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4.1 Solving Systems of Linear Equations in Two Variables Systems of Linear Equations A system of equations consists of two or more equations. The solution of a system of two equations in two variables is an ordered pair (x, y) that makes both equations true. Example Determine whether (–3, 1) is a solution of the system. x–y=–4 2x + 10y = 4 Replace x with –3 and y with 1 in both equations. First equation: –3 – 1 = – 4 True Second equation: 2(–3) + 10(1) = – 6 + 10 = 4 True Since the point (–3, 1) produces a true statement in both equations, it is a solution of the system. Example Determine whether (4, 2) is a solution of the system. 2x – 5y = – 2 3x + 4y = 4 Replace x with 4 and y with 2 in both equations. First equation: 2(4) – 5(2) = 8 – 10 = – 2 True Second equation: 3(4) + 4(2) = 12 + 8 = 20 ≠ 4 False Since the point (4, 2) produces a true statement in only one equation, it is NOT a solution. Solving Systems of Equations by Graphing Since a solution of a system of equations is a solution common to both equations, it is also a point common to the graphs of both equations. To find the solution of a system of two linear equations, we graph the equations and see where the lines intersect. Example Solve the system of equations by graphing. 2x – y = 6 x + 3y = 10 First, graph 2x – y = 6. Second, graph x + 3y = 10. The lines APPEAR to intersect at (4, 2). Y 10 Y Y 109 10 Y 10 989 9 878 8 767 7 656 6 545 5 434 4 323 3 212 2 X 1 1 X X 10 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1-1 1 2 3 4 5 6 7 8 9 10 X 0 1 12 23 34 45 56 67 78 89 10 -10-10 -9 -9 -8 -8 -7 -7 -6 -6 -5 -5 -4 -4 -3 -3 -2 -2 -1-10-1-1 10 0 1 2 3 4 5 6 7 8 9910 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1-2 -1 -2-3-2 -2 -3-4-3 -3 -4-5-4 -4 -5-6-5 -5 -6-7-6 -6 -7-8-7 -7 -8-9-8 -8 -9 -9 -10 -9 -10-10 -10 continued Example (cont) Although the solution to the system of equations appears to be (4, 2), you still need to check the answer by substituting x = 4 and y = 2 into the two equations. First equation: 2(4) – 2 = 8 – 2 = 6 True Second equation: 4 + 3(2) = 4 + 6 = 10 True The point (4, 2) checks, so it is the solution of the system. Helpful Hint Neatly drawn graphs can help when “guessing” the solution of a system of linear equations by graphing. Example Solve the system of equations by graphing. –x + 3y = 6 3x – 9y = 9 First, graph – x + 3y = 6. Second, graph 3x – 9y = 9. The lines APPEAR to be parallel. Y 10 Y 10 Y 10 9 9 9 8 8 8 7 7 7 6 6 6 5 5 5 4 4 4 3 3 3 2 2 2 1 X 1 X 1 X -10 -9 -8 -7 -6 -5 -4 -3 -2 -1-10 1 2 3 4 5 6 7 8 9 10 0 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1-10 1 2 3 4 5 6 7 8 9 10 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1-1 1 2 3 4 5 6 7 8 9 10 -2 -2 -2 -3 -3 -3 -4 -4 -4 -5 -5 -5 -6 -6 -6 -7 -7 -7 -8 -8 -8 -9 -9 -9 -10 -10 -10 continued Example (cont) Although the lines appear to be parallel, we need to check their slopes. –x + 3y = 6 3y = x + 6 1 3 y= x+2 3x – 9y = 9 –9y = –3x + 9 1 y= x–1 3 First equation Add x to both sides. Divide both sides by 3. Second equation Subtract 3x from both sides. Divide both sides by –9. 1 Both lines have a slope of , so they are parallel and do 3 not intersect. Hence, there is no solution to the system. Example Solve the system of equations by graphing. x = 3y – 1 2x – 6y = –2 First, graph x = 3y – 1. Second, graph 2x – 6y = –2. The lines APPEAR to be identical. YY Y 10 10 10 99 9 88 8 77 7 66 6 55 5 44 4 33 3 22 2 11 XX 1 X 00 11 22 33 44 55 66 77 88 9910 -10 -10-9-9-8-8-7-7-6-6-5-5-4-4-3-3-2-2-1-1 -1-1 10 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1-10 1 2 3 4 5 6 7 8 9 10 -2-2 -2 -3-3 -3 -4-4 -4 -5-5 -5 -6-6 -6 -7-7 -7 -8-8 -8 -9-9 -9 -10 -10 -10 continued Example (cont) Although the lines appear to be identical, we need to check that their slopes and y-intercepts are the same. x = 3y – 1 First equation 3y = x + 1 Add 1 to both sides. y = 1x + 1 3 Divide both sides by 3. 