn how to write and use a linear sstem as a real-life model Use an of the 3 Methods for Solving Linear Sstems of Equations: 1. Graph. Substitute 3. Combine Example 1: Sam sells gloves at a department store. At the end of the da, Sam sold 0 pair of gloves. Of the 3 tpes, Sam onl sold knit gloves at $7.95 and cloth gloves at $ None of the leather gloves at $7.50 sold. If Sam sold a total of $1.5 for the da, how man gloves of each tpe did Sam sell? k = knit gloves c = cloth gloves EQ #1: k + c = 0 EQ #: 7.95k c = 1.5 Use the Substitution Method for this problem Revised EQ #1: k = 0 c EQ #: 7.95(0 c) c = c c = c = c = 53.5 c = 15 Mr. Noes, Akimel A-al Middle School 1 Heath Algebra 1 - An Integrated Approach

13 7. Problem Solving Using Linear Sstems Revised EQ #1: k = 0 c k = 0 (15) k = 5 You sold 15 cloth gloves and 5 knit gloves. Check: EQ #: 7.95k c = (5) (15) = = = 1.5 Example : Two planes leave the same airport but fl in opposite directions. If plane # flies 50 mph faster than plane #1, but starts one-half hour later, how fast is each plane fling when after hours the are 1,85 miles apart? *remember D = rt Verbal Model: EQ #1: Plane # speed = Plane #1 speed + 50 mph EQ #: Plane #1 speed time + Plane # speed time = Distance Apart Labels: plane #1 s speed = x plane # s speed = x + 50 plane #1 s time = t = hours plane # s time = 0.5 = 1.5 hours (remember, 30 minutes is 0.5 hours) Mr. Noes, Akimel A-al Middle School Heath Algebra 1 - An Integrated Approach

14 7. Problem Solving Using Linear Sstems Algebraic Model: EQ #1: plane # speed = x + 50 EQ #: x + 1.5(x + 50) = 185 x + 1.5x + 75 = x + 75 = x = 1750 x = 500 x + 50 = 550 Plane 1 s speed is 500 mph Plane s speed is 550 mph Check: x + 1.5(x + 50) = 185 (500) + 1.5( ) = (550) = = = 185 Mr. Noes, Akimel A-al Middle School 3 Heath Algebra 1 - An Integrated Approach

15 7. Problem Solving Using Linear Sstems Example 3: (See problem # on page 37 of textbook) x + 5 = 5 x = x = ( 5 + 5) = = = = 8 x + 5 = 5 x + 5(8) = 5 x + 0 = x = 5 Solution: (5, 8) Check: x + 5 = 5 x = (5) + 5(8) = 5 (5) (8) = = = 5 = 5 = Graph: = mx + b x + 5 = 5 x = x x x x 5 = x + 5 = x = 1 5 x + 9 = x (5, 8) x Mr. Noes, Akimel A-al Middle School Heath Algebra 1 - An Integrated Approach

16 7.5 Special Tpes of Linear Sstems Objectives: Learn how to visualize the solutions possible for linear sstems Learn how to identif a linear sstem as having man solutions A farmer keeps track of his cows and chickens b counting legs and heads. If he counts 78 legs and 35 heads, how man cows and chickens does he have? How man was can ou solve this? Solve b using sstems of equations, substitution method. a = chickens c = cows a + c = 35 c = 35 a a + c = 78 a + (35 a) = 78 a + 10 a = 78 a = 6 a = c = 35 c = There are cows and 31 chickens. Mr. Noes, Akimel A-al Middle School 1 Heath Algebra 1 - An Integrated Approach

17 7.5 Special Tpes of Linear Sstems Special Linear Sstems: Intersecting Parallel Same Line one solution no solution (an ordered pair) (a false statement) Man (infinite) solutions (the same line) (x, ) 0 = 0 = 0 6 = 6 x x x When ou solve each sstem, ou either get an ordered pair, a false statement or both sides are equal. Example #1: Solve b substitution or combination then graph. 3x = 3 6x + = 6 Tr combination -- Multipl top equation b 6x = 6 6x = 6 0 = 0 What does this mean? It means that there are man solutions because each equation describes the same line. Mr. Noes, Akimel A-al Middle School Heath Algebra 1 - An Integrated Approach

18 7.5 Special Tpes of Linear Sstems Graph to check: Change to = mx + b EQ #1: = 3 x 3 6 EQ #: = 3 x 3 What do ou notice? same slope same intercept same line! MANY SOLUTIONS 6 6 x 6 Example #: Solve b substitution or combination then graph. x + = 8 x + = 1 Tr substitution: Revised EQ #1: = x + 8 EQ #: x + ( x + 8) = 1 x x + 8 = 1 8 = 1 (This statement clearl isn t true!) No solution, this sstem of equations describe parallel lines. Mr. Noes, Akimel A-al Middle School 3 Heath Algebra 1 - An Integrated Approach

