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5.6 Daily Warm-Up 7.1 Is the Ordered Pair a Solution of the equation 2x – 3y = 5? 1. (1, 0) 2. (-1, 1) 3. (1, -1) 4. (4, 1) 7.1 Solving Linear Systems by Graphing Objective: Solve a system of linear equations by graphing What is a System of Linear Equations? A system of linear equations is simply two or more linear equations using the same variables. We will only be dealing with systems of two equations using two variables, x and y. If the system of linear equations is going to have a solution, then the solution will be an ordered pair (x , y) where x and y make both equations true at the same time. We will be working with the graphs of linear systems and how to find their solutions graphically. How to Use Graphs to Solve Linear Systems y Consider the following system: x – y = –1 x + 2y = 5 Using the graph to the right, we can see that any of these ordered pairs will make the first equation true since they x lie on the line. (1 , 2) We can also see that any of these points will make the second equation true. However, there is ONE coordinate that makes both true at the same time… The point where they intersect makes both equations true at the same time. How to Use Graphs to Solve Linear Systems y Consider the following system: x – y = –1 x + 2y = 5 We must ALWAYS verify that your coordinates actually satisfy both equations. x (1 , 2) To do this, we substitute the coordinate (1 , 2) into both equations. x – y = –1 x + 2y = 5 (1) – (2) = –1 (1) + 2(2) = Since (1 , 2) makes both equations 1+4=5 true, then (1 , 2) is the solution to the system of linear equations. Graphing to Solve a Linear System Solve the following system by graphing: 3x + 6y = 15 Start with 3x + 6y = 15 –2x + 3y = –3 Subtracting 3x from both sides yields 6y = –3x + 15 While there are many different Dividing everything by 6 gives us… ways to graph these equations, we 1 5 will be using the slope – intercept y= - 2 x+ 2 form. Similarly, we can add 2x to both sides and then divide everything by To put the equations in slope 3 in the second equation to get intercept form, we must solve both y= 2 x- 1 3 equations for y. Now, we must graph these two equations Graphing to Solve a Linear System Solve the following system by graphing: y 3x + 6y = 15 –2x + 3y = –3 Using the slope intercept forms of these equations, we can graph them carefully x on graph paper. (3 , 1) y = - 1 x+ 5 2 2 2 y= 3 x- 1 Label the Start at the y – intercept, then use the slope. solution! Lastly, we need to verify our solution is correct, by substituting (3 , 1). Since 3(3)+ 6(1)= 15 and - 2(3)+ 3(1)= - 3, then our solution is correct! Graphing to Solve a Linear System Let's summarize! There are 4 steps to solving a linear system using a graph. Step 1: Put both equations in Solve both equations for y, so that slope – intercept form each equation looks like y = mx + b. Step 2: Graph both equations Use the slope and y – intercept for on the same coordinate plane each equation in step 1. Be sure to use a ruler and graph paper! Step 3: Estimate where the This is the solution! LABEL the graphs intersect. solution! Step 4: Check to make sure your Substitute the x and y values into both solution makes both equations true. equations to verify the point is a solution to both equations. Graphing to Solve a Linear System Let's do ONE more…Solve the following system of equations by graphing. 2x + 2y = 3 y x – 4y = –1 LABEL the solution! Step 1: Put both equations in slope – intercept form (1 , 1 ) 2 y = - x+ 32 1 1 y= 4 x+ 4 x Step 2: Graph both equations on the same coordinate plane Step 3: Estimate where the graphs intersect. LABEL the solution! 2(1)+ 2(1 )= 2 + 1 = 3 Step 4: Check to make sure your 2 solution makes both equations true. 1- 4 ( 1 ) = 1- 2 = - 1 2 Graphing to Solve a Linear System It’s your turn! … Solve the following system of equations by graphing. 