* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download 3.1
Schrödinger equation wikipedia , lookup
Debye–Hückel equation wikipedia , lookup
Unification (computer science) wikipedia , lookup
Two-body problem in general relativity wikipedia , lookup
Perturbation theory wikipedia , lookup
Itô diffusion wikipedia , lookup
Maxwell's equations wikipedia , lookup
BKL singularity wikipedia , lookup
Equation of state wikipedia , lookup
Calculus of variations wikipedia , lookup
Derivation of the Navier–Stokes equations wikipedia , lookup
Euler equations (fluid dynamics) wikipedia , lookup
Navier–Stokes equations wikipedia , lookup
Equations of motion wikipedia , lookup
Schwarzschild geodesics wikipedia , lookup
Differential equation wikipedia , lookup
Chapter 3 Systems of Linear Equations § 3.1 Systems of Linear Equations in Two Variables Systems of Equations EXAMPLE Determine whether (3,2) is a solution of the system 3 x 6 y 10 5 x 8 y 2 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. ? 3x 6 y 10 33 62 10 ? ? 9 12 10 3 10 false 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. Blitzer, Intermediate Algebra, 4e – Slide #3 Systems of Equations Solve Systems of Two Linear Equations in Two Variables, x and y, by Graphing 1) Graph the first equation. 2) Graph the second equation on the same set of axes. 3) If the lines representing the two graphs intersect at a point, determine the coordinates of this point of intersection. The ordered pair is the solution to the system. 4) Check the solution in both equations. NOTE: In order for this method to be useful, you must graph the lines very accurately. Blitzer, Intermediate Algebra, 4e – Slide #4 Systems of Equations EXAMPLE Solve by graphing: 4x y 4 3x y 3 SOLUTION 1) Graph the first equation. I first rewrite the equation in slope-intercept form. 4x y 4 y 4 x 4 m = -4 = -4/1, b = 4 Now I can graph the equation. Blitzer, Intermediate Algebra, 4e – Slide #5 Systems of Equations CONTINUED 2) Graph the second equation on the same set of axes. I first rewrite the equation in slope-intercept form. 3x y 3 y 3 x 3 y 3x 3 m = 3 = 3/1, b = -3 Now I can graph the equation. Blitzer, Intermediate Algebra, 4e – Slide #6 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, 4e – Slide #7 Systems of Equations CONTINUED 4) Check the solution in both equations. 4x y 4 3x y 3 41 0 4 31 0 ? 3 ? ? 40 4 4 4 true ? 3 0 3 3 3 true Because both equations are satisfied, (1,0) is the solution and {(1,0)} is the solution set. Blitzer, Intermediate Algebra, 4e – Slide #8 Substitution Method Solving Linear Systems by Substitution 1) Solve either of the equations for one variable in terms of the other. (If one of the equations is already in this form, you can skip this step) 2) Substitute the expression found in step 1 into the other equation. This will result in an equation in one variable! (Hence, it’s called the “Substitution Method”) 3) Solve the equation containing one variable. (the resultant equation from step 2) 4) Back-Substitute the value found in step 3 into one of the original equations. Simplify and find the value of the remaining variable. 5) Check the proposed solution in both of the system’s given equations. Blitzer, Intermediate Algebra, 4e – Slide #9 Substitution Method EXAMPLE Solve by the substitution method: 4x y 4 3x y 3 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. 4x y 4 y 4 x 4 Solve for y by subtracting 4x from both sides Blitzer, Intermediate Algebra, 4e – Slide #10 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: 3x y 3 3x 4x 4 3 Replace y with -4x+4 3) Solve the resulting equation containing one variable. 3x 4x 4 3 7x 4 3 7x 7 x 1 Distribute Add like terms Add 4 to both sides Divide both sides by 7 Blitzer, Intermediate Algebra, 4e – Slide #11 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. 4x y 4 41 y 4 4 y 4 y0 Replace x with 1 Multiply Subtract 4 from both sides Therefore, the potential solution is (1,0). Blitzer, Intermediate Algebra, 4e – Slide #12 Substitution Method CONTINUED 5) Check. Now we will show that (1,0) is a solution for both of the original equations. 4x y 4 3x y 3 ? 41 0 4 31 0 ? 3 ? ? 40 4 3 0 3 4 4 true 3 3 true Because both equations are satisfied, (1,0) is the solution and {(1,0)} is the solution set. Blitzer, Intermediate Algebra, 4e – Slide #13 Addition (Elimination) Method Solving Linear Systems by Addition 1) If necessary, rewrite both equations in the form Ax + By = C. 2) If necessary, multiply either equation or both equations by appropriate nonzero numbers so that the sum of the x-coefficients or the sum of the ycoefficients is 0. 3) Add the equations in step 2. The sum should be an equation in one variable. 4) Solve the equation in one variable (the result of step 3). 5) Back-substitute the value obtained in step 4 into either of the given equations and solve for the other variable. 6) Check the solution in both of the original equations. 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, 4e – Slide #14 Addition (Elimination) Method EXAMPLE Solve by the addition method: SOLUTION 4x = -2y + 4 -y = -3x + 3 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, 4e – Slide #15 Addition (Elimination) Method CONTINUED 2) If necessary, multiply either equation or both equations by appropriate numbers so that the sum of the xcoefficients 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. 4x + 2y = 4 3x - y = 3 No Change Multiply by 2 4x + 2y = 4 6x - 2y = 6 Blitzer, Intermediate Algebra, 4e – Slide #16 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, 4e – Slide #17 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 -y = -3(1) + 3 Replace x with 1 -y = -3 + 3 Multiply -y = 0 Add y=0 Multiply by -1 Therefore, the potential solution is (1,0). Blitzer, Intermediate Algebra, 4e – Slide #18 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 ? 0 =? -3 + 3 4(1) = -2(0) + 4 4=0+4 4=4 true 0=0 true Because both equations are satisfied, (1,0) is the solution and {(1,0)} is the solution set. Blitzer, Intermediate Algebra, 4e – Slide #19 Solving Systems of Equations Comparing Solution Methods Method Advantages You can see the solutions. Graphing Substitution Addition Disadvantages If the solutions do not involve integers or are too large to be seen on the graph, it’s impossible to tell exactly what the solutions are. Gives exact solutions. Solutions cannot be seen. Easy to use if a variable is Introduces extensive work with on one side by itself. fractions when no variable has a coefficient of 1 or -1. Gives exact solutions. Easy to use if a variable has a coefficient of 1 or 1. Solutions cannot be seen. Blitzer, Intermediate Algebra, 4e – Slide #20 Solving Systems of Equations The Number of Solutions to a System of Two Linear Equations Number of Solutions What This Means Graphically Exactly one ordered-pair solution The two lines intersect at one point. No solution The two lines are parallel. Infinitely Many Solutions The two lines are identical. Blitzer, Intermediate Algebra, 4e – Slide #21 Graphical Examples 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, 4e – Slide #22 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 model N = -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 model N = 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, 4e – Slide #23 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, 4e – Slide #24 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, 4e – Slide #25 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 1800 = 30p 60 = p Add 25p to both sides Subtract 6000 from both sides Divide both sides by 30 Blitzer, Intermediate Algebra, 4e – Slide #26 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, 4e – Slide #27 Substitution Method CONTINUED 5) Check. Now we will show that (60,6300) is a solution for both of the original equations. N = -25p + 7800 6300 =? -25(60) + 7800 6300 =? -1500 + 7800 6300 = 6300 true N = 5p + 6000 6300 =? 5(60) + 6000 6300 =? 300 + 6000 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, 4e – Slide #28