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Math 150 – Fall 2015 Section 7A & 7B 1 of 4 Section 7A – Systems of Linear Equations Geometry of Solutions The standard form for a system of two linear equations in two unknowns is ax + by = c dx + f y = g where the constants a, b, c, d, f, and g are known numbers. A solution of this system is a pair of numbers x0 and y0 which are solutions to both equations. This pair of numbers is commonly written as a point (x0 , y0 ) and interpreted as a point in the Euclidean plane R2 . Example 1. Are the points (4, 0) or (−3, 2) solutions to the system of equations 2x + 7y = 8 −x + y = 5 Example 2. The graph below contains both lines from the previous example. Each point on a line is a solution to the equation of that line. The point where the lines cross is the only solution to both lines. Theorem. When solving a system of two linear equations, there are three possibilities for the number of solutions. 1. There is one solution, a single point where the two lines cross. 2. There are no solutions. The two lines are parallel and never cross. 3. There are an infinite number of solutions (the two equations are the exact same line when graphed). Math 150 – Fall 2015 Section 7A & 7B 2 of 4 Example 3. The three following examples demonstrate all three cases. There is one solution where the lines cross, the point (1, 0). x−y =1 x+y =1 There are no solutions, the lines never cross. 2x + y = 1 2x + y = 3 The two lines are exactly the same. Every point on the line solves both equations, and there are infinitely many solutions. 3x + 2y = 6 6x + 4y = 12 Example 4. Graph the solution set of the system, and from your plot describe the solution set to this system. x+y =2 2x − 3y = 3 x − y = 4. Algebraic Methods – Substitution Substitution Method: To solve a system of linear equations by substitution, perform the following steps: 1. Solve one of the equations for one of the unknowns in terms of the other unknown (i.e., if the unknowns are x and y, solve one of the equations for y.). 2. Substitute the expression for the unknown in the remaining equations. 3. Repeat this process until one of the equations has been reduced to an equation in only one unknown. 4. Solve this equation for the unknown. 5. Use this value to determine the values of the other unknowns. Math 150 – Fall 2015 Section 7A & 7B 3 of 4 Example 5. Solve the following equations 1. x − 3y = 6 2x + 5y = 7 2. 4x − 3y = 5 8x − 6y = 10 Algebraic Methods – Elimination Elimination To solve a system of linear equations by elimination, perform the following steps: 1. Multiply one of your equations by a number so that two equations have the opposite number of the same variable (i.e., if one equation has 6x, and the second has 2x, multiply the second equation by −3. ) 2. Add the two equations together (use the new equation after multiplying by a constant). 3. Now you have a new equation with one less variable. 4. Continue this process eliminating variables, until you have an equation with only one variable left. 5. Solve this equation for the variable. 6. Use this value to determine the value of the other variables. Example 6. Solve the following systems of equations: 1. 3x − 5y = 16 x+y =2 2. 2x − 6y = 5 −x + 3y = 4 Math 150 – Fall 2015 Section 7A & 7B 4 of 4 Example 7. Suppose a plane flies a round trip between two cities. The flight from the first city is into a strong headwind and takes 1 hr and 30 minutes. The return flight is with the wind and takes 50 minutes. If the cities are 100 miles apart, what is the aircraft’s speed and the winds speed. Assume that both the aircraft’s and wind’s speeds are constant. Example 8. Suppose a lab worker needs to make a 15% acid solution, but the lab only has 10% and 45% acid solutions. How many milliliters of the 15% and 45% solutions should the worker use to get 350 mL of a 15% solution? 7B – Non-linear Systems of Equations To solve a non-linear system of equations, we use similar methods as before. Try to use either the substitution method or elimination method. Example 9. Solve the following non-linear systems of equations. 1. (x − 2)2 + y 2 = 9 y − (x − 2)2 = −3 2. x2 + y 2 = 5 x x2 2 6 − 6 +y =4