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7 January 2011

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7 January 2011

Precalculus

- (5,3)
- (-3,5)
- (5,-3)
- (3,5)
- (-3,-5)

360

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Only two options are available if you have a system of equations with three variables.

Three Variables

Only two options are available if you have a system of equations with three variables.

- Elimination

Three Variables

Only two options are available if you have a system of equations with three variables.

- Elimination

- Substitution

Three Variables

Solve using elimination

x â€“ 2y + z = 15

2x + 3y â€“ 3z = 1

4x + 10y â€“ 5z = -3

Elimination

x â€“ 2y + z = 15

2x + 3y â€“ 3z = 1

4x + 10y â€“ 5z = -3

*HINT* the first equation has x by itself

use that to your advantage

Use Elimination here

Elimination

x â€“ 2y + z = 15

2x + 3y â€“ 3z = 1

4x + 10y â€“ 5z = -3

*HINT* the first equation has x by itself

use that to your advantage

-2(x â€“ 2y + z = 15) =

Elimination

x â€“ 2y + z = 15

2x + 3y â€“ 3z = 1

4x + 10y â€“ 5z = -3

*HINT* the first equation has x by itself

use that to your advantage

-2(x â€“ 2y + z = 15) =

Elimination

x â€“ 2y + z = 15

2x + 3y â€“ 3z = 1

4x + 10y â€“ 5z = -3

*HINT* the first equation has x by itself

use that to your advantage

-2(x â€“ 2y + z = 15) = -2x + 4y â€“ 2z = -30

Elimination

x â€“ 2y + z = 15

2x + 3y â€“ 3z = 1

4x + 10y â€“ 5z = -3

*HINT* the first equation has x by itself

use that to your advantage

-2(x â€“ 2y + z = 15) = -2x + 4y â€“ 2z = -30

2x + 3y â€“ 3z = 1__

Elimination

x â€“ 2y + z = 15

2x + 3y â€“ 3z = 1

4x + 10y â€“ 5z = -3

*HINT* the first equation has x by itself

use that to your advantage

-2(x â€“ 2y + z = 15) = -2x + 4y â€“ 2z = -30

2x + 3y â€“ 3z = 1__

7y â€“ 5z = -29

Elimination

x â€“ 2y + z = 15

2x + 3y â€“ 3z = 1

4x + 10y â€“ 5z = -3

*HINT* the first equation has x by itself

use that to your advantage

-2(x â€“ 2y + z = 15) = -2x + 4y â€“ 2z = -30

2x + 3y â€“ 3z = 1__

7y â€“ 5z = -29

we will use this later

Elimination

x â€“ 2y + z = 15

2x + 3y â€“ 3z = 1

4x + 10y â€“ 5z = -3

*HINT* the first equation has x by itself

use that to your advantage

-4(x â€“ 2y + z = 15

Elimination

Use Elimination here

x â€“ 2y + z = 15

2x + 3y â€“ 3z = 1

4x + 10y â€“ 5z = -3

*HINT* the first equation has x by itself

use that to your advantage

-4(x â€“ 2y + z = 15) =

Elimination

x â€“ 2y + z = 15

2x + 3y â€“ 3z = 1

4x + 10y â€“ 5z = -3

*HINT* the first equation has x by itself

use that to your advantage

-4(x â€“ 2y + z = 15) = -4x + 8y â€“ 4z = -60

Elimination

x â€“ 2y + z = 15

2x + 3y â€“ 3z = 1

4x + 10y â€“ 5z = -3

*HINT* the first equation has x by itself

use that to your advantage

-4(x â€“ 2y + z = 15) = -4x + 8y â€“ 4z = -60

4x + 10y â€“ 5z = -3

Elimination

x â€“ 2y + z = 15

2x + 3y â€“ 3z = 1

4x + 10y â€“ 5z = -3

*HINT* the first equation has x by itself

use that to your advantage

-4(x â€“ 2y + z = 15) = -4x + 8y â€“ 4z = -60

4x + 10y â€“ 5z = -3

18y â€“ 9z = -63

Elimination

x â€“ 2y + z = 15

2x + 3y â€“ 3z = 1

4x + 10y â€“ 5z = -3

*HINT* the first equation has x by itself

use that to your advantage

-4(x â€“ 2y + z = 15) = -4x + 8y â€“ 4z = -60

4x + 10y â€“ 5z = -3

18y â€“ 9z = -63

we will use this later

Elimination

Line up the two equations you already found

7y â€“ 5z = -29

18y â€“ 9z = -63

Elimination

Multiply the first equation by whatever number is in front of the z term of the second equation

7y â€“ 5z = -29

18y â€“ 9z = -63

Elimination

Multiply the first equation by whatever number is in front of the z term of the second equation

7y â€“ 5z = -29

18y â€“ 9z = -63

-9(7y â€“ 5z = -29) = -63y + 45z = 261

Elimination

Multiply the second equation by the opposite of whatever number is in front of the z term of the first equation

7y â€“ 5z = -29

18y â€“ 9z = -63

-9(7y â€“ 5z = -29) = -63y + 45z = 261

5(18y â€“ 9z = -63) = 90y â€“ 45z = -315

Elimination

Multiply the second equation by the opposite of whatever number is in front of the z term of the first equation

