6.1 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 Step 2. Substitute 5 6 in 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 , : 7. The population y in year x of Long Beach and new Orleans is approximated by the equations : 9.92 2 722 : 3.87 558 Where x = 0 corresponds to 1980 and y in thousands. In what year do the two cities have the same population? Solution: We multiply both sides of equation (2) by ‐2 so that the coefficient of y in the two equations are negatives of one another. 9.92 2 722 7.74 2 1116 17.66
394 22.31 The year 1980 +22 = 2002 the two cities will have the same population. 8. An apparel shop sells skirts for $45 and blouses for $35. Its entire stock is worth $51,750, but sales are slow and only half the skirts and two‐thirds of the blouses are sold, for a total of $30,600. How many skirts and blouses are left in the store? x = number of skirts in the store y = number of blouses in the store 45 35 51,750 45 1 2 35 2 3 30,600 45 35 51,750 45 70 30,600 2 3
Multiply equation (1) by 1 2
45 45
2 2
35 70
2 3
25,875 30,600
35 6 4725 810 45 35 810 51,750 45 28,350 51,750 45 23,400 520 Half the skirts are left (they sold half) and one‐third of the blouses are left (they sold two‐thirds) Therefore, there are 260 skirts left and 270 blouses left in the store. 9. A company purchases two models of bicycles: model 201 and model 301. Model 201 requires 2 hours of assembly time, and model 301 requires 3 hours of assembly time. The parts for model 201 cost $25 per bike, and the parts for model 301 cost $30 per bike. If the company has a total of 34 hours of assembly time and $365 available per day for these two models, how many of each can be made in a day? x = number of 201 models y = number of 301 models 2 3 34 25 30 365 20 30 340 25 30 365 5
25 5
2 5 3 34 10 3 34 3 24 8 5 201 models and 8 301 models can be produced in one day.