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sistem persamaan linear dan kuadrat dalam bahasa inggris

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sistem persamaan linear dan kuadrat dalam bahasa inggris

1. 1. 1 LINEAR AND QUADRATIC EQUATION SYSTEM A. Linear Equation System of Two Variables 1. Definition of Linear Equation System of Two Variables A general linear equation system of two variables x and y can be written as ax + by = c where a,b, and c R Definition Linear equation system of two variables can be written as follow. a1x + b1y = c1 a2x + b2y = c2 where a1 ,a2, b1, b2, c1, and c2 are real numbers. 2. Solution of Linear Equation System of Two Variables A pair of x and y which of satisfy the equation ax + by = c are defined as solution of linear equation system. To determine a solution set of linear equation system, we can use the following methods. a. Graphic method b. Elimination method c. Substitution method d. Mixed method (elimination and substitution) a) Graphic Method
2. 2. 2 To determine the solution of linear equation system of two variables a1x+b1y=c1 and a2x + b2y = c2 by graphic method, we can use the following steps. 1) Graph the straight lines of the equations in Cartesian system of coordinates. 2) Intersection point of those equations is the solution of the linear equation system. Example ; Determine the solution set of 2x + 3y = 6 and 2x + y = -2 by graphic method! Answers : In equations 2x + 3y = 6 for, x 0 3 y 2 0 Thus, graphic 2x + 3y = 6 passes through the points (0,2) and (3,0) In equation 2x + y = -2 for, x 0 -1 y -2 0 Thus ,graphic 2x + y = -2 passes through the points (0,-2) and (-1,0).
3. 3. 3 From the graphic above, the two straight lines of the equations intersect in one point, which is (-3,4). Then, the solution set is {(-3,4)} b) Elimination Method To determine the solution of linear equation system of two variables by elimination method, we can use the following steps. 1) Choose the variables you want to eliminate first, and then match the coefficients by multplying the equations by the appropriate numbers 2) Eliminate one of the variables by adding or subtracting the equations. Example : Determine the solution set of the following linear equation system x + 3y = 1 and 2x – y = 9 by elimination. Answer: x + 3y = 1 x 1 x + 3y = 1 x 2 2x + 6y = 2 2x – y = 9 x 3 2x – y = 9 x 1 2x – y = 9 Thus, the solution set is {(4-1,)} c) Substitution method The idea of substitution method is tosolve one of the equations for one of the variables, and plug this into the other equation. The steps are as follows : 1) Solve one of the equations (you choose which one) for one of the variables (you choose which one). + 7x = 28 x = 4 + 7 y = -7 y = -1
4. 4. 4 2) Plug the equation in step 1 back into the other equation, substitute for the chosen variable and solve for the other. Then you back-solve for the first variable. Example : Determine the solution set the following linear equation system 3x + y = 7 and 2x - 5y = 33 by substitution method! Answer: 3x + y = 7 → y = 7 - 3x y = 7 - 3x substituted into 2x - 5y = 33 → 2x -5 (7-3x) = 33 → 2x - 35 + 15 x = 33 → 2x + 15x - 35 = 33 → 17x = 33 + 35 → 17x = 68 → x = 68/17 → x = 4 x = 4 substituted into y = 7 – 3x y = 7 - 3x y = 7-3 (4) y = 7-12 y = -5 Thus, the solution set is {(4, -5)} d) Mixed Method (Elimination and Substitution) The mixed method is conducted by eliminating one of the variables (you choose which one), and then substituting the result into one of the equations. This method is regarded as the most effective one in solving linear equation system.