3 2x – 6y = – 2 –6y = – 2x – 2 y 1 = 3 x+ 1 3 Second equation Subtract 2x from both sides. Divide both sides by -6. Any ordered pair that is a solution of one equation is a solution of the other. This means that the system has an infinite number of solutions. Identifying Special Systems of Linear Equations There are three possible outcomes when graphing two linear equations in a plane. One point of intersection—one solution Parallel lines—no solution Coincident lines—infinite number of solutions If there is at least one solution, the system is considered to be consistent. If the system defines distinct lines, the equations are independent. Possible Solutions of Linear Equations Graph Number of Solutions Type of System (3, 5) If the lines intersect, the Consistent system of equations has The equations are one solution given by the independent. point of intersection. Two lines intersect at one point. Parallel lines Lines coincide If the lines are parallel, then the system of equations has no solution because the lines never intersect. Inconsistent The equations are independent. If the lines lie on top of each other, then the system has infinitely many solutions. The solution set is the set of all points on the line. Consistent The equations are dependent. The Substitution Method Another method that can be used to solve systems of equations is called the substitution method. To use the substitution method, we first need an equation solved for one of its variables. Then substitute that new expression for the variable into the other equation and solve for the other variable. Example Solve the system using the substitution method. 3x – y = 6 – 4x + 2y = –8 The first equation can be easily solved for y. 3x – y = 6 –y = –3x + 6 y = 3x – 6 Subtract 3x from both sides. Multiply both sides by –1. Now we substitute this value for y in the second equation. –4x + 2y = –8 –4x + 2(3x – 6) = –8 –4x + 6x – 12 = –8 2x – 12 = –8 2x = 4 x=2 Replace y with result from first equation. Use the distributive property. Simplify. Add 12 to both sides. Divide both sides by 2. continued Example (cont) Substitute x = 2 into the first equation, which has already been solved for y. y = 3x – 6 = 3(2) – 6 = 6 – 6 = 0 The solution of the system is the ordered pair (2, 0). Check the point (2, 0) in the original equations. First equation: 3x – y = 6 3(2) – 0 = 6 True Second equation: –4x + 2y = –8 –4(2) + 2(0) = –8 True The solution of the system is (2, 0). The Substitution Method Solving a System of Two Equations Using the Substitution Method Step 1: Solve one of the equations for one of its variables. Step 2: Substitute the expression for the variable found in Step 1 into the other equation. Step 3: Find the value of one variable by solving the equation from Step 2. Step 4: Find the value of the other variable by substituting the value found in Step 3 into the equation from Step 1. Step 5: Check the ordered pair solution in both original equations. Example Solve the system: y = 2x – 5 8x – 4y = 20 Since the first equation is already solved for y, substitute this value into the second equation. 8x – 4y = 20 8x – 4(2x – 5) = 20 Replace y with result from first equation. 8x – 8x + 20 = 20 Use distributive property. 20 = 20 Simplify. continued Example (cont) A true statement such as 20 = 20 indicates that the two linear equations actually represent the same line. There are an infinite number of solutions for this system. Any solution of one equation would automatically be a solution of the other equation. This represents a consistent system and the linear equations are dependent equations. Example Solve the following system of equations: 3x – y = 4 6x – 2y = 4 Solve the first equation for y. 3x – y = 4 –y = –3x + 4 Subtract 3x from both sides. y = 3x – 4 Multiply both sides by –1. Substitute this value for y into the second equation. 6x – 2y = 4 6x – 2(3x – 4) = 4 Replace y with the result from the first equation. 6x – 6x + 8 = 4 Use the distributive property. 