* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

# Download Section 6.1

Unification (computer science) wikipedia , lookup

Debye–Hückel equation wikipedia , lookup

Schrödinger equation wikipedia , lookup

Two-body problem in general relativity wikipedia , lookup

BKL singularity wikipedia , lookup

Perturbation theory wikipedia , lookup

Maxwell's equations wikipedia , lookup

Itô diffusion wikipedia , lookup

Equation of state wikipedia , lookup

Calculus of variations wikipedia , lookup

Euler equations (fluid dynamics) wikipedia , lookup

Derivation of the Navier–Stokes equations wikipedia , lookup

Navier–Stokes equations wikipedia , lookup

Equations of motion wikipedia , lookup

Schwarzschild geodesics wikipedia , lookup

Differential equation wikipedia , lookup

Chapter 6 College Algebra ! Are there any questions about our previous section’s HW? ! Are there any course questions before we begin today? College Algebra 2 Section 6.1 College Algebra ! Solve a system of two linear equations in two variables by graphing ! Solve a system of two linear equations in two variables using the substitution method and the elimination method ! Use systems of two linear equations to solve applied problems College Algebra 4 ! A system of linear equations consists of two or more linear equations. " 1st equation ! Looks like # $2nd equation ! A solution of a system of linear equations in two variables is an ordered pair € that is a solution of both equations in the system. College Algebra 5 3x – 2y = 2 x + 2y = 6 no b) (2,2) yes a) (0, -1) College Algebra 6 ! Since a solution of a system of equations is a solution common to both equations, it is also a point common to the graphs of both equations. ! To find the solution of a system of two linear equations: (steps) 1. Graph each equation 2. Identify the intersection College Algebra 7 Solve the system of equations by graphing. # 2x − y = 6 $ % x + 3y = 10 First, graph 2x – y = 6. €Second, graph x + 3y = 10. The lines APPEAR to intersect at (4, 2). College Algebra 8 Although the solution to the system of equations appears to be (4, 2), you still need to check the answer by substituting x = 4 and y = 2 into the two equations. First equation: 2(4) – (2) = 8 – 2 = 6 True Second equation: (4) + 3(2) = 4 + 6 = 10 True The point (4, 2) checks, so it is the solution of the system. College Algebra 9 ! Neatly drawn graphs can help when guessing the solution, but you must check to make sure. College Algebra 10 Solve the system of equations by graphing. # −x + 3y = 6 $ % 3x − 9y = 9 First, graph -x + 3y = 6. (0, 2) (-6, 0) €Second, graph 3x – 9y = 9. (0, -1) (3, 0) The lines are parallel. No solution College Algebra 11 Solve the system of equations by graphing. # x = 3y −1 $ %2x − 6y = −2 First, graph x = 3y – 1. (-1, 0) (0, ⅓) € Second, graph 2x – 6y = -2. The lines are coincident. {(x,y) | 2x – 6y = -2} College Algebra 12 ! There are three possible outcomes when graphing two linear equations in a plane. ! One point of intersection " one solution ! Parallel lines " no solution ! Coincident lines " infinitely many solutions ! If there is at least one solution, the system is considered to be consistent. ! If the system defines distinct lines, the equations are independent. College Algebra 13 Graph Number of Solutions Type of System intersecting lines one solution Consistent The equations are independent. parallel lines no solutions Inconsistent The equations are independent. coincident lines (same line) infinitely many solutions Consistent The equations are dependent. Two lines intersect at one point. Parallel lines Lines coincide College Algebra 14 ! Steps for Substitution: 1. Solve one of the equations for one variable (try to solve for the variable with a coefficient of one) 2. Substitute the expression from step 1 into the other equation and solve this new equation 3. Substitute the value from step 2 into one of your original equations to complete the ordered pair 4. Check the ordered pair solution to make sure it satisfies the entire system College Algebra 15 Solve the system using the substitution method. 