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Definition Linear equation system of two variables can be written as follow a 1 x + b 1 y = c 1 a 2 x + b 2 y = c 2 Where a 1, a 2, b 1, b 2, c 1, and.

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Presentation on theme: "Definition Linear equation system of two variables can be written as follow a 1 x + b 1 y = c 1 a 2 x + b 2 y = c 2 Where a 1, a 2, b 1, b 2, c 1, and."— Presentation transcript:

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2 Definition Linear equation system of two variables can be written as follow a 1 x + b 1 y = c 1 a 2 x + b 2 y = c 2 Where a 1, a 2, b 1, b 2, c 1, and c 2 are real numbers.

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5 Determine the solution set of 2x + 3y = 6 and 2x + y = -2 by graphic method! Answer: In equation 2x + 3y = 6 Forx = 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 Forx = 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 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 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 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. 1)Solve one of the equations ( you choose which one) for one of the variables (you choose which one). 2)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 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  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 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  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.  a 1 x + b 1 y + c 1 z = d 1  a 2 x + b 2 y + c 3 z = d 3  a 3 x + b 3 y + c 3 z = d 3  Where a 1, a 2, a 3, b 1, b 2,, b 3, c 1, c 2, c 3, d 1, d 2, and d 3 are real numbers.

13  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)

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22 Y= x 2 + 3x - 18 Y = -x 2 + x + 6


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