5. 5. 5 Example; Determine the solution set the following linear equation system 3x + 7y = -1 and x - 3y = 5 by mixed method (elimination and substitution)! Answer: 3x + 7y = -1 x 1 3x + 7y = -1 x - 3y = 5 x 3 3x – 9y = 15 then, substitute y = -1 into the second equation x - 3y = 5 to get the x value. x -3(-1) = 5 Thus the solution set is {(2,-1)} B. Linear Equation System of Three Variables A general linear equation system of three variables x,y, and z can be written as. ax + by + cz = d where a, b, c, and d R Definition Linear equation system of three variables can be written as follow a1x + b1y + c1z = d1 a2x + b2y + c2z = d2 a3x + b3y + c3z = d3 where a1, a2, a3, b1, b2, b3, c1, c2, c3, d1, d2, and d3 are real numbers Solving a linear equation system of three variables can be done by using the same methods as used in solving linear equation of two variables. However ,mixed - - 16 y =-16 y =-1
6. 6. 6 method (elimination and substitution) is the easiest method in solving problems of linear equation system of three variables. Example: Determine the solution set of the linear equation system below by mixed method! 122 112 1 zyx zyx zyx Answer: 122 112 1 zyx zyx zyx )3.....( )2.....( )1.....( From equations (1) dan (2),eliminate variable z, so we have x + y – z = 1 2x + y +z = 11 _ 3x + 2y = 12 ….. (4) From equations (2) dan (3),eliminate variable z, so we have 2x + y +z = 11 x + 2y +z = 12 _ x - y = -1 ….. (5) From equations (4) dan (5),eliminate variable y, so we have 1-y-x 122y3x 2 1 x x 222 1223 yx yx 5x = 10
7. 7. 7 x = 2 x = 2 substitute into the equation (5) x – y = -1 2 – y = -1 -y = -1 – 2 y = 3 x = 2, y = 3 substitute into the equation (1) x + y – z = 1 2 + 3– z = 1 -z = 1 – 5 z = 4 thus,the solution set is {(2, 3, 4)} C. Quadratic Equation Systems Definition A general quadratic equation system can be written as D. y = ax2 + bx + c E. y = px2 + qx + r where a, b, c, p, q, and r are real numbers , a ≠ 0, and p ≠ 0. Geometrically, the element of the solution set is any intersection point between the parabola y = ax2 + bx + c and then parabola y = px2 + qx + r. The number of elements of solution set is determined by discriminant D = (b - q)2 – 4(a - p)(c - r). a. If D > 0, then quadratic equation system has two elements of solution set. It is also shown by the intersection of two parabolas at two different points on the graph
8. 8. 8 b. If D = 0, then quadratic equation system has one elements of solution set. It is also shown by the intersection of two parabolas at only one point on the graph. c. D < 0, then quadratic equation system has no elements of solution set or {}. It is also shown by the intersection of two parabolas at any point on the graph. Quadratic equations system can use the following methods with elimination, substitution and mixed methods(elimination and substitution). Example : Solve the following quadratic equation system! y = x2 - 5x + 6 y = x2 – 3x + 2 answer ; y = x2 - 5x + 6 ...(1) y = x2 – 3x + 2 ...(2) substitute the equations (1) into the equation (2) x2 - 5x + 6 = x2 – 3x + 2 x = 2 y = (2)2 - 3(2) + 2 = 0 Thus the solution set is {(2,0)}
9. 9. 9 D. Mixed Equation System ; Linear and Quadratic Definition A general mixed equation system-linear and quadratic, can be written as. y = ax + b y = px2 + qx + r where a, b, p, q, and r are real numbers, a ≠ 0, and p ≠ 0. We can use substitution methods to solve mixed equation system-linear and quadratic. By substituting the linear equation y = ax + b into the quadratic equation y = px2 + qx + r, then we have ax + b = px2 + qx + r px2 + qx – ax + r – b = 0 px2 + (q – a)x + r – b = 0 The equation px2 + (q – a)x + r – b = 0 is quadratic equation which has discriminant D = (q- a)2 – 4q(r - b). The number of elements of solution set is determined by the discriminant. a. If D > 0, then mixed equation system-linear and quadratic-has two elements of solution set. It is also shown by the intersection of line and parabola at the two different points on the graph b. If D = 0, then mixed equation system-linear and quadratic- has one elements of solution set. It is also shown by the intersection of line and parabola at only one point on the graph. c. D < 0, then mixed equation system-linear and quadratic- has no elements of solution set or {}. It is also shown by no intersection of line and parabola at any points on the graph.
10. 10. 10 Example: Find the solution set of the following equation system! 2x – y = 2 y = x2 + 5x – 6 answer: 2x – y = 2 ...(1) y = x2 + 5x – 6 ...(2) substitute the equation (1) into the equation (2). 2x – y = x2 + 5x – 6 x2 + 5x – 2x – 6 = 0 x2 + 3x – 4 = 0 (x - 1)(x + 4) = 0 x1 = 1 or x2 = -4 for x1 = 1 y1 = 2(1) – 2 = 0 x2 = -4 y2 = 2(-4) – 2 = -10 thus, the solution set is {(1,0), (-4, -10)
11. 11. 11 REFERENCES Mulyati, Yati, dkk. 2008. Matematika untuk SMA dan MA Kelas X. Jakarta: Penerbit PT Piranti Darma Kalokatama. Marwanta,dkk. 2009. Mathemathics For Senior High School Year X. Jakarta:Yudistira. http://www.slideshare.net/ridharakhmi/sistem-persamaan-linear-dan-kuadrat http://www.belajar-matematika.com/matematika- smp/Sistem%20Persamaan%20Linear%20Dua%20Variabel.pdf http://id.wikipedia.org/wiki/Persamaan_linear
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