A Linear Equation System of Two Variables

1. ALinear Equation System of Two Variables

2. 1. Definition of Linear Equation Systems of Two VariablesA general linear equation system of two variables x and y can be written asax + by = c where a, b, and c ЄR 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.

3. 2. Solution of Linear Equation Systems of Two VariablesA pair of x and y which 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 methodb. Elimination methodc. Substitution methodd. Mixed method ( elimination and subtitution)

4. a. Solving Lincar Equation Systems of Two Variables by Graphic MethodTo determine the solution of linear equation system of two variables a1 x + b1 y = c1 and a2 x + b2y = c2 by graphic method, we can use the following steps.1) Graph the straight lines of the equations in Carstesian system of coordinates.2) Intersection point of those equations is the solutions of the linear equation system.

5. Example 3.1 Determine the solution set of 2x + 3y = 6 and 2x + y = -2 by graphic method! Answer: In equation 2x + 3y = 6 For x = 0 → y = 2 y = 0 → x = 3 Thus, graphic 2x + 3y = 6 passes through the points (0, 2) and (3, 0). In equation 2x + 2y = -2 For x = 0 → y = -2 y = 0 → x = -1 Thus, graphic 2x + y = -2 passes through the points (0, -2) and (-1,0). 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)}.

6. b. Solving Linear Equation Systems of Two Variables by 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 variable you want to eliminate first, and then match the coefficients by multiplying the equations by the appropriate number. 2)Eliminate one of the variables by adding or subtracting the equations. To be able to determine the solution of linear equation system of two variables by elimination method, study the following example.

7. Example 3.2 Determine the solution set of the following linear equation system X + 3x = 1 2x – y = 9 by elimination method ! x + 3y = 1| x2 | = 2x + 6y = 2 2x – y = 9| x1 | = 2x – y = 9 7y = -7 Answer: y = -1 x + 3y = 1 | x1 | = x + 3y = 1 2x – y = 9 | x3 | = 6x – 3y = 27 7x = 28 x = 4 Thus, the solution set is {(4, -1)}

8. c. Solving the Linear Equation System of Two Variables by Subtitution Method The idea of subtitution method is to solve one of the equations for one of the variables, and plug this into the other equation. The steps are as follows. Solve one of the equations ( you choose which one) for one of the variables (you choose which one). Plug the equation in step 1 bavk into the other equation, subtitute for the chosen variable and solve for the other. Then you back-solve for the first variable. To understand the solution of linear equation system of two variables by subtitution method, study the following example.

9. Example 3.3 Determine the solution set of the following linear equation system 2x – 3y = -7 3x + 5y = -1 by subtitution method! Answer: 2x – 3y = -7 → 3y = 2x + 7 y = 2x + 7 3 Then, subtitute y = 2x + 7 into the second equation 3x + 5y = -1 to get the x value. 3 3x + 5 (2x + 7) = -1 3x + 10x + 35 = -1 3 3 ↔ 9x + 10 x + 35 = -3 ↔ 19x = -38 9 ↔ x = -2 Now we can subtitute this x value back into the expression y = 2x + 7, so we get Y= 2(-2) + 7 = 1 3 3 Thus solution set is {(-2, 1)}

10. d. Solving the Linear Systems of Two Variables by Mixed Method ( Elimination and Subtitution) The mixed method is conducted by eliminating one the variables (you choose which one), and then subtituting the result into one of the equations. This method is regarded as the most effective one is solving linear equation system. To understand the method of finding the solution, study the following example.

11. Example 3.4 Determine the solution set of the following linear equation system 3x + 7x = -1 X – 3y = 5 by mixed method (elimination and subtitution)! Answer: 3x + 5y = -1 | x1 | = 3x + 7y = -1 X – 3y = 5 | x3 | = 3x – 9y = 15 16y = -16 ↔ y = -1 Then, subtitute y = -1 into the second equation x – 3y = 5 to get the x value. x – 3(-1) = 5 ↔ x + 3 = 5 ↔ x = 2 Thus, the solution set is {(2, -1)}

12. 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 Є R. Definition Linear equation system of three variables can be written as follow. a1x + b1y + c1z = d1 a2x + b2y + c3z = d3 a3x + b3y + c3z = d3 Where a1, a2, a3,b1, b2,, b3,c1, c2, c3, d1, d2, and d3 are real numbers.