3x – y = 6 -4x + 2y = –8 The first equation can be easily solved for y. -y = –3x + 6 y = 3x – 6 Subtract 3x from both sides Multiply both sides by –1 Now, substitute this value for y in the second equation. -4x + 2(3x – 6) = –8 -4x + 6x – 12 = –8 2x – 12 = –8 2x = 4 x=2 College Algebra Replace y with result from first equation Distribute Simplify Add 12 to both sides Divide both sides by 2 16 Substitute x = 2 into the first equation, which has already been solved for y. y = 3x – 6 = 3(2) – 6 = 6 – 6 = 0 The solution of the system is the ordered pair (2, 0). Check the point (2, 0) in the original equations. First equation: 3x – y = 6 3(2) – 0 = 6 True Second equation: -4x + 2y = –8 -4(2) + 2(0) = –8 True The solution of the system is (2, 0). College Algebra 17 Solve the system using the substitution method. y = 2x – 5 8x – 4y = 20 The first equation is already solved for y. Now, substitute this value for y in the second equation. 8x – 4(2x – 5) = 20 Replace y with result from first equation 8x – 8x + 20 = 20 Distribute 20 = 20 Simplify This is a true statement, so these lines are the same. Hence, the system has infinitely many solutions. College Algebra {(x,y) | 8x – 4y = 20} 18 Solve the system using the substitution method. 3x – y = 4 6x – 2y = 4 The first equation can be easily solved for y. -y = -3x + 4 Subtract 3x from both sides y = 3x – 4 ultiply both sides by –1 M Now, substitute this value for y in the second equation. 6x – 2(3x – 4) = 4 Replace y with result from first equation 6x – 6x + 8 = 4 Distribute 8=4 Simplify This is a false statement, so these lines are the parallel. Hence, the system has no solutions. College Algebra No solution 19 ! Steps for elimination: 1. Make sure system s equations are written in “pseudo-standard” form 2. Make one of the variables have opposite coefficients (multiply by a constant if necessary) 3. Add the equations together and solve for the remaining variable 4. Substitute the value from step 3 into one of the original equations to complete the ordered pair 5. Check the ordered pair solution to make sure it satisfies the entire system College Algebra 20 Solve the following system by elimination 3x = y + 4 6x – 2y = 4 1 3x – y = y – y + 4 6x – 2y = 4 2 -2(3x – y = 4) (6x – 2y = 4) College Algebra 3 -6x + 2y = -8 6x – 2y = 4 0 + 0 = -4 0 = ≠ -4 False! No Solution 21 Solve the following system by elimination 3 8x + 3y = -18 ⅔x + ¼y = -3/2 6x – 3y = -24 ½x – ¼y = -2 14x + 0 = -42 14x = -42 2 1 14 14 x = -3 12(⅔x + ¼y = -3/2) 12(½x – ¼y = -2) College Algebra 22 4 Use x = -3 to find y 1st equation: 5 8x + 3y = -18 ⅔(-3) + ¼(2) = -3/2 8(-3) + 3y = -18 -2 + ½ = -3/2 True -24 + 3y = -18 ½(-3) – ¼(2) = -2 +24 +24 -3/2 – ½ = -2 3y = 6 True 3 3 y = 2 So our solution is College Algebra (-3, 2) ? (-3, 2) 23 Solve the following system by elimination 3 -8x + 4y = -20 8x – 4y = 20 y = 2x – 5 0 + 0 = 0 8x – 4y = 20 0 = = 0 -2x + y = 2x – 2x – 5 True! 1 8x – 4y = 20 2 4(-2x + y = -5) (8x – 4y = 20) College Algebra {(x,y) | 8x – 4y = 20} 24 General Strategy for Problem Solving 1. U NDERSTAND the problem. • Read and reread the problem • Choose two variables to represent the two unknowns • Construct a drawing, whenever possible 2. MODEL the problem with a system. 3. SOLVE the system of equations. 