3x + 2y = 2 y x = –1 LABEL the solution! Step 1: Put both equations in slope – intercept form (-1, 2 ) 5 3 y = - 2 x+ 1 x= - 1 x Step 2: Graph both equations on the same coordinate plane Step 3: Estimate where the graphs intersect. LABEL the solution! 3(-1)+ 2( 2 ) = - 3 + 5 = 2 5 Step 4: Check to make sure your solution makes both equations true. -1 = - 1 Summary 1. When you are asked to solve the linear equation system by graphing, you only need to follow the four-step: Step 1: Put both equations in slope – intercept form: y = mx + b Step 2: Graph both equations on the same coordinate plane: use the y- intercept as the first point and then use slope to create the second point. Step 3: Estimate where the graphs intersect, and also label the intersection which is the solution of the linear equation system. Step 4: Check to make sure your solution makes both equations true because the graphing may contain some error. Substitute the x and y values into both equations to verify the point is a solution to both equations. 2. Don’t forget the special line!!! Guided Practice – L 7.1 DHQ PAGE: 401 in Textbook Numbers: 8-10 on graph paper! * ALL ANSWERS ARE ORDERED PAIRS! Tonight’s Homework P. WORKSHEET 7.1 #’s 1-10 ALL 5.6 Daily Warm-Up 7.2 Is (-3, 2) a solution of the equation? 1. 3x – y = 4 2. x + 2y = 1 3. Is (4,0) a solution of the system of equations? 7.2 Solving Systems of Equations using Substitution Objective: Use substitution method to solve a linear system Solving Systems of Equations using Substitution Steps: 1. Solve one equation for one variable (y = ; x = ; a =) 2. Substitute the expression from step one into the other equation. 3. Simplify and solve the equation. 4. Substitute back into either original equation to find the value of the other variable. 5. Check the solution in both equations of the system. Example #1: y = 4x 3x + y = -21 Step 1: Solve one equation for one variable. y = 4x (This equation is already solved for y.) Step 2: Substitute the expression from step one into the other equation. 3x + y = -21 3x + 4x = -21 Step 3: Simplify and solve the equation. 7x = -21 x = -3 y = 4x 3x + y = -21 Step 4: Substitute back into either original equation to find the value of the other variable. y = 4x This substitution is easier y = 4(-3) = -12 or 3x + y = -21 3(-3) + y = -21 -9 + y = -21 y = -12 Solution to the system is (-3, -12). y = 4x 3x + y = -21 Step 5: Check the solution in both equations. Solution to the system is (-3,-12). y = 4x 3x + y = -21 -12 = 4(-3) 3(-3) + (-12) = -21 -12 = -12 -9 + (-12) = -21 -21= -21 Example #2: x + y = 10 5x – y = 2 Step 1: Solve one equation for one variable. x + y = 10 y = –x +10 Step 2: Substitute the expression from step one into the other equation. 5x – y = 2 5x – (–x +10) = 2 x + y = 10 5x – y = 2 Step 3: Simplify and solve the equation. 5x – (–x + 10) = 2 5x + x – 10 = 2 6x – 10 = 2 6x = 12 x=2 x + y = 10 5x – y = 2 Step 4: Substitute back into either original equation to find the value of the other variable. x + y = 10 2 + y = 10 y=8 Solution to the system is (2,8). x + y = 10 5x – y = 2 Step 5: Check the solution in both equations. Solution to the system is (2, 8). x + y =10 5x – y = 2 2 + 8 =10 5(2) - (8) = 2 10 =10 10 – 8 = 2 2=2 You try Exercise 1 Solve the system using the substitution method. y = –63 – 7x 3x + 7y = –73 A) (8, –8) B) (–8, –7) C) (–7, –8) D) (–7, 9) YOU TRY: Solve by substitution: 2a 3b 7 2a b 5 Summary 1. When solving the linear systems, it is very important to remember the 5 steps and follow them in the order. 2. When in a certain step the equation reduces to a contradiction, you could conclude that the system is inconsistent and has no solution. 3. When in a certain step the equation reduces to an identity, you could conclude that the system is dependent and has many solutions. Guided Practice – L 7.2 DHQ PAGE: 408 in Textbook Numbers: 7-8 • ALL ANSWERS ARE ORDERED PAIRS! • MUST SHOW WORK Tonight’s Homework P. 408-410 #’s 17-24, 54-59 5.6 Daily Warm-Up 7.3 Give the OPPOSITE of each number. 