7y â€“ 5z = -29

18y â€“ 9z = -63

-9(7y â€“ 5z = -29) = -63y + 45z = 261

5(18y â€“ 9z = -63) = 90y â€“ 45z = -315

27y = -54

y = -2

Elimination

Using one of the equations you already found, substitute -2 for y

7y â€“ 5z = -29

7(-2) â€“ 5z = -29

y = -2

Elimination

Using the equations you already found, substitute -2 for y

7y â€“ 5z = -29

7(-2) â€“ 5z = -29

-14 â€“ 5z = -29

y = -2

Elimination

Using the equations you already found, substitute -2 for y

7y â€“ 5z = -29

7(-2) â€“ 5z = -29

-14 â€“ 5z = -29

- 5z = -15

y = -2

Elimination

Using the equations you already found, substitute -2 for y

7y â€“ 5z = -29

7(-2) â€“ 5z = -29

-14 â€“ 5z = -29

- 5z = -15

z = 3

y = -2, z = 3

Elimination

Going all the way back to one of the original three equations, substitute in the two values you know

x â€“ 2y + z = 15

y = -2, z = 3

Elimination

Going all the way back to one of the original three equations, substitute in the two values you know

x â€“ 2y + z = 15

x â€“ 2(-2) + 3 = 15

y = -2, z = 3

Elimination

Going all the way back to one of the original three equations, substitute in the two values you know

x â€“ 2y + z = 15

x â€“ 2(-2) + 3 = 15

x + 4 + 3 = 15

y = -2, z = 3

Elimination

Going all the way back to one of the original three equations, substitute in the two values you know

x â€“ 2y + z = 15

x â€“ 2(-2) + 3 = 15

x + 4 + 3 = 15

x + 7 = 15

y = -2, z = 3

Elimination

Going all the way back to one of the original three equations, substitute in the two values you know

x â€“ 2y + z = 15

x â€“ 2(-2) + 3 = 15

x + 4 + 3 = 15

x + 7 = 15

x = 8

y = -2, z = 3

Elimination

Going all the way back to one of the original three equations, substitute in the two values you know

x â€“ 2y + z = 15

x â€“ 2(-2) + 3 = 15

x + 4 + 3 = 15

x + 7 = 15

x = 8

x = 8, y = -2, z = 3

Elimination

x = 8, y = -2, z = 3

Final Answer ïƒ (8, -2, 3)

(Yay!! Weâ€™re done with 1 problem!!)

Elimination

x = 4, y = 5, z = -2

Final Answer ïƒ (4,5,-2)

(Yay!!! Two questions answered!!!)

Substitution

Solve for an easy variable

4x + 8z = 0 ïƒŸ this is easy

3x â€“ 2y + z = 0

-2x + y â€“ z = -1ignore

these two

Substitution

Solve for an easy variable

4x + 8z = 0 ïƒŸ this is easy

Substitution

Solve for an easy variable

4x + 8z = 0 ïƒŸ this is easy

4x = -8z

Substitution

Solve for an easy variable

4x + 8z = 0 ïƒŸ this is easy

4x = -8z

x = -2z

Substitution

Replace every x with -2z in the original equations

3xâ€“ 2y + z = 0

-2x + y â€“ z = -1

Substitution

Replace every x with -2z in the original equations

3(-2z) â€“ 2y + z = 0

-2(-2z) + y â€“ z = -1

Substitution

Simplify both equations

3(-2z) â€“ 2y + z = 0 ïƒ -6z â€“ 2y + z = 0

-2(-2z) + y â€“ z = -1

Substitution

Simplify both equations

3(-2z) â€“ 2y + z = 0 ïƒ -6z â€“ 2y + z = 0

- 2y â€“ 5z = 0

-2(-2z) + y â€“ z = -1

Substitution

Simplify both equations

3(-2z) â€“ 2y + z = 0 ïƒ -6z â€“ 2y + z = 0

- 2y â€“ 5z = 0

-2(-2z) + y â€“ z = -1 ïƒ 4z + y â€“ z = -1

Substitution

Simplify both equations

3(-2z) â€“ 2y + z = 0 ïƒ -6z â€“ 2y + z = 0

- 2y â€“ 5z = 0

-2(-2z) + y â€“ z = -1 ïƒ 4z + y â€“ z = -1

y + 3 = -1

Substitution

Solve y + 3z = -1 for y

y = -3z â€“ 1

Substitute y = -3z â€“ 1 into -2y â€“ 5z = 0

-2y â€“ 5z = 0

-2(-3z â€“ 1) â€“ 5z = 0

6z + 2 â€“ 5z = 0

z = -2

z = -2

Substitution

Substitute z = -2 into the following two equations

y = -3z â€“ 1x = -2z

y = -3(-2) â€“ 1

z = -2

Substitution

Substitute z = -2 into the following two equations

y = -3z â€“ 1x = -2z

y = -3(-2) â€“ 1

y = 6 â€“ 1

z = -2

Substitution

Substitute z = -2 into the following two equations

y = -3z â€“ 1x = -2z

y = -3(-2) â€“ 1

y = 6 â€“ 1

y = 5

y = 5, z = -2

Substitution

Substitute z = -2 into the following two equations

y = -3z â€“ 1x = -2z

y = -3(-2) â€“ 1x = -2(-2)

y = 6 â€“ 1

y = 5

y = 5, z = -2

Substitution

Substitute z = -2 into the following two equations

y = -3z â€“ 1x = -2z

y = -3(-2) â€“ 1x = -2(-2)

y = 6 â€“ 1x = 4

y = 5

y = 5, z = -2

Substitution

Substitute z = -2 into the following two equations

y = -3z â€“ 1x = -2z

y = -3(-2) â€“ 1x = -2(-2)

y = 6 â€“ 1x = 4

y = 5

x = 4, y = 5, z = -2

Substitution