Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
5.1 Systems of Linear Equations An equation is a linear equation in two variables x and y. Similarly, is a linear equation in three variables x, y, and z. We may also consider linear equations in four, five, or any number of variables. A system of linear equations is collection of two or more linear equations each containing one or more variables. In this section we shall consider only systems of two linear equations in two variables. Systems involving more than two variables are discussed in the next section. We can view the problem of solving a system of two linear equation containing two variables as a geometry problem. The graph of each equation in such as system is a line. So, a system of two equations containing two variables represents a pair of lines. 1. intersecting lines; system 2. Parallel lines; system 3. Lines coincide; system has exactly one solution has no solution. has infinitely many (system is inconsistent) solutions (system is dependent) If the system has infinitely many solutions (dependent system) the system is solved by finding the general form of the solution set. The two methods of solving a system of linear equations that will be discussed in this section are the substitution method and the elimination method. Steps For Solving By Substitution 1. Pick one of the equations and solve for one of the variables in terms of the other variable. 2. Substitute this quantity in the remaining equation and solve. Find the value of the remaining variable by back Solve by Substitution. 3 2 5 1. 5 6 Solution: Step 1 The easiest to solve is y in equation 2. 5 6 in 5 6 Step 2. Substitute 6 5 equation 1. 3 2 5 Step 2. Substitute 5 6 in equation 1. 3 2 5 3 2 5 6 5 3 10 12 5 7 12 5 7 7 1 5 6 5 1 5 1 The solution is 1, 1 3 5 2. 2 3 8 Step 1 The easiest to solve is x in equation 1. 3 5 5 3 Step 2. Substitute 5 3 in equation 2. 2 3 8 2 5 3 3 8 10 6 3 8 10 9 8 9 18 2 5 3 5 3 2 1 The solution is 1,2 2 4 3. 4 2 3 Step 1 The easiest to solve is y in equation 1 2 4 4 2 Step 2. Substitute 4 2 in equation 2. 4 2 3 4 2 4 2 3 4 8 4 3 8 3 The System has no solution. The system is inconsistent. Steps For Solving By Elimination 1. Multiply both sides of one (or both) equation(s) by the appropriate nonzero numbers so that when the equations are added together one of the variables will be eliminated. 2. Solve this equation for the remaining variable. 3. Find the value of the remaining variable by back substitution. Solve by Elimination 2 3 1 4. 3 Solution: We multiply both sides of equation (2) by 2 so that the coefficient of x in the two equations are negatives of one another. 2 3 1 2 2 6 5 5 1 Now find the value of the remaining variable by back substitution. 2 3 1 2 3 1 1 2 3 1 2 4 2 The solution is 2, 1 . 3 6 2 5. 5 4 1 Solution: The smallest common multiple between 6 and 4 is 12, so we multiply both sides of equation (1) by 2 and both sides of equation (2) by 3 so that the coefficient of y in the two equations are negatives of one another. 6 12 4 15 12 3 21 7 1 3 Now find the value of the remaining variable by back substitution. 3 6 2 3 13 6 2 1 6 2 6 1 1 6 The Solution to the system is 1 3 , 1 6 . 2 4 2 4 8 Solution: We multiply both sides of equation (1) by ‐2 so that the coefficient of x in the two equations are negatives of one another. 2 4 8 2 4 8 6. 0 0 The system is a dependent system of linear equations and thus has infinitely many solutions. To determine the general form of the solution set, we solve for one of the variables in either equation. 2 4 4 2 The General Form of the solution set is 4 2 , : 2 5 7. 2 2 Solution: Pick any pair of equations and eliminate a variable. We will choose equations 1 and 2 and eliminate z. 2 2 5 3 2 7 Now pick another pair of equations and eliminate the same variable, namely z. We will choose equations 2 and 3. 2 5 2 3 3 1 3 2 7 3 1 2 7 2 4 2 2 1 2 z 2 3 2 1 Solution 1,2, 1 4 3 2 5 15 8. 3 2 4 14 Solution: Pick equations 1 and 3 and eliminate y. 4 3 2 2 4 14 2 7 12 Pick equations 1 and 2 and eliminate y by multiplying equation 1 by 5. 5 4 3 5 2 20 5 15 10 20 5 15 10 3 5 15 23x 14z 5 2 7 12 23x 14z 5 Eliminate z by multiplying equation 1 by ‐2 2 2 7 2 12 4 14 24 4 14 24 23x 14z 5 19 19 1 2 7 12 2 1 7 12 7 14 2 2 4 14 2 1 4 2 14 2 8 14 10 14 4 Solution 1,4,2 5.1 Applications of Systems of Linear Equations 1. One Kung Pau chicken and two Big Macs provide 2620 calories. Two Kung Pao chickens and one Big Mac provides 3740 calories. Find the caloric content of each item. 2 2620 2 3740 Solution One Kung Pao chicken has 1620 calories and one Big Mac has 500 calories 2. A hotel has 200 rooms. Those with kitchen facilities rent for $100 per night and those without rent for $80 per night. On a night when the hotel was completely occupied, revenues were $17,000. How many of each type of room does the hotel have? 200 100 80 17,000 Solution: 50 rooms with kitchens and 150 without 3. A rectangular lot whose perimeter is 360 feet is fenced along three sides. An expensive fencing along the lot’s length cost $20 per foot, and an inexpensive fencing along the two side widths cost only $8 per foot. The total cost of the fencing along the three sides is $3280. What are the lot’s dimensions? 2 2 360 20 8 2 3280 Solution: The length L is 100 feet and the width W is 80 feet 4. At a college play, 400 tickets were sold. The ticket prices were $8, $10, and $12, and the total revenue from ticket sales was $3700. How many tickets of each type were sold if the combined number of $8 and $10 tickets was 7 times the number of $12 tickets? 400 8 10 12 3700 7 Solution: 200 of the $8 tickets, 150 of the $10 tickets, and 50 of the $12 tickets were sold 5. A certain brand of razor blades comes in packages of 6, 12, and 24 blades costing $2, $3, and $4 per package, respectively. A store sold 12 packages containing a total of 162 razor blades and took in $35. How many packages of each type were sold? 12 6 12 24 162 2 3 4 35 Solution: 5 packs of the 6 blades, 3 packs of the 12 blades, and 4 packs of the 24 blades were sold 6. Find the quadratic function 2,7 , 1, 2 , 2,3 . whose graph passes through the points 2 1 2 4 2 2 7 2 3 2, 1, Or 4 Solution: 2 7 2 3 1 2 3 . The function is 2 3