1. 3 2. 1 3. -1/2 Simplify. 4. 3(2x – 7y) 5. -2(5x + y) 7.3 Solving Linear Systems by Linear Combinations (Elimination) Method Objective: Use linear combinations to solve a system of linear equations. What is a System? A system of linear equations is: a. A set of parabolas b. A set of two or more lines c. A stereo component Linear Combinations (Elimination) Method Linear Combinations (Elimination) method is used when it appears easy to eliminate one variable from the system through transformation. Remember that linear transformations do not change the solutions of a system. Linear Combinations (Elimination) Method Step 1 Make sure the variables are lined up properly by their names -2x + y = 4 -6x + y = 0 Step 2 Make the coefficients of one of variables opposites (Multiply one or both equations by appropriate numbers so that the coefficients in one variable are opposites). Correctly applying the distributive property is very important!!! In this question, notice that the y coefficients are 1, therefore we can multiply either equation by -1 and add the system, thus eliminating the y variable. Linear Combination (Elimination) Method Step 2 Let’s transform the second equation: -1(-6x + y = 0) 6x – y = 0 Step 3 Add the new two equations up: -2x + y = 4 + 6x – y = 0 4x =4 Linear Combination (Elimination) Method Step 4 Solving the resulted one variable equation for x 4x = 4 yields x = 1. Step 5 Back substitute to one of the equation in the system to find the value for the other variable. Once we have the x value, we can plug it into either of our original equations and solve for y: -2x + y = 4 -6x + y = 0 Linear Combination (Elimination) Method Step 5 Plugging x = 1 into the first equation yields: -2 (1) + y = 4 -2 + y = 4, y=6 Step 6 Check the answer. -2(1) + 6 = 4 -6(1) + 6 = 0 So our solution is (1, 6). Linear Combination (Elimination) Method Now you try Use the elimination method to solve the system: x – 5y = –2 3x + 2y = 11 The first step is: a. Add the equations together b. Transform the first equation by multiplying it by –3 c. I’m not sure. I need to review the elimination method. Linear Combination (Elimination) Method Yes, notice that if we multiply the first equation by -3, we obtain additive inverses for the x coefficient. Our system is now: -3x + 15y = 6 3x + 2y = 11 The next step is: a. Add the two equations b. Solve the second equation for y c. I’m not sure. I need to review. Linear Combination (Elimination) Method Yes, by adding the two equations we get: – 3x + 15y = 6 + 3x + 2y = 11 17y = 17 Solving for y, we get y = 1. The next step is: a. Plug y = 1 into either equation and solve for x b. Plug x = 1 into either equation and solve for y c. I’m not sure. I need to review. Linear Combination (Elimination) Method We found that y=1, not x=1. So we must plug y=1 into either equation to solve for x not for y. Linear Combination (Elimination) Method Yes, now we can plug y = 1 into either of the original equations, or the transformed equation. Let’s choose the first original equation: x – 5y = –2 x – 5(1) = –2 x – 5 = –2 x=3 So our solution is (3, 1), which must be the only solution to a system of two lines. Next, we must check the solved order pair is the solution to the system. (omitted here) You Try Solve the system using elimination method: 2x + 5y = 7 3x + y = -9 The solution is: a. (3, -4) b. (-4, -3) c. (-4, 3) d. (-3, -4) Linear Combination (Elimination) Method Example Use any method to solve the system: 3g – 24 = –4h –2 + 2h = g The first step is(are): a. Add the equations together b. Use the subsitution method, replace the first variable g by the expression in the second. c. Line up the variables. d. I’m not sure. I need to review the both methods. Any Method (Substitution) Use any method to solve the system: 3g – 24 = –4h –2 + 2h = g Plug the second variable g into the g in