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Chapter 3 Systems of Linear Equations

Chapter 3 Systems of Linear Equations. § 3.1. Systems of Linear Equations in Two Variables. Systems of Equations. We know that an equation of the form Ax + By = C is a line when graphed. Two such equations is called a system of linear equations.

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Chapter 3 Systems of Linear Equations

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  1. Chapter 3Systems of Linear Equations

  2. §3.1 Systems of Linear Equations in Two Variables

  3. Systems of Equations We know that an equation of the form Ax + By = C is a line when graphed. Two such equations is called a system of linear equations. A solution of a system of linear equations is an ordered pair that satisfies both equations in the system. For example, the ordered pair (2,1) satisfies the system 3x + 2y = 8 4x – 3y = 5 Blitzer, Intermediate Algebra, 5e – Slide #3 Section 3.1

  4. Systems of Equations EXAMPLE Determine whether (3,2) is a solution of the system SOLUTION Because 3 is the x-coordinate and 2 is the y-coordinate of the point(3,2), we replace x with 3 and y with 2. ? Since the result is false, (3,2) is NOT a solution for the system. Also, I need not check the other equation since the first one failed. ? ? false Blitzer, Intermediate Algebra, 5e – Slide #4 Section 3.1

  5. Systems of Equations Since two lines may intersect in exactly one point, may not intersect at all, or may intersect in every point; it follows that a system of linear equations will have exactly one solution, will have no solution, or will have infinitely many solutions. Blitzer, Intermediate Algebra, 5e – Slide #5 Section 3.1

  6. Solving Systems of Equations Blitzer, Intermediate Algebra, 5e – Slide #6 Section 3.1

  7. Systems of Equations NOTE: In order for this method to be useful, you must graph the lines veryaccurately. Blitzer, Intermediate Algebra, 5e – Slide #7 Section 3.1

  8. Systems of Equations EXAMPLE Solve by graphing: SOLUTION 1) Graph the first equation. I first rewrite the equation in slope-intercept form. m = -4 = -4/1, b = 4 Now I can graph the equation. Blitzer, Intermediate Algebra, 5e – Slide #8 Section 3.1

  9. Systems of Equations CONTINUED 2) Graph the second equation on the same set of axes. I first rewrite the equation in slope-intercept form. m = 3 = 3/1, b = -3 Now I can graph the equation. Blitzer, Intermediate Algebra, 5e – Slide #9 Section 3.1

  10. Systems of Equations CONTINUED 3) Determine the coordinates of the intersection point. This ordered pair is the system’s solution. Using the graph below, it appears that the solution is the point (1,0). We won’t know for sure until after we check this potential solution in the next step. Blitzer, Intermediate Algebra, 5e – Slide #10 Section 3.1

  11. Systems of Equations CONTINUED 4) Check the solution in both equations. ? ? ? ? true true Because both equations are satisfied, (1,0) is the solution and {(1,0)} is the solution set. Blitzer, Intermediate Algebra, 5e – Slide #11 Section 3.1

  12. Substitution Method Blitzer, Intermediate Algebra, 5e – Slide #12 Section 3.1

  13. Substitution Method EXAMPLE Solve by the substitution method: SOLUTION 1) Solve either of the equations for one variable in terms of the other. We’ll isolate the variable y from the first equation. Solve for y by subtracting 4x from both sides Blitzer, Intermediate Algebra, 5e – Slide #13 Section 3.1

  14. Substitution Method CONTINUED 2) Substitute the expression from step 1 into the other equation. -4x + 4 is the ‘expression from step 1’ and ‘the other equation’ is 3x– y = 3. Therefore: Replace y with -4x+4 3) Solve the resulting equation containing one variable. Distribute Add like terms Add 4 to both sides Divide both sides by 7 Blitzer, Intermediate Algebra, 5e – Slide #14 Section 3.1

  15. Substitution Method CONTINUED 4) Back-substitute the obtained value into one of the original equations. We back-substitute 1 for x into one of the original equations to find y. Let’s use the first equation. Replace x with 1 Multiply Subtract 4 from both sides Therefore, the potential solution is (1,0). Blitzer, Intermediate Algebra, 5e – Slide #15 Section 3.1

  16. Substitution Method CONTINUED 5) Check. Now we will show that (1,0) is a solution for both of the original equations. ? ? ? ? true true Because both equations are satisfied, (1,0) is the solution and {(1,0)} is the solution set. Blitzer, Intermediate Algebra, 5e – Slide #16 Section 3.1

  17. Addition (Elimination) Method NOTE: As you now know, there is more than one method to solve a system of equations. The reason for learning more than one method is because sometimes one method will be preferable or easier to use over another method. Blitzer, Intermediate Algebra, 5e – Slide #17 Section 3.1

  18. Addition (Elimination) Method EXAMPLE Solve by the addition method: 4x = -2y + 4 -y = -3x + 3 SOLUTION 1) Rewrite both equations in the form Ax + By = C. We first arrange the system so that variable terms appear on the left and constants appear on the right. We obtain 4x + 2y = 4 Add 2y to both sides 3x - y = 3 Add 3x to both sides Blitzer, Intermediate Algebra, 5e – Slide #18 Section 3.1