19 7.5 Special Tpes of Linear Sstems Change each to slope-intercept form: Revised EQ #1: = x + 8 Revised EQ #: = x 1 What do ou notice? same slope different -intercept parallel lines! NO SOLUTIONS x 6 8 Given a graph, write a sstem of equations to match = x 1 = x + = x = x 6 6 = 3 x + x = x 6 same slope same -intercept same line! MANY SOLUTIONS 6 6 x 6 same slope different -intercept parallel lines! NO SOLUTIONS 6 6 x 6 different slope different intercepts intersecting lines! ONE SOLUTION Mr. Noes, Akimel A-al Middle School Heath Algebra 1 - An Integrated Approach

20 7.6 Solving Sstems of Linear Inequalities Objectives: Learn how to solve a sstem of linear inequalities b graphing Learn how to model a real-life situation using a sstem of linear inequalities When Graphing Inequalities... Greater Than (>) and Less Than (>) smbols get dashed ( ) lines Greater Than or Equal To ( ) and Less Than or Equal To ( ) smbols get solid ( ) lines Pick a point and shade true side Graph the following sstem of Inequalities: < x 1 > x Hint: Use colored pencils when shading inequalities Graph < graph x 1 graph > x dotted line where =, solid line at x = 1, shade shade down right Dashed line with a - intercept at, and slope 1. Shade up and left x x x 3 5 Mr. Noes, Akimel A-al Middle School 1 Heath Algebra 1 - An Integrated Approach

21 7.6 Solving Sstems of Linear Inequalities Combining all three graphs, we find an area where all three regions intersect. This area is the solution to the sstem of inequalities. x Solution is the area where the inequalities intersect Graph and Shade: (ou ma want to use colored pencils) Find the solution for this sstem: 0 x + 1 x 0 x Mr. Noes, Akimel A-al Middle School Heath Algebra 1 - An Integrated Approach

22 7.6 Solving Sstems of Linear Inequalities Solve: 1 x + 5 x + 1 x + 5 x x Mr. Noes, Akimel A-al Middle School 3 Heath Algebra 1 - An Integrated Approach

23 7.7 Exploring Data: Linear Programming Objectives: Learn how to solve a linear programming problem Learn how to model a real-life situation using linear programming The process of Optimization means finding the minimum and/or maximum value of some quantit... one method for doing this is called Linear Programming. A Linear Programming problem consists of two parts: 1. An Objective Quantit - the numeric value of a quantit (expressed in terms of an equation) we want to minimize and/or maximize. Constraints limiting factors expressed in terms of a sstem of linear inequalities To find the minimum or maximum value of the objective quantit, find all the vertices of the graph of the constraint inequalities and then evaluate the objective quantit at each vertex. The smallest value is the minimum, and the largest value is the maximum. Example 1: Find the minimum value and maximum value of C = 3x + Subject to the following constraints: x 0 0 x + Objective Quantit Constraints Mr. Noes, Akimel A-al Middle School 1 Heath Algebra 1 - An Integrated Approach

24 7.7 Exploring Data: Linear Programming Solution: The three vertices are (0, 0), (, 0), and (0, ). To find the minimum and maximum values of C, evaluate C = 3x + at each of the three vertices: At (0, 0): C = 3(0) + (0) = 0 (Minimum Value of C) At (, 0): C = 3() + (0) = 1 (Maximum Value of C) At (0, ): C = 3(0) + () = 8 The minimum value of C is 0 and it occurs when x = 0 and = 0. The maximum value is 1, when x = and = 0. When evaluating C at other points in the graph, the value of C will alwas fall between 0 and and 1. Example : Finding the Maximum Profit Suppose ou own a skateboard manufacturing compan that makes a premium skateboard b using two ver different processes. The hours of skilled labor and machine time per skateboard can be found in the matrix below. You can use up to 1796 hours of skilled labor and up to 3500 hours of machine time. How man skateboards should be made b each process to maximize our profits? Process Assembl hours A B Skilled labor 3 1 Machine Time 1 3 Skateboards made b Process A cost $35 each, and skateboards made b process B cost $15 each. Mr. Noes, Akimel A-al Middle School Heath Algebra 1 - An Integrated Approach

25 7.7 Exploring Data: Linear Programming Solution: Let a and b represent the number of skateboards assembled with each process. Because ou want to maximize profit, P, the objective quantit is: P = 35a + 15b Profit factors for each process The constraints are as follows: 3a + b 1796 Skilled labor: Up to 1796 hours a + 3b 3500 Machine time: Up to 3500 hours a 0, b 0 Cannot produce negative amounts At (0, 1166 ⅔): P = 35(0) + 15(1166 ⅔) = $17,500 At (36, 1088): P = 35(36) + 15(1088) = $,580 At (598 ⅔, 0): P = 35(598 ⅔) + 15(0) = $0,953 At (0, 0): P = 35(0) + 15(0) = $0 Maximum Profit The maximum profit is obtained b making 36 skateboards with Process A and 1088 skateboards with Process B. Mr. Noes, Akimel A-al Middle School 3 Heath Algebra 1 - An Integrated Approach

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