the first equation: 3(-2 + 2h) – 24 = -4h -6 + 6h – 24 = -4h 6h – 30 = -4h 10h – 30 = 0 10h = 30 h=3 Go to the second equation to find out g: g = -2 + 2(3) = -2 + 6 = 4 The solution is (4, 3). Any Method (Linear Combination/Elimination) Use the any method to solve the system: 3g – 24 = –4h –2 + 2h = g Line up the variable: 3g + 4h = 24 g – 2h = -2 Multiply 2 on both sides of the second equation and add onto the first equation: 2g – 4h = -4 + 3g + 4h = 24 5g = 20 g=4 Plug into second equation to find h: 4 – 2h = -2 -2h = -6 h=3 The solution is (4, 3) Summary 1. The key concepts of the linear combinations (elimination) method are to (a) variables are lined up properly (b) find out if there are any opposite coefficients for the same variable. If yes, directly add two equations up and eliminate one variable. If not, (c) multiply (or divide) some number(s) to equation(s) so that two opposite coefficients for the same variable show up. Then follow (b). 2. Some problems can be applied both substitution and linear combination (elimination) methods. However, it is better to use linear combination (elimination) method than to use the substitution method to some other problems. Choosing an appropriate method is important. Guided Practice – L 7.3 DHQ PAGE: 414 in Textbook Numbers: 4-5 • ALL ANSWERS ARE ORDERED PAIRS! • MUST SHOW WORK Tonight’s Homework P. 414 #’s 8-13, 17-19 + QUIZ REVIEW WORKSHEET Daily Warm-Up 7.5 Rewrite the equations to Slope-Intercept Form ( y = mx + b ) 1. 5( x 1) y 3 2. 6 x 3 y 9 Graph the equations 3. 5( x 1) y 3 4. 6 x 3 y 9 7.5 Special Types of Liner Systems Objectives 1) Identify linear system algebraically as having one solution, no solution, and infinite many solutions. Slope-Intercept Form 2) Identify linear system geometrically as having one solution, no solution, and infinite many solutions. Special Types of Linear Systems Example 1 Solve the linear equation system 3x 2 y 5 (1) x 2y 7 (2) Solution 1 Substitution 3(3) 2(2) 5 (1) From (2) x = –2y + 7 3 2(2) 7 (2) To (1) 3(–2y + 7) – 2y = 5 The solution is (3, 2). –6y + 21 – 2y = 5 –8y = –16 y=2 then x = –2(2) + 7 = 3 Special Types of Linear Systems Example 1 Solve the linear equation system 3x 2 y 5 (1) x 2y 7 (2) rise 3 Solution 2 m Graphing run 2 3x 2 y 5 2 y 3x 5 3 5 3 5 y x y x 2 2 2 2 x 2y 7 2 y x 7 (3 , 2) 1 7 1 7 y x y x 2 2 2 2 rise 1 m run 2 NUMBER OF SOLUTIONS OF A LINEAR SYSTEM y Lines intersect one solution x A Linear System with No Solution Show that this linear 2x y 5 Equation 1 METHOD 1: SUBSTITUTION system has no solution. 2x y 1 Equation 2 Because Equation 2 can be revised to y –2 x 1, you can substitute –2 x 1 for y in Equation 1. 2x y 5 Write Equation 1. 2 x (–2 x 1) 5 Substitute –2 x 1 for y. 15 Simplify. False statement! Once the variables are eliminated, the statement is not true regardless of the values of x and y. This tells you the system has no solution. A Linear System with No Solution Show that this linear system 2x y 5 Equation 1 METHOD 2: GRAPHING has no solution. 2x y 1 Equation 2 Rewrite each equation y –2 x 5 Revised Equation 1 in slope-intercept form. y –2 x 1 Revised Equation 2 Graph the linear system. 6 5 y 2x 5 4 The lines are parallel; they 3 have the same slope but 2 different y-intercepts. Parallel y 2x 1 1 lines never intersect, so the system has no solution. 5 4 3 2 1 0 1 2 3 4 5 1 Special Types of Linear Systems • When solving a system of linear equations, if the resulting equation in certain step reduces to an inconsistent equation (contradiction), then the system has no solution and is inconsistent. • Geometrically speaking, the two equations in the system represent two parallel lines which never intersect. IDENTIFYING THE NUMBER OF SOLUTIONS NUMBER OF SOLUTIONS OF A LINEAR SYSTEM y Lines are parallel no solution x Practice 2 Use the following method to solve the equation system (1) Substitution (2) Graphing 2x – 4y = -2 x – 2y = 0 Special Types of