  19. Addition (Elimination) Method CONTINUED 2) If necessary, multiply either equation or both equations by appropriate numbers so that the sum of the x-coefficients is 0. We can eliminate the y’s by multiplying the second equation by 2. Or we can eliminate the x’s by multiplying the first equation by -3 and the second equation by 4. Let’s use the first method. No Change 4x + 2y = 4 4x + 2y = 4 3x - y = 3 Multiply by 2 6x - 2y = 6 Blitzer, Intermediate Algebra, 5e – Slide #19 Section 3.1

  20. Addition (Elimination) Method CONTINUED 3) Add the equations. 4x + 2y = 4 6x - 2y = 6 Add: 10x + 0y = 10 10x = 10 4) Solve the equation in one variable. Now I solve the equation 10x = 10. 10x = 10 x = 1 Blitzer, Intermediate Algebra, 5e – Slide #20 Section 3.1

  21. Addition (Elimination) Method CONTINUED 5) Back-substitute and find the value of the other variable. Now we will use one of the original equations and replace x with 1 to determine y. I’ll use the second equation. -y = -3x + 3 Replace x with 1 -y = -3(1) + 3 -y = -3 + 3 Multiply -y = 0 Add y = 0 Multiply by -1 Therefore, the potential solution is (1,0). Blitzer, Intermediate Algebra, 5e – Slide #21 Section 3.1

  22. Addition (Elimination) Method CONTINUED 6) Check. I now check the potential solution (1,0) in both original equations. -y = -3x + 3 4x = -2y + 4 ? ? -(0) = -3(1) + 3 4(1) = -2(0) + 4 ? ? 0 = -3 + 3 4 = 0 + 4 0 = 0 true true 4 = 4 Because both equations are satisfied, (1,0) is the solution and {(1,0)} is the solution set. Blitzer, Intermediate Algebra, 5e – Slide #22 Section 3.1

  23. Solving Systems of Equations Blitzer, Intermediate Algebra, 5e – Slide #23 Section 3.1

  24. Solving Systems of Equations NOTE: It is extremely helpful to understand these relationships as well as any other relationship between an equation and it’s graph. NOTE: To determine that a system has exactly one solution, solve the system using one of the methods. A single solution will occur as in the previous examples. NOTE: To determine that a system has no solution, solve the system using one of the methods. Eventually, you’ll get an obviously false statement, like 3 = 4. NOTE: To determine that a system has infinitely many solutions, solve the system using one of the methods. Eventually, you’ll get an obviously true statement, like -2 = -2. Blitzer, Intermediate Algebra, 5e – Slide #24 Section 3.1

  25. Solving Systems of Equations EXAMPLE At a price of p dollars per ticket, the number of tickets to a rock concert that can be sold is given by the demand modelN = -25p + 7800. At a price of p dollars per ticket, the number of tickets that the concert’s promoters are willing to make available is given by the supply modelN = 5p + 6000. (a) How many tickets can be sold and supplied for $50 per ticket? (b) Find the ticket price at which supply and demand are equal. At this price, how many tickets will be supplied and sold? Blitzer, Intermediate Algebra, 5e – Slide #25 Section 3.1

  26. Solving Systems of Equations CONTINUED SOLUTION (a) How many tickets can be sold and supplied for $50 per ticket? The number of tickets that can be sold for $50 per ticket is found using the demand model: N = -25(50) + 7800 = -1250 + 7800 = 6550 tickets sold. The number of tickets that can be supplied for $50 per ticket is found using the supply model: N = 5(50) + 6000 = 250 + 6000 = 6250 tickets supplied. Blitzer, Intermediate Algebra, 5e – Slide #26 Section 3.1

  27. Solving Systems of Equations CONTINUED (b) Find the ticket price at which supply and demand are equal. At this price, how many tickets will be supplied and sold? 1) Solve either of the equations for one variable in terms of the other. The system of equations already has N isolated in both equations. N = -25p + 7800 N = 5p + 6000 Blitzer, Intermediate Algebra, 5e – Slide #27 Section 3.1

  28. Substitution Method CONTINUED 2) Substitute the expression from step 1 into the other equation. I’ll replace the N in the second equation with -25p + 7800. -25p + 7800 = 5p + 6000 3) Solve the resulting equation containing one variable. -25p + 7800 = 5p + 6000 7800 = 30p + 6000 Add 25p to both sides 1800 = 30p Subtract 6000 from both sides Divide both sides by 30 60 = p Blitzer, Intermediate Algebra, 5e – Slide #28 Section 3.1

  29. Substitution Method CONTINUED 4) Back-substitute the obtained value into one of the original equations. We back-substitute 60 for p into one of the original equations to find N. Let’s use the first equation. N = -25p + 7800 N = -25(60) + 7800 Replace p with 60 N = -1500 + 7800 Multiply N = 6300 Add Blitzer, Intermediate Algebra, 5e – Slide #29 Section 3.1

  30. Substitution Method CONTINUED 5) Check. Now we will show that (60,6300) is a solution for both of the original equations. N = -25p + 7800 N = 5p + 6000 ? ? 6300 = -25(60) + 7800 6300 = 5(60) + 6000 ? ? 6300 = -1500 + 7800 6300 = 300 + 6000 6300 = 6300 true 6300 = 6300 true Therefore, the solution is (60,6300). Therefore, supply and demand will be equal when the ticket price is $60 and 6300 tickets are sold. Blitzer, Intermediate Algebra, 5e – Slide #30 Section 3.1

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