6.1Systems of Linear Equations in Two Variables


 Walter Palmer
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1 6.Systems of Linear Equations in Two Variables Groups of equations, called systems, serve as a model for a wide variety of applications in science and business. In these notes, we will be concerned only with groups of equations that are linear, that is, equations that may be written in the general form Ax + By = C. The following are examples of systems of linear equations in two variables. Notice that the equations are linear although they may not be written in general form. + y = 7 8y = y = 7 y = + 9 y = + ( x 5) y ( y + 4) = 8 Solution to a System of Equations: A solution of a system of two linear equations in two variables is an ordered pair that satisfies both equations, that is, it makes both equations in the system TRUE. Example: Is the ordered pair (4, ) a solution to the system x y = 6 y? Solution: To determine if the ordered pair is a solution we must substitute the values of the xcoordinate and ycoordinate into both equations and determine if the resulting equations are true. x y = 6 4 ( ) = = 6 6 = 6 y (4) ( ) Because both equations are true, the ordered pair is a solution to the system of linear equations. Example: Is the ordered pair (, ) a solution to the system + y = + 4y? Solution: To determine if the ordered pair is a solution we must substitute the values of the xcoordinate and ycoordinate into both equations and determine if the resulting equations are true. + y = 3() + = 3 = = + 4y () + 4( ) 4 Because the ordered pair only satisfies the first equation but not the second equation, it is not a solution to the system.
2 Types of Solutions & Graphing: A system of linear equations in two variables may have any one of three different solution types: one solution, no solution, or an infinite number of solutions. To justify these solution types, it will help to look at the graph of a system. Consistent & Independent System: Example: Graph the system x + y = 5 + y = 3 and determine the solution. Solution: Graph both equations on the same coordinate plane using any acceptable method (table, intercepts, slopeintercept form). By examining the graph, it is clear that the equations intersect at a single point. This point is an ordered pair that lies on both lines. Consequently, the point of intersection is the ordered pair that is a solution to both equations and is therefore a solution to the system. A system with at least one solution is called a consistent system. A consistent system with exactly one solution is an independent system. Inconsistent System: Example: Graph the system + y = 5 y = 4 and determine the solution. Solution: Graph both equations on the same coordinate plane using any acceptable method (table, intercepts, slopeintercept form). By examining the graph, it is clear that the equations will never intersect because the lines are parallel. Consequently, because there is no ordered pair that satisfies both equations, there is no solution to the system. A system with no solutions is called an inconsistent system.
3 Consistent & Dependent System: Example: Graph the system + y = 0 y = and determine the solution. Solution: Graph both equations on the same coordinate plane using any acceptable method (table, intercepts, slopeintercept form). Although it appears as though there is only one line, there are actually two. Both lines lie directly on top of the other. Because every ordered pair on one line also lies on the other, there are an infinite number of solutions. A system with at least one solution is called a consistent system. A consistent system with exactly one solution is a dependent system. A quick way to determine the type of solutions is to compare the slope and yintercept of each line.. If the slopes are different the system has one solution and is independent.. If the slopes are the same, then compare the yintercepts: a. If the yintercepts are different, then there is no solution (inconsistent system.) b. If the yintercepts are the same, then there are an infinite number of solutions (dependent system). Example: By examining the slope and yintercept, determine the type of solutions to the system. a. y = 3 y = b. y = 3 y = c. y = + y = + Solution: a. Slopes are different, exactly one solution (independent system). b. Slopes are the same, yintercepts are different, so no solution (inconsistent system.) c. Slopes are the same; yintercepts are the same, so infinite solutions (dependent system.) Example: Determine if the system y = 5 y = 3 x + 3 is independent, dependent or inconsistent. Solution: Write the first equation id slope intercept form to obtain y x 5. Because the slopes are the same but the yintercepts are different, the system is inconsistent. There are no solutions.
4 The Substitution Method: To solve a system of linear equations by substitution:. Solve one equation for either variable.. Substitute the expression equal to the solved variable into the other equation. Solve the resulting equation for the remaining variable. 3. Substitute this value into either equation to solve for the other variable. y = 3 y = by using substitution. Solution: Since the st equation is already solved for y, substitute this expression in for y in the other equation. y = ( 3) = + 3 = = 5 x = 5 Now solve for y. y = 3 y = ( 5) 3 y = 3 The ordered pair (5, 3) is the solution to the system. y = 5 x + 3y = 5 by using substitution. Solution: Solve the nd equation for x. This gives the system: y = 5 x = 3y Substitute the expression for x into the st equation. y = 5 ( 3y ) y = 5 6y + 0 y = 5 y = 5 y = 5 Solve for x. x = 3y x = 3( 5) x = 0 The ordered pair (0, 5) is the solution to the system.
5 y = 8 = 5 Solution: Solve the first equation for y. by using substitution. y = + 8 = 5 Now substitute the expression for y into the nd equation. = 5 5( + 8) = 5 40 = 5 40 = 5 This is a contradiction because A contradiction is a statement that is never true. Therefore, there is no solution (inconsistent system.) 3y = + + 6y = 4 by using substitution Solution: Solve the first equation for y to obtain the system. y = 5 x y = 4 Substitute this expression in for y in the nd equation. + 6y = 4 4 = 4 ( 5 x + 4) + 6 = = 4 This is an identity. An identity is any equation that is always true. Consequently, there are an infinite number of solutions (consistent, dependent system.) The Elimination Method: The elimination method is sometimes called the addition method. The procedure for solving a system by elimination is as follows:. Write both equations in General Form Ax + By = C. Choose the variable to eliminate 3. If necessary, multiply one or both equations by appropriate constants so that the coefficients of the variable to be eliminated are opposites. 4. Add the equations. Then solve the resulting equation for the remaining variable. 5. Substitute the value found in the previous step in either of the original equations and solve for the other variable.
