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## Chapter 3: Systems of Linear Equations

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**Two or more linear equations form a system of linear**equations.Example: • A solution to a system of linear equationsis an ordered pair that is a solution to each individual linear equation.(–2, –4) is a solution to the system above because it is a solution to each equation.**A solution to a system of two linear equations may be**interpreted graphically as a point of intersection between the two lines.**Two lines drawn in a rectangular coordinate system may**intersect at exactly one point, no point, or infinitely many points (the lines are coinciding).**A consistent system of linear equations has one or more**solutions. • An inconsistent system has no solution. • A dependent system has infinitely many solutions (both equations in the system represent the same line). • An independent system is one in which the two equations represent different lines.**To solve a system of linear equations by graphing, graph**both equations and finding the point(s) of intersection. It can be helpful to first write the equations in slope-intercept form. • Different slopes: lines are different and nonparallel, thus must cross in exactly one point • Same slopes, different y-intercepts: lines are parallel and do not intersect • Same slopes, same y-intercepts: equations represent the same line**Exercise 1**Which point is a solution to the system? 10x + 9y = 154 –x + 10y = 50 A) (10, 10) B) (10, 6) C) (6, 10) D) (6, 6)**Exercise 2**Which of the following describes the system of equations whose graph is shown? A) Consistent and Independent B) Consistent and Dependent C) Inconsistent and Independent D) Inconsistent and Dependent**Exercise 3**Solve the system using the graphing method. x + y = 13 3x + y = 35 A) (12, –1) B) (11, 2) C) (2, 11) D) No solution**3.2 Solving Systems of Equations by Using the Substitution**Method**The substitution method is another technique used to solve a**system of equations. • This method can be used when the system includes nonlinear equations.**Solving a System of Equations by the Substitution Method**• Isolate one of the variables from one equation. • Substitute the quantity found in step 1 into the other equation. • Solve the resulting equation. • Substitute the value found in in step 3 back into the equation in step 1 to find the value of the remaining variable. • Check the solution in both equations, and write the answer as an ordered pair.**ExampleSolve the system by using the substitution**method. 3x – 2y = –7 6x + y = 6 Step 1: Solve the second equation for y.**Example (continued)Solve the system by using the**substitution method. 3x – 2y = –76x + y = 6 Step 2: Substitute the quantity –6x + 6 for y in the other equation.**Example (continued)Solve the system by using the**substitution method. 3x – 2y = –76x + y = 6 Step 3: Solve for x.**Example (continued)Solve the system by using the**substitution method. 3x – 2y = –76x + y = 6 Step 4: Substitute x = into the expression y = –6x + 6.**Example (continued)Solve the system by using the**substitution method. 3x – 2y = –76x + y = 6 Step 5: Check the ordered pair in each original equation. The solution is .**When using the substitution method to solve a system of**linear equations, if the equation in step 3 reduces to a contradiction, then the system has no solution and is inconsistent. • If the equation in step 3 reduces to an identity, the system is dependent and the solution may be represented by the set of all solutions to either equation in the system.**Exercise 4**Solve the system using the substitution method. y = –63 – 7x 3x + 7y = –73 A) (8, –8) B) (–8, –7) C) (–7, –8) D) (–7, 9)**Exercise 5**Solve the system using the substitution method. y = 7x + 8 –3y + 21x = –6 A) (8, –6) B) (0, 8) C) There are infinitely many solutions. D) There is no solution.**Exercise 6**Solve the system using the substitution method. 6x + 10y = 6 30y = 18 – 18x A) (–10, 7) B) (7, –10) C) {(x, y) | 6x + 10y = 6} D) There is no solution.**3.3 Solving Systems of Equations by Using the Addition**Method**Solving a System of Equations by the Addition Method**• Write both equations in standard form Ax + By = C. • Clear fractions or decimals (optional). • Multiply one or both equations by nonzero constants to create opposite coefficients for one of the variables. • Add the equations from step 3 to eliminate one variable. • Solve for the remaining variable. • Substitute the known value found in step 5 into one of the original equations to solve for the other variable. • Check the ordered pair in both equations.**ExampleSolve the system by using the addition method. 4x +**5y = 23x = 1 – 4y Step 1: Write both equations in standard form. 4x + 5y = 2 3x + 4y = 1 Step 2: There are no fractions or decimals.**Example (continued)Solve the system by using the addition**method. 4x + 5y = 23x = 1 – 4y Step 3: Multiply the first equation by 3. Multiply the second equation by –4. 4x + 5y = 2 12x + 15y = 6 3x + 4y = 1 –12x – 16y = –4 Step 4: Add the equations. 12x + 15y = 6 –12x – 16y = –4 –y = 2**Example (continued)Solve the system by using the addition**method. 