4. INTERPRET the result. • Check proposed solution in original problem. • State your conclusion. College Algebra 25 Hilton University Drama club sold 311 tickets for a play. Student tickets cost 50 cents each; non-student tickets cost $1.50. If the total receipts were $385.50, find how many tickets of each type were sold. 1. UNDERSTAND Since we are looking for two numbers, we let s = the number of student tickets n = the number of non-student tickets College Algebra 26 2. TRANSLATE Hilton University Drama club sold 311 tickets for a play. s + n = 311 total receipts were $385.50 Admission for students 0.50s College Algebra Admission for non-students + 1.50n Total receipts = 385.50 27 3. SOLVE We are solving the system s + n = 311 0.50s + 1.50n = 385.50 We should use the elimination method since the equations are in pseudo-standard form, and we need to multiply to clear the decimals anyway. s + n = 311 –2(0.50s + 1.50n) = –2(385.50) College Algebra s + n = 311 –s – 3n = –771 –2n = –460 n = 230 28 Now we substitute 230 for n into the first equation to solve for s. s + n = 311 s + 230 = 311 s = 81 4. I NTERPRET Check: Substitute s = 81 and n = 230 into both of the equations. s + n = 311 First Equation 81 + 230 = 311 0.50s + 1.50n = 385.50 0.50(81) + 1.50(230) = 385.50 40.50 + 345 = 385.50 True Second Equation True State: There were 81 student tickets and 230 non student tickets sold. College Algebra 29 Terry Watkins can row about 10.6 kilometers in 1 hour downstream and 6.8 kilometers upstream in 1 hour. Find how fast he can row in still water, and find the speed of the current. 1. UNDERSTAND Remember that d = r • t (or r • t = d) But what about the water? Since we are looking for two rates, we let r = the rate of the rower in still water w = the rate of the water current So, when traveling downstream, add the speeds " r + w when traveling upstream, subtract the water s speed " r – w College Algebra 30 2. TRANSLATE rate downstream (r + w) time downstream • rate upstream (r – w) College Algebra 1 distance downstream = time upstream • 1 10.6 distance upstream = 6.8 31 3. SOLVE We are solving the system r + w = 10.6 r – w = 6.8 We should use the elimination method since the two equations are in standard form. r + w = 10.6 r – w = 6.8 2r = 17.4 r = 8.7 College Algebra 32 Now we substitute 8.7 for r into the first equation. 4. INTERPRET r + w = 10.6 8.7 + w = 10.6 w = 1.9 Check: Substitute r = 8.7 and w = 1.9 into both equations. (r + w)1 = 10.6 (8.7 + 1.9)1 = 10.6 (r – w)1 = 1.9 First equation True Second equation (8.7 – 1.9)1 = 6.8 True State: Terry s rate in still water is 8.7 km/hr and the rate of the water current is 1.9 km/hr. College Algebra 33 A Candy Barrel shop manager mixes M&M s worth $2.00 per pound with trail mix worth $1.50 per pound. How many pounds of each should she use to get 50 pounds of a party mix worth $1.80 per pound? 1. UNDERSTAND To find out the cost of any quantity of items we use the formula price per unit • number of units = price of all units Since we are looking for two quantities, we let x = the amount of M&M s y = the amount of trail mix College Algebra 34 2. TRANSLATE Fifty pounds of party mix x + y = 50 price per unit • Price of M&M s 2x College Algebra number of units = Price of mixture Price of trail mix + 1.5y price of all units = 1.8(50) 90 35 3. SOLVE We are solving the system x + y = 50 2x + 1.50y = 90 We should use the elimination method since the equations are in pseudo-standard form, and we need to multiply to clear the decimals anyway. 3(x + y) = 3(50) –2(2x + 1.50y) = –2(90) College Algebra 3x + 3y = 150 –4x – 3y = –180 –x = –30 x = 30 36 Now we substitute 30 for x into the first equation. x + y = 50 30 + y = 50 y = 20 4. INTERPRET Check: Substitute x = 30 and y = 20 into both of the equations. x + y = 50 First equation 30 + 20 = 50 T rue 2x + 1.50y = 90 S econd equation 2(30) + 1.50(20) = 90 60 + 30 = 90 T rue State: The store manager needs to mix 30 pounds of M&M s and 20 pounds of trail mix to get the mixture at $1.80 a pound. College Algebra 37 When should the substitution method be used? a coefficient is one or an equation is already solved for a variable When should the elimination method be used? College Algebra coefficients are not one and/or opposite signs 38