6 x + y = 6 x y = 0 by the elimination method. Solution:. Both equations are already in general form.. Because the y variables have opposite coefficients, the y should be eliminated. 3. Not necessary 4. Add the two equations together. x + y = 6 x y = 0 + 0y x = 5. Solve for the remaining variable using either of the original equations. x + y = 6 The solution is the ordered pair (, 8). + y = 6 y = 8 x 3y = 8 + y = by the elimination method. Solution:. Both equations are already in general form.. Because the coefficient of the x variable in the st equation is, the x variable should be eliminated. 3. Multiply the st equation by (5). x 3y = 8 + y = ( 5)( x 3y) = ( 8)( 5) + y = y = 90 + y = 4. Add the two equations together. y = 90 + y = 0x + 7 y = 0 7y = 0 y = 6 5. Solve for the remaining variable using either of the original equations. x 3y = 8 x 3(6) = 8 x = 0 The solution is the ordered pair (0, 6).
7 7 y = 4 = 6 by the elimination method. Solution:. Both equations are already in general form.. Eliminate the x variable. Note: This choice of variable is completely arbitrary. 3. Multiply the st equation by (3) and the nd equation by (5). 7 y = 4 = 6 ( 3)( 7 y) = (4)( 3) (5)( ) = (6)(5) + y = 7 5y = Add the two equations together. + y = 7 5y = 80 0x 4y = 8 y = 5. Solve for the remaining variable using either of the original equations. = 6 5( ) = 6 x = The solution is the ordered pair (, ). 6 y + = 0 3y by the elimination method. Solution:. Rewrite the st equation in general form. 6 y + = 0 3y. Eliminate the y variable. 3. Multiply the nd equation by () 6y = 3y 6y = 3y 6 y = ( )( 3y) = ( 4)( ) 6y = + 6 y = 8 4. Add the two equations together. 6y = + 6 y = 8 0x + 0y 0 5. This is an inconsistent system with no solutions.
8 Example: Three times a first number decreased by a second number is. The first number increased by twice the second number is. Find the numbers. Solution: Let x = the first number and y = the second number. The following system of equations may be created. y = x + y = Using substitution, solve the first equation for y. y = Substitute this value into the other equation and solve for y. x + y = x + ( ) = x + 6x = 7x = 4 x = Solve for the remaining variable. y = y () y = 6 y = 5 The first number is and the second number is 5. Example: The Phoenix Zoo has different admission prices for adults and children. When Mr. and Mrs. Coughlin went with their 5 children, the bill was $ If Mrs. Weaver and her 3 children got in for $8.50, then what is the price of an adults ticket and what is the price of a child s ticket? Solution: Let c = the cost of a child s ticket and a = the cost of an adult ticket. Then we can create the following equations. a c 3 Solve for c by using elimination. a + 3c = 8.50 a c 3 a + 3c = 8.50 a c 3 ( )( a + 3c) = (8.50)( ) a c 3 a 6c = 37 c = 4 Solve for a using substitution. a + 3c = 8.50 a + 3(4) = 8.50 a = 6.50 Therefore, the cost of an adult ticket is $6.50 and a child s ticket is $4.00.
9 Example: A math teacher has a choice of two plans for renting furniture for his new office. Under plan A he pays $800 plus $50 per month, while under plan B he pays $00 plus $00 per month. Create a cost function for each plan. For what number of months do the two plans cost the same? Which plan costs less in the long run? Solution: Let m = the number of months and C = the cost of the plan for m months. Then we can create the equations. C = 50m Solve for m using substitution. C 50m = 600 m = = 00m m + 00 = 50m Solve for C using substitution. Either equation may be used. C C C = 50m = 50() = 600 The breakeven point is m = months and the cost at this time is $600. This is the point of intersection of the two graphs and represents where the two equations (plans) are equal. To determine which plan is best in the long run look at the graph. After the breakeven point the function C = 50m 800 is lower than the other function. + Therefore, the best deal in the long run will be plan A.
10 Example: One plan for federal income tax reform is to tax an individual s income in excess of $5,000 at a 7% rate. Another plan is to institute a national retail sales tax of 5%. If Elyssa spends 75% of her income in retail stores where it is taxed at 5%, then for what income would the amount of tax be the same under either plan? Solution: Let x = income and let T = Tax plan. We can then create the following equations. T T = 0.7( x 5,000) = 0.5(0.7) Solve for x using substitution. T = T 0.7( x 5,000) = 0.5(0.7) 0.7x 550 = = 550 x = 44,347.8 Solve for T using substitution. Either equation may be used. T T = 0.5(0.7) = The tax will be the same at an income of x = $44,347.8 and the tax on this amount is $4,989.. This is the point of intersection of the two graphs and represents where the two equations (plans) are equal.
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