Linear Systems Practice 2 Solve the linear equation system 2 x 4 y 2 (1) x 2y 0 (2) Solution 1 Substitution From (2) x = 2y To (1) 2(2y) – 4y = –2 4y – 4y = –2 0 = –2 This is an inconsistent equation So the system has no solution. Special Types of Linear Systems Practice 2 Solve the linear equation system 2 x 4 y 2 (1) x 2y 0 (2) rise 1 Solution 2 m run 2 Graphing 2 x 4 y 2 4 y 2 x 2 2 2 1 1 y x y x 4 4 2 2 x 2y 0 2 y x 0 1 0 1 y x y x0 2 2 2 rise 1 m run 2 A Linear System with Many Solutions METHOD 1:this linear system Show that LINEAR COMBINATIONS –2x y 3 Equation 1 has infinitely many solutions. – 4 x 2y 6 Equation 2 You can multiply Equation 1 by –2. 4x – 2y – 6 Multiply Equation 1 by –2. – 4 x 2y 6 Write Equation 2. 0 0 Add Equations. True statement! The variables are eliminated and you are left with a statement that is true regardless of the values of x and y. This result tells you that the linear system has infinitely many solutions. A Linear System with Many Solutions Show that this linear system METHOD 2: GRAPHING –2x y 3 Equation 1 has infinitely many solutions. – 4 x 2y 6 Equation 2 Rewrite each equation y 2x 3 Revised Equation 1 in slope-intercept form. y 2x 3 Revised Equation 2 Graph the linear system. 6 5 –2x y 3 – 4x 2y 6 4 From these graphs you 3 can see that the equations 2 represent the same line. Any point on the line is 1 a solution. 5 4 3 2 1 0 1 2 3 4 5 1 Special Types of Linear Systems • When solving a system of linear equations, if the resulting equation in certain step reduces to an identity equation, then the system has infinite many solutions and is dependent. • Geometrically speaking, the two equations in the system represent two identical lines which always intersect. Any point on the first line (the solution to the first equation) is also on the second line (the solution to the second equation). The two identical lines have infinitely many intersection. IDENTIFYING THE NUMBER OF SOLUTIONS NUMBER OF SOLUTIONS OF A LINEAR SYSTEM y Lines coincide infinitely many solutions (the coordinates of every point on the line) x Practice 3 Use the following method to solve the equation system (1) Linear Combination (2) Graphing 3x – 4y = -4 -6x + 8y = 8 Special Types of Linear Systems Practice 3 Solve the linear equation system 3x 4 y 4 (1) 6 x 8 y 8 (2) Solution 1 Linear Combination Multiply 2 to (1) 2(3x – 4y)= 2(– 4) 6x – 8y = –8 (1) -6x + 8y = 8 (2) 0=0 This is an identity. So the equation system has many solutions. Special Types of Linear Systems Practice 3 Solve the linear equation system 3x 4 y 4 (1) 6 x 8 y 8 (2) Solution 2 rise 3 m Graphing run 4 3x 4 y 4 4 y 3x 4 3 4 3 y x y x 1 4 4 4 6 x 8 y 8 8 y 6x 8 6 8 3 y x y x 1 8 8 4 rise 3 m run 4 IDENTIFYING THE NUMBER OF SOLUTIONS NUMBER OF SOLUTIONS OF A LINEAR SYSTEM y Lines intersect one solution x IDENTIFYING THE NUMBER OF SOLUTIONS NUMBER OF SOLUTIONS OF A LINEAR SYSTEM y Lines are parallel no solution x IDENTIFYING THE NUMBER OF SOLUTIONS NUMBER OF SOLUTIONS OF A LINEAR SYSTEM y Lines coincide infinitely many solutions (the coordinates of every point on the line) x IDENTIFYING THE NUMBER OF SOLUTIONS CONCEPT NUMBER OF SOLUTIONS OF A LINEAR SYSTEM SUMMARY y y y x x x Lines intersect Lines are parallel Lines coincide one solution no solution infinitely many solutions Independent Inconsistent Dependent Summary Similar to the situations when solving the one variable linear equations, solving the linear systems has the similar 3 situations: (a) One solution -- two lines intersect -- the system is independent (b) No solution -- two lines are parallel -- the system is inconsistent -- the system contains a contradiction (c) Many solution -- two lines are coincide -- the system is dependent -- the system contains an identity Guided Practice – L 7.3 DHQ PAGE: 429 in Textbook Numbers: 6-8 • USE ANY METHOD • ALL ANSWERS ARE ORDERED PAIRS! Tonight’s Homework P. 429-430 #’s 12-21, 24-27 • MUST SHOW WORK