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Systems of Equations Solving a system of linear equations means to determine all the points that are common to all the lines in that system of equations. In Grade 10 students learned how to solve a 2x2 system of linear equations by several methods. These methods were: a) comparison b) graphing c) substitution In Grade 11, students will expand these methods to include addition and subtraction (elimination) and the inverse matrix. Before introducing these new methods, a review of the previously learned methods is necessary. Examples: 1. Solve the following system of linear equations by the comparison method. x 4 y 1 4 x 5 y 15 Solution: Solve each equation in terms of the same variable. Usually students will solve the equations in terms of the variable y since they have done this when determining the equation of a line and when graphing. Express each equation in the form y mx b x 4 y 1 x x 4 y x 1 4y x 1 4 y 1x 1 1 1 1 4 y x 1 4 1 1 y x 4 4 4 1 1 y x 4 4 4 x 5 y 15 4 x 4 x 5 y 4 x 15 5 y 4 x 15 5 4 15 y x 5 5 5 4 y x3 5 Since both equations are equal to y, they are equal to each other. Set both mx + b parts equal to each other and solve the linear equation. 1 1 4 x x3 4 4 5 20 1 x 20 1 20 4 x 203 4 4 5 5x 5 16x 60 5x 16x 5 16x 16x 60 11x 5 60 11x 5 5 60 5 11x 55 11 55 x 11 11 x 5 Substitute this value into one of the equations to determine the value of y. 4 x3 5 4 y 5 3 5 y 4 3 y The common point or the solution is (-5,-1) y 1 2. Solve the following system of linear equations by graphing. 2 2 x 5 y 5 y x 1 5 Put each equation in the form y mx b 4 x 2 y 22 y 2 x 11 Graph both lines on the same axis: The point of intersection of the two lines is the common point or the solution. The solution is (5, 1). 3. Solve the following system of linear equations by using the substitution method. 2 x 3 y 27 4 x y 5 Solution: Solve one of the equations in terms of one of the variables. The process of substitution will involve fewer and simpler computations if one of the variables has a coefficient of 1 or -1. Since the coefficient of y is -1, solve for that variable. 4x y 5 4 x 4 x y 4 x 5 y 4 x 5 1 4 5 y x 1 1 1 y 4x 5 Substitute this expression for y into the other equation wherever the variable y appears. Solve the linear equation. 2 x 3 y 27 2 x 34 x 5 27 2 x 12 x 15 27 14 x 15 27 14 x 15 15 27 15 14 x 42 14 42 x 14 14 x3 Substitute this value into the equation to determine the value of y. y 4x 5 y 43 5 y 12 5 y7 The common point or the solution is (3, 7) Exercises: 1. Solve the following system of linear equations by using the method of comparison. x 3 y 15 0 7 y 3x 21 2. Solve the following system of linear equations by using the method of substitution. x 3 y 11 3x 2 y 30 3. Solve the following system of linear equations by using the method of graphing. 2 x y 7 0 x 2 y 1 0 Solving a system of linear equations by adding and subtracting (elimination). To solve a system of linear equations by elimination requires one of the variables, in both equations, to have opposite numerical coefficients. This can be achieved by multiplying one or both of the equations by an appropriate value. When this is done, the equations are then combined. This will eliminate the variable with the opposite coefficients and will leave one variable in the answer equation. This equation must be solved. The solution will then be substituted into one of the given equations of the system to determine the value of the variable that was eliminated. To facilitate this procedure, the system should be aligned as a1 x b1 y c1 a2 x b2 y c2 Example1: Solve the following system of linear equations by using the method of elimination. 