13. Example 3.5 Determine the solution set of the linear equation system below! 2x – 3y + 2z = -1 3x + 2y – z = 10 -4x – y – 3z = -3 Answer: 2x – 3y + 2z = -1 … (1) 3x + 2y – z = 10 … (2) -4x – y – 3z = -3 … (3) From equations (1) and (3), eliminate x, so we have 2x – 3y + 2z = -1 | x2 | 4x – 2y +4z = -2 3x + 2y – z = 10 | x1 | -4x – y – 3z = -3 -3y + z = -5 … (4)

14. From equations (2) and (3), eliminate variable x, so we have3x + 2y – z = 10 | x4 | 12x + 8y – 4z = 40-4x – y – 3z = -3 | x3 | -12x – 3y – 9z = -9 5y – 13z = 31 … (5)then, eliminate variable y from equations (4) and (5) o get the x value.-3y + z = -5 | x5 | -15y + 5z = -255y – 13z = 31 | x3 | 15y – 39z = 93 -34z = 68 z = -2subtitute, value z= -2 and y= 1 into the equation (1), so we have2x – y +2z = -12x – 1 + 2(-2) =-12x – 1 – 4 = -12x = 4x = 2thus, the solution set is {(2, 1, -2)}

15. C. Application of Linear Equation System of Two and Three Variables

16. Example 3.7MrYudi paid Rp 10,250.00 for 2 tickets for adult and 3 tickets for kid. While Joko paid Rp 9,250.00 for 3 tickets for adult and 1 ticket for kid. If Andhika gives Rp 10,000.00 for 1 ticket for adult and 1 ticket for kid, how much is the change received by Andhika?Answer:suppose that the price of a ticket for adult is x rupiahs, and the price of a ticket for kid is y rupiahsthen, we have equation system below2x + 3y = Rp 10,250.00 ... (1)3x + y = Rp 9,250.00 ... (2)Eliminate the equation (1) and (2) to get the y value. 2x + 3y = Rp 10,250.00 | x3 | 6x + 9y = 30,7503x + y = Rp 9,250.00 | x2 | 6x + 2y = 18,500 7y = 12,250 y = 1,750Subtitute the equations (3) into the equation (1) to get the x value.2x + 3y =- 10,2502x + 3(1,750) = 10,2502x = 5,000x = 2,500thus change received by Andika is= Rp 10,000.00 – (x+y)= Rp 10,000.00 – 2,500 – 1,750= Rp 5,750.00

17. D. Mixed Equation System; Linear and QuadraticDefinitionA general mixed equation system-linear and quadratic, can be written as. y = ax + b y = px2 + qx + rWhere a, b, p, q, and r are real numbers, a≠0, and p≠0

18. We can use subtitution methods to solve mixed equation system-linear and quadratic. By subtituting 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-q)x+r – b = 0the equation px2 + (q-q)x+r – b = 0 is quadratic equation which has discriminant D = (q-a)2 -4p(r-b).The number of elements of solution set is determined by the discriminant.1. 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.2. 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 points on the graph.3. If D < 0, 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.

19. a. 2x – y = 2y = x2 + 5x – 6Answer:a. 2x – y = 2 y = 2x - 2 … (1) y = x2 + 5x – 6 … (2)subtitute the equation (1) into the equation (2).2x – 2 = x2 + 5x – 6x2 + 5x – 2x – 6 + 2 = 0x2 +3x – 4 = 0(x-1)(x+4) = 0For x1 = 1 or x2 = -4For x 1 = 1 y1 = 2(1) – 2 = 0x2 = -4 y2 = 2(-4) – 2 = -10Thus, the solution set is {(1, 0), (-4, -10)}

20. E. Quadratic Equation Systems (Enrichment Subject)DefinitionA general quadratic equation system can be written asy = ax2 + bx + cy = px2 + qx + rWhere a, b, c, p, q, and r are real numbers, a≠0 and p≠0

21. 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 the discriminant D = (b-q)2 – 4(a-p)(c-r).1. If D > 0, then quadratic equation system has two elements of solution set. it is also shown by the intersection of two parabolas at the two different points on the graph.2. If D =0, then quadratic equation system has one elements of solution set. it is also shown by the intersection of two parabolas at the one points on the graph.3. If D < 0, then quadratic equation system has no elements of solution set or { }. it is also shown by no intersection of two parabolas at any points on the graph.

22. Example 3.10a. y = x2 + 3x – 18 y = -x2 + x + 6Answer:a. y = x2 + 3x – 18 … (1) y = -x2 + x + 6 … (2)Subtitute the equation (1) into the equation (2) x2 + 3x – 18 = -x2 + x + 6 x2 +x2 +3x – x – 18 – 6 = 0 2x2 + 2x – 24 = 0 Y= x2 + 3x - 18 Y = -x2 + x + 6