example 1
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example 1 Solution by Elimination Chapter 7.1 Solve the system 2 3 1 2 3 3 9 x y z x y z x y z 2009 PBLPathways
description
example 1. Solution by Elimination. Chapter 7.1. Solve the system. 2009 PBLPathways. Solve the system. Solve the system. If necessary, interchange two equations or use multiplication to make the coefficient of x in the first equation a 1. E1 E2. Solve the system. - PowerPoint PPT Presentation
Transcript of example 1
example 1 Solution by Elimination
Chapter 7.1
Solve the system2 3 1
2 33 9
x y zx y zx y z
2009 PBLPathways
2009 PBLPathways
Solve the system2 3 1
2 33 9
x y zx y zx y z
2009 PBLPathways
Solve the system2 3 1
2 33 9
x y zx y zx y z
1. If necessary, interchange two equations or use multiplication to make the coefficient of x in the first equation a 1.
2 32 3 13 9
x y zx y zx y z
E1 E22 3 12 3
3 9
x y zx y zx y z
2009 PBLPathways
Solve the system2 3 1
2 33 9
x y zx y zx y z
2. Add a multiple of the first equation to each of the following equations so that the coefficients of x in the second and third equations become 0.
2 33 5
3 9
x y zy z
x y z
2 32 3 13 9
x y zx y zx y z
-2 R1 + R2 R2
2 2 4 62 3 1
3 5
x y zx y z
y z
2009 PBLPathways
Solve the system2 3 1
2 33 9
x y zx y zx y z
2. Add a multiple of the first equation to each of the following equations so that the coefficients of x in the second and third equations become 0.
2 33 5
3 9
x y zy z
x y z
2 32 3 13 9
x y zx y zx y z
-2 E1 + E2 E2
2 2 4 62 3 1
3 5
x y zx y z
y z
2009 PBLPathways
Solve the system2 3 1
2 33 9
x y zx y zx y z
2. Add a multiple of the first equation to each of the following equations so that the coefficients of x in the second and third equations become 0.
2 33 5
3 9
x y zy z
x y z
2 32 3 13 9
x y zx y zx y z
-2 E1 + E2 E2
2 2 4 62 3 1
3 5
x y zx y z
y z
2009 PBLPathways
Solve the system2 3 1
2 33 9
x y zx y zx y z
2. Add a multiple of the first equation to each of the following equations so that the coefficients of x in the second and third equations become 0.
-3 R1 + R3 R3
3 3 6 93 9
4 7 18
x y zx y z
y z
2 33 5
3 9
x y zy z
x y z
2 33 5
4 7 18
x y zy zy z
2009 PBLPathways
Solve the system2 3 1
2 33 9
x y zx y zx y z
2. Add a multiple of the first equation to each of the following equations so that the coefficients of x in the second and third equations become 0.
-3 E1 + E3 E3
3 3 6 93 9
4 7 18
x y zx y z
y z
2 33 5
3 9
x y zy z
x y z
2 33 5
4 7 18
x y zy zy z
2009 PBLPathways
Solve the system2 3 1
2 33 9
x y zx y zx y z
2. Add a multiple of the first equation to each of the following equations so that the coefficients of x in the second and third equations become 0.
-3 E1 + E3 E3
3 3 6 93 9
4 7 18
x y zx y z
y z
2 33 5
3 9
x y zy z
x y z
2 33 5
4 7 18
x y zy zy z
2009 PBLPathways
Solve the system2 3 1
2 33 9
x y zx y zx y z
2. Add a multiple of the first equation to each of the following equations so that the coefficients of x in the second and third equations become 0.
-3 E1 + E3 E3
3 3 6 93 9
4 7 18
x y zx y z
y z
2 33 5
3 9
x y zy z
x y z
2 33 5
4 7 18
x y zy zy z
2009 PBLPathways
Solve the system2 3 1
2 33 9
x y zx y zx y z
3. Multiply (or divide) both sides of the second equation by a number that makes the coefficient of y in the second equation equal to 1.
2 33 5
4 7 18
x y zy zy z
2 33 5
4 7 18
x y zy zy z
-1 R2 R2
2009 PBLPathways
Solve the system2 3 1
2 33 9
x y zx y zx y z
3. Multiply (or divide) both sides of the second equation by a number that makes the coefficient of y in the second equation equal to 1.
2 33 5
4 7 18
x y zy zy z
2 33 5
4 7 18
x y zy zy z
-1 E2 E2
2009 PBLPathways
Solve the system2 3 1
2 33 9
x y zx y zx y z
3. Multiply (or divide) both sides of the second equation by a number that makes the coefficient of y in the second equation equal to 1.
2 33 5
4 7 18
x y zy zy z
2 33 5
4 7 18
x y zy zy z
-1 E2 E2
2009 PBLPathways
Solve the system2 3 1
2 33 9
x y zx y zx y z
4. Add a multiple of the (new) second equation to the (new) third equation so that the coefficient of y in the newest third equation becomes 0.