8=4 Simplify. continued Example (cont) The false statement 8 = 4 indicates that this system has no solution. The graph of the linear equations in the system would be a pair of parallel lines. This represents an inconsistent system, even though the linear equations are independent. Solving a System Using Elimination Another method that can be used to solve systems of equations is called the addition or elimination method. You multiply both equations by numbers that will allow you to combine the two equations and eliminate one of the variables. Example Solve the following system x+y=7 x–y=9 x+y= 7 x–y= 9 2x = 16 x=8 Add the equations to eliminate y. Divide both sides by 2. continued Example (cont) Substitute 8 for x into one of the original equations and solve for y. x+y=7 8+y=7 y = –1 Replace the x value with 8 in the first equation. Add 8 to both sides of the equation. Our computations have produced the point (8, –1). continued Example (cont) Check the point in the original equations. x+y= 7 8 + (–1) = 7 True x–y= 9 8 – (–1) = 9 True The solution of the system is (8, –1). Example Solve the following system of equations 6x – 3y = –3 4x + 5y = –9 Multiply both sides of the first equation by 5 and the second equation by 3. 5(6x – 3y) = 5(–3) 3(4x + 5y) = 3(–9) 30x – 15y = –15 12x + 15y = –27 42x = –42 x = –1 Add the equations. Solve for y. Divide both sides by 42. continued Example (cont) Substitute –1 for x into one of the original equations and solve for y. 6x – 3y = –3 6(–1) – 3y = –3 Replace the x value in the first equation. –6 – 3y = –3 Simplify. –3y = 3 Add 6 to both sides. y = –1 Divide both sides by –3. Our computations have produced the point (–1, –1). continued Example (cont) Check the point in the original equations. First equation, 6x – 3y = –3 6(–1) – 3(–1) = –3 True Second equation, 4x + 5y = –9 4(–1) + 5(–1) = –9 True The solution of the system is (–1, –1). The Elimination Method Solving a System of Two Linear Equations Using the Elimination Method Step 1: Rewrite each equation in standard form, Ax + By = C. Step 2: If necessary, multiply one or both equations by some nonzero number so that the coefficients of a variable are opposites of each other. Step 3: Add the equations. Step 4: Find the value of one variable by solving the equation from Step 3. Step 5: Find the value of the second variable by substituting the value found in Step 4 into either of the original equations. Step 6: Check the proposed solution in both original equations. Example Solve the system of equations using the elimination method. 2 1 3 x y 3 4 2 1 1 x y 2 2 4 First multiply both sides of the equations by a number that will clear the fractions out of the equations. continued Example (cont) Multiply both sides of each equation by 12. (Note: You do not have to multiply each equation by the same number, but in this case it will be convenient to do so.) First equation, 2 1 3 x y 3 4 2 2 1 3 12 x y 12 4 3 2 8x 3y 18 Multiply both sides by 12. Simplify. continued Example (cont) Second equation, 1 1 x y 2 2 4 1 1 12 x y 12 2 4 2 6 x 3 y 24 Multiply both sides by 12. Simplify. Add the two resulting equations. 8x + 3y = – 18 6x – 3y = – 24 14x = – 42 x = –3 Divide both sides by 14. continued Example (cont) Substitute –3 for x into one of the original equations. 8x + 3y = –18 8(–3) + 3y = –18 –24 + 3y = –18 3y = –18 + 24 = 6 y=2 We need to check the ordered pair (–3, 2) in both equations of the original system. continued Example (cont) Check the point in the original equations. First equation: Second equation: 2 1 3 x y 3 4 2 1 1 x y 2 2 4 2 1 3 (3) (2) 3 4 2 1 1 (3) (2) 2 2 4 1 3 2 2 2 3 1 2 2 2 True The solution is (–3, 2). True Special Cases In a similar fashion to what you found in the last section, use of the addition method to combine two equations might lead you to results like … 5 = 5 (which is always true, thus indicating that there are infinitely many solutions, since the two equations represent the same line), or 0 = 6 (which is never true, thus indicating that there are no solutions, since the two equations represent parallel lines).