3x 5 y 12 2 x 10 y 4 The system is properly aligned. Choose one variable and perform the necessary multiplication to produce opposite coefficients for that variable in both equations. Solution: 23x 5 y 12 6x – 10y = 24 This equation was multiplied by 2 in 2 x 10 y 4 2x + 10y = 4 order to change the coefficient of y to a negative 10. This makes the 8x = 28 coefficients of y opposites. When the 8 8 equations were combined (added), the y’s were eliminated. x = 3.5 Substitute this value into one of the given equations of the system to determine the value of y. 2(3.5) + 10y = 4 7 + 10y = 4 7 – 7 + 10y = 4 – 7 10y = -3 10 10 y = – 0.3 The common point or the solution is (3.5, – 0.3) Example 2: Solve the following system of linear equations by using the method of elimination. 2 x 9 y 1 4 x y 15 22 x 9 y 1 4 x y 15 4 x 18 y 2 4 x y 15 17 y 17 17 17 y 17 17 y 1 Substitute this value into one of the given equations of the system to determine the value of x. 4 x y 15 4 x 1 15 4 x 1 15 4 x 1 1 15 1 The common point or the solution is (4,-1) 4 x 16 4 16 x 4 4 x4 Example 3: Solve the following system of linear equations by using the method of elimination. 3x 2 y 9 0 4x 3y 5 3x 2 y 9 0 3x 2 y 9 9 0 9 4 x 3 y 3 y 3 y 5 4 x 3 y 5 3x 2 y 9 4x 3y 5 3 x 2 y 9 4x 3y 5 The system is properly aligned. 33 x 2 y 9 24 x 3 y 5 9 x 6 y 27 8 x 6 y 10 17 x 17 17 x 17 17 17 x 1 Substitute this value into one of the given equations of the system to determine the value of y. 3x 2 y 9 0 3 1 2 y 9 0 3 2y 9 0 2y 6 0 2y 6 6 0 6 The common point or the solution is (-1,-3) 2 y 6 2 6 y 2 2 y 3 Example 4: Solve the following system of linear equations by using the method of elimination. 2 3 20 3 x 20 2 y 202 4 x 5 y 2 4 5 1 3 113 1 3 113 x y 14 x 14 y 14 7 7 2 7 2 7 215 x 8 y 40 152 x 21y 226 30 x 16 y 80 30 x 315 y 3390 331y 3310 331 3310 y 331 331 y 10 15 x 8 y 40 2 x 21y 226 Substitute this value into one of the given equations of the system to determine the value of x. 3 2 x 10 2 4 5 3 x4 2 4 3 x44 24 4 3 x6 4 4 3 x 46 4 3 x 24 The common point or the solution is (8,10) 3 24 x 3 3 x8 Exercises: Solve the following system of linear equations by using the method of elimination. 16 x y 181 0 1. 19 x y 214 3 26 x y 4. 5 5 4 y 61 7 x x 69 6 y 2. 3x 4 y 45 x 7 y 38 3. 14 y x 46 Solutions: x 3 y 15 0 1. 7 y 3x 21 Comparison Method x 3 y 15 0 x x 3 y 15 0 x 3 y 15 15 x 15 7 y 3x 21 7 y 3x 3x 3x 21 7 y 3x 21 7 3 21 y x 7 7 7 3 y x3 7 3 y x 15 3 1 15 y x 3 3 3 1 y x5 3 1 3 x5 x3 3 7 1 3 21 x 215 21 x 213 3 7 7 x 105 9x 63 7 x 105 105 9x 63 105 7 x 9x 42 7 x 9x 9x 9x 42 2x 42 2 42 x 2 2 x 21 1 x5 3 1 y 21 5 3 y 75 y 12 y The solution is (21, 12) x 3 y 11 2. 3x 2 y 30 x 3 y 11 x 3 y 3 y 3 y 11 x 3 y 11 Substitution Method x 3 y 11 x 3 9 11 x 27 11 x 16 3x 2 y 30 3 3 y 11 2 y 30 9 y 33 2 y 30 7 y 33 30 7 y 33 33 30 33 7 y 63 7 63 y 7 7 y 9 The solution is (16,-9). 2 x y 7 0 3. x 2 y 1 0 2x y 7 0 2x 2x y 7 0 2x y 7 2 x y 7 7 2 x 7 y 2 x 7 x 2y 1 0 x x 2y 1 0 x 2 y 1 x 2 y 1 1 x 1 2 y x 1 2 1 1 y x 2 2 2 1 1 y x 2 2 Graphing Method Following is the graph of y 2 x 7 and y 1 1 x . 2 2 The solution is the point of intersection (3, 1). Solutions: Elimination Method 16 x y 181 0 1. 19 x y 214 16 x y 181 0 16 x y 181 181 0 181 16 x y 181 16 x y 181 19 x y 214 16 x y 181 19 x y 214 3x 33 3 33 x 3 3 x 11 The solution is (11, -5). 1(16 x y 181) 19 x y 214 16 x y 181 16(11) y 181 176 y 181 176 176 y 181 176 1 5 y 1 1 y 5 x 69 6 y 2. 3x 4 y 45 x 69 6 y 3 x 4 y 45 x 6 y 69 6 y 6 y 3 x 4 y 45 4 y x 6 y 69 3 x 4 y 45 x 6 y 69 3 x 4 y 45 3( x 6 y 69) 3 x 4 y 45 3x 18 y 207 3x 4 y 45 14 y 252 14 252 y 14 14 y 18 x 69 6 y x 69 6(18) x 69 108 x 39 The solution is (-39, -18) x 7 y 38 3. 14 y x 46 x 7 y 38 14 y x 46 x 7 y 7 y 7 y 38 14 y x x x 46 x 7 y 38 x 14 y 46 x 7 y 38 x 14 y 46 2( x 7 y 38) x 14 y 46 2 x 14 y 76 x 14 y 46 3x 30 3 30 x 3 3 x 10 x 7 y 38 10 7 y 38 10 38 7 y 38 38 28 7 y 28 7 y 7 7 4 y The solution is (10, -4). 3 26 x y 4. 5 5 4 y 61 7 x 3 26 y 5 5 3 26 5 x 5 y 5 5 5 5 x 3 y 26 x 5 x 3 y 26 7 x 4 y 61 4 y 61 7 x 4 y 7 x 61 7 x 7 x 7 x 4 y 61 4(5 x 3 y 26) 3(7 x 4 y 61) 20 x 12 y 104 21x 12 y 183 41x 287 41 287 x 41 41 x7 The solution is (7, 3) 4 y 61 7 x 4 y 61 7(7) 4 y 61 49 4 y 12 4 12 y 4 4 y3