2 33 5
19 38
x y zy z
z
2 33 5
4 7 18
x y zy zy z
-4 R2 + R3 R3
4 12 204 7 18
19 38
y zy z
z
2009 PBLPathways
Solve the system2 3 1
2 33 9
x y zx y zx y z
4. Add a multiple of the (new) second equation to the (new) third equation so that the coefficient of y in the newest third equation becomes 0.
2 33 5
19 38
x y zy z
z
2 33 5
4 7 18
x y zy zy z
-4 E2 + E3 E3
4 12 204 7 18
19 38
y zy z
z
2009 PBLPathways
Solve the system2 3 1
2 33 9
x y zx y zx y z
4. Add a multiple of the (new) second equation to the (new) third equation so that the coefficient of y in the newest third equation becomes 0.
2 33 5
19 38
x y zy z
z
2 33 5
4 7 18
x y zy zy z
-4 E2 + E3 E3
4 12 204 7 18
19 38
y zy z
z
2009 PBLPathways
Solve the system2 3 1
2 33 9
x y zx y zx y z
4. Add a multiple of the (new) second equation to the (new) third equation so that the coefficient of y in the newest third equation becomes 0.
2 33 5
19 38
x y zy z
z
2 33 5
4 7 18
x y zy zy z
-4 E2 + E3 E3
4 12 204 7 18
19 38
y zy z
z
2009 PBLPathways
Solve the system2 3 1
2 33 9
x y zx y zx y z
5. Multiply (or divide) both sides of the third equation by a number that makes the coefficient of z in the third equation equal to 1. This gives the solution for z in the system of equations.
2 33 5
2
x y zy z
z
2 33 5
19 38
x y zy z
z
E3 E3119
2009 PBLPathways
Solve the system2 3 1
2 33 9
x y zx y zx y z
5. Multiply (or divide) both sides of the third equation by a number that makes the coefficient of z in the third equation equal to 1. This gives the solution for z in the system of equations.
2 33 5
2
x y zy z
z
2 33 5
19 38
x y zy z
z
E3 E3119
2009 PBLPathways
Solve the system2 3 1
2 33 9
x y zx y zx y z
6. Use the solution for z to solve for y in the second equation. Then substitute values for y and z to solve for x in the first equation.
3 2 5
6 5
1
y
y
y
1 2 2 3
5 3
2
x
x
x
2 33 5
2
x y zy z
z
2009 PBLPathways
Solve the system2 3 1
2 33 9
x y zx y zx y z
6. Use the solution for z to solve for y in the second equation. Then substitute values for y and z to solve for x in the first equation.
3 2 5
6 5
1
y
y
y
1 2 2 3
5 3
2
x
x
x
2 33 5
2
x y zy z
z
2009 PBLPathways
Solve the system2 3 1
2 33 9
x y zx y zx y z
6. Use the solution for z to solve for y in the second equation. Then substitute values for y and z to solve for x in the first equation.
3 2 5
6 5
1
y
y
y
1 2 2 3
5 3
2
x
x
x
2 33 5
2
x y zy z
z
2009 PBLPathways
Solve the system2 3 1
2 33 9
x y zx y zx y z
6. Use the solution for z to solve for y in the second equation. Then substitute values for y and z to solve for x in the first equation.
3 2 5
6 5
1
y
y
y
1 2 2 3
5 3
2
x
x
x
2 33 5
2
x y zy z
z
2009 PBLPathways
Solve the system2 3 1
2 33 9
x y zx y zx y z
6. Use the solution for z to solve for y in the second equation. Then substitute values for y and z to solve for x in the first equation.
3 2 5
6 5
1
y
y
y
1 2 2 3
5 3
2
x
x
x
2 33 5
2
x y zy z
z
2009 PBLPathways
Solve the system2 3 1
2 33 9
x y zx y zx y z
6. Use the solution for z to solve for y in the second equation. Then substitute values for y and z to solve for x in the first equation.
3 2 5
6 5
1
y
y
y
1 2 2 3
5 3
2
x
x
x
2 33 5
2
x y zy z
z
2009 PBLPathways
Solve the system2 3 1
2 33 9
x y zx y zx y z
6. Use the solution for z to solve for y in the second equation. Then substitute values for y and z to solve for x in the first equation.
3 2 5
6 5
1
y
y
y
1 2 2 3
5 3
2
x
x
x
2 33 5
2
x y zy z
z
2009 PBLPathways
Solve the system2 3 1
2 33 9
x y zx y zx y z
6. Use the solution for z to solve for y in the second equation. Then substitute values for y and z to solve for x in the first equation.
3 2 5
6 5
1
y
y
y
1 2 2 3
5 3
2
x
x
x
2 33 5
2
x y zy z
z
2009 PBLPathways
Solve the system2 3 1
2 33 9
x y zx y zx y z
6. Use the solution for z to solve for y in the second equation. Then substitute values for y and z to solve for x in the first equation.
3 2 5
6 5
1
y
y
y
1 2 2 3
5 3
2
x
x
x
2 33 5
2
x y zy z
z
2009 PBLPathways
Does the solution solve the system?
Solve the system2 3 1
2 33 9
x y zx y zx y z
2 2 3 1 2 1
2 1 2 2 3
3 2 1 2 9