4x + 5y = 23x = 1 – 4y Step 5: Solve for y. –y = 2 y = –2 Step 6: Substitute y = –2 back into one of the original equations and solve for x. 4x + 5y = 2 4x + 5(–2) = 2 4x – 10 = 2 4x = 12 x = 3**Example (continued)Solve the system by using the addition**method. 4x + 5y = 23x = 1 – 4y Step 7: Check the ordered pair (3,–2) in both original equations. The solution is (3,–2).**If both variables are eliminated when the equations are**added, the two original equations are equivalent and the system is dependent. The solution may be represented by the set of all solutions to either equation in the system. • If the equations reduce to a contradiction when added, the system has no solution. The system is inconsistent and the two equations represent parallel lines.**Exercise 7**Which step would successfully eliminate y from the system below? 6x + 8y = 27 x + 24y = 54 A) Add the 2 equations. B) Multiply the first equation by 3 and add the equations. C) Multiply the second equation by 6 and add the equations. D) Multiply the first equation by –3 and add the equations.**Exercise 8**Solve the system using the addition method. 5x + y = 33 7x – y = 27 A) (5, 8) B) (2, 9) C) (8, 5) D) There is no solution.**Exercise 9**Solve the system of equations using the addition method. 10x – 8y = 0 20x = 16y + 0 A) {(x, y): 10x – 8y = 0} B) (0, 0) C) (10, 8) D) There is no solution.**3.4 Applications of Systems of Linear Equations in Two**Variables**When an application has more than one unknown, it may be**necessary to use multiple variables. • When two variables are present, the goal is to set up a system of two independent equations. • To solve the system, use the graphing method, the substitution method, or the addition method.**There are several types of application problems that can be**solved using a system of linear equations in two variables: • Cost applications • Mixture applications • Applications involving principal and interest • Applications involving distance, rate, and time • Geometry applications**Steps to Solve Applications**• Label two variables. • Construct two equations in words. • Write two equations. • Solve the system. • Write the answer.**Exercise 10**At one store, 5 pairs of jeans and 2 sweatshirts costs $166, while 3 pairs of jeans and 4 sweatshirts costs $164. Find the cost of one sweatshirt. A) $23 B) $24 C) $20 D) $26**Exercise 11**A plane can fly 1,488 miles with the wind in 3 hours, while it takes the same plane 4 hours to fly the same distance against the wind. Find the speed of the plane in still air and the speed of the wind. A) Plane: 496 mph, Wind: 62 mph B) Plane: 992 mph, Wind: 124 mph C) Plane: 372 mph, Wind: 124 mph D) Plane: 434 mph, Wind: 62 mph**Exercise 12**A new sports car, the Revolution, can be leased for $3,000 down and $300 per month, so that the cost of leasing is given by y = 3000 + 300x, where x is the length of the lease in months. The same dealership offers the Yachtsman, a luxury car, for $5,000 down and $260 per month. After how many months would the total cost of leasing the two cars be equal? A) 12 months B) 36 months C) 45 months D) 50 months**3.5 Systems of Linear Equations in Three Variables and**Applications**A linear equation in three variables can be written in the**form Ax + By + Cz = D, where A, B, and C are not all zero. • Solutions to such an equation are ordered triples of the form (x, y, z) that satisfy the equation.**The set of all ordered triples that are solutions to a**linear equation in three variables may be represented graphically by a plane in space.**A solution to a system of linear equations in three**variables is an ordered triple that satisfies each equation. • Geometrically, a solution is a point of intersection of the planes represented by the equations in the system. • A system of linear equations in three variables may have one unique solution, infinitely many solutions, or no solution.**One unique solution (planes intersect at one point)**• The system is consistent. • The system is independent.**No solution(the three planes do not all intersect)**• The system is inconsistent. • The system is independent.**Infinitely many solutions(planes intersect at infinitely**many points) • The system is consistent. • The system is dependent.**Solving a System of Three Linear Equations in Three**Variables • Write each equation in standard form Ax + By + Cz = D. It is often helpful to label the equations. • Choose a pair of equations and eliminate one of the variables by using the addition method. • Choose a different pair of equations and eliminate the same variable.**Solving a System of Three Linear Equations in Three**Variables • After step 3 there should be two equations in two variables. Solve this system using the addition method or the substitution method. • Substitute the values of the variables found in step 4 into any of the three original equations that contain the third variable. Solve for the third variable. • Check the ordered triple in each of the original equations.**ExampleSolve the system.**Step 1:The equations are already in standard form. Label the equations. A B C