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7 January 2011. Precalculus. Solve the system of equations x + 3y = 18 and –x + 2y = 7. (5,3) (-3,5) (5,-3) (3,5) (-3,-5). 360. 0 of 30. Notes: Solving Systems of Equations 1/7. Only two options are available if you have a system of equations with three variables. Three Variables.

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7 january 2011

7 January 2011

Precalculus


Solve the system of equations x 3y 18 and x 2y 7
Solve the system of equationsx + 3y = 18 and –x + 2y = 7

  • (5,3)

  • (-3,5)

  • (5,-3)

  • (3,5)

  • (-3,-5)

360

0 of 30


Notes solving systems of equations 1 7
Notes: Solving Systems of Equations 1/7

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

Three Variables


Notes solving systems of equations 1 71
Notes: Solving Systems of Equations 1/7

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

- Elimination

Three Variables


Notes solving systems of equations 1 72
Notes: Solving Systems of Equations 1/7

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

- Elimination

- Substitution

Three Variables


Notes solving systems of equations 1 73
Notes: Solving Systems of Equations 1/7

Solve using elimination

x – 2y + z = 15

2x + 3y – 3z = 1

4x + 10y – 5z = -3

Elimination


Notes solving systems of equations 1 74
Notes: Solving Systems of Equations 1/7

x – 2y + z = 15

2x + 3y – 3z = 1

4x + 10y – 5z = -3

*HINT* the first equation has x by itself

use that to your advantage

Use Elimination here

Elimination


Notes solving systems of equations 1 75
Notes: Solving Systems of Equations 1/7

x – 2y + z = 15

2x + 3y – 3z = 1

4x + 10y – 5z = -3

*HINT* the first equation has x by itself

use that to your advantage

-2(x – 2y + z = 15) =

Elimination


Notes solving systems of equations 1 76
Notes: Solving Systems of Equations 1/7

x – 2y + z = 15

2x + 3y – 3z = 1

4x + 10y – 5z = -3

*HINT* the first equation has x by itself

use that to your advantage

-2(x – 2y + z = 15) =

Elimination


Notes solving systems of equations 1 77
Notes: Solving Systems of Equations 1/7

x – 2y + z = 15

2x + 3y – 3z = 1

4x + 10y – 5z = -3

*HINT* the first equation has x by itself

use that to your advantage

-2(x – 2y + z = 15) = -2x + 4y – 2z = -30

Elimination


Notes solving systems of equations 1 78
Notes: Solving Systems of Equations 1/7

x – 2y + z = 15

2x + 3y – 3z = 1

4x + 10y – 5z = -3

*HINT* the first equation has x by itself

use that to your advantage

-2(x – 2y + z = 15) = -2x + 4y – 2z = -30

2x + 3y – 3z = 1__

Elimination


Notes solving systems of equations 1 79
Notes: Solving Systems of Equations 1/7

x – 2y + z = 15

2x + 3y – 3z = 1

4x + 10y – 5z = -3

*HINT* the first equation has x by itself

use that to your advantage

-2(x – 2y + z = 15) = -2x + 4y – 2z = -30

2x + 3y – 3z = 1__

7y – 5z = -29

Elimination


Notes solving systems of equations 1 710
Notes: Solving Systems of Equations 1/7

x – 2y + z = 15

2x + 3y – 3z = 1

4x + 10y – 5z = -3

*HINT* the first equation has x by itself

use that to your advantage

-2(x – 2y + z = 15) = -2x + 4y – 2z = -30

2x + 3y – 3z = 1__

7y – 5z = -29

we will use this later

Elimination


Notes solving systems of equations 1 711
Notes: Solving Systems of Equations 1/7

x – 2y + z = 15

2x + 3y – 3z = 1

4x + 10y – 5z = -3

*HINT* the first equation has x by itself

use that to your advantage

-4(x – 2y + z = 15

Elimination

Use Elimination here


Notes solving systems of equations 1 712
Notes: Solving Systems of Equations 1/7

x – 2y + z = 15

2x + 3y – 3z = 1

4x + 10y – 5z = -3

*HINT* the first equation has x by itself

use that to your advantage

-4(x – 2y + z = 15) =

Elimination


Notes solving systems of equations 1 713
Notes: Solving Systems of Equations 1/7

x – 2y + z = 15

2x + 3y – 3z = 1

4x + 10y – 5z = -3

*HINT* the first equation has x by itself

use that to your advantage

-4(x – 2y + z = 15) = -4x + 8y – 4z = -60

Elimination


Notes solving systems of equations 1 714
Notes: Solving Systems of Equations 1/7

x – 2y + z = 15

2x + 3y – 3z = 1

4x + 10y – 5z = -3

*HINT* the first equation has x by itself

use that to your advantage

-4(x – 2y + z = 15) = -4x + 8y – 4z = -60

4x + 10y – 5z = -3

Elimination


Notes solving systems of equations 1 715
Notes: Solving Systems of Equations 1/7

x – 2y + z = 15

2x + 3y – 3z = 1

4x + 10y – 5z = -3

*HINT* the first equation has x by itself

use that to your advantage

-4(x – 2y + z = 15) = -4x + 8y – 4z = -60

4x + 10y – 5z = -3

18y – 9z = -63

Elimination


Notes solving systems of equations 1 716
Notes: Solving Systems of Equations 1/7

x – 2y + z = 15

2x + 3y – 3z = 1

4x + 10y – 5z = -3

*HINT* the first equation has x by itself

use that to your advantage

-4(x – 2y + z = 15) = -4x + 8y – 4z = -60

4x + 10y – 5z = -3

18y – 9z = -63

we will use this later

Elimination


Notes solving systems of equations 1 717
Notes: Solving Systems of Equations 1/7

Line up the two equations you already found

7y – 5z = -29

18y – 9z = -63

Elimination


Notes solving systems of equations 1 718
Notes: Solving Systems of Equations 1/7

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

7y – 5z = -29

18y – 9z = -63

Elimination


Notes solving systems of equations 1 719
Notes: Solving Systems of Equations 1/7

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

7y – 5z = -29

18y – 9z = -63

-9(7y – 5z = -29) = -63y + 45z = 261

Elimination


Notes solving systems of equations 1 720
Notes: Solving Systems of Equations 1/7

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

7y – 5z = -29

18y – 9z = -63

-9(7y – 5z = -29) = -63y + 45z = 261

5(18y – 9z = -63) = 90y – 45z = -315

Elimination


Notes solving systems of equations 1 721
Notes: Solving Systems of Equations 1/7

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

7y – 5z = -29

18y – 9z = -63

-9(7y – 5z = -29) = -63y + 45z = 261

5(18y – 9z = -63) = 90y – 45z = -315

27y = -54

y = -2

Elimination


Notes solving systems of equations 1 722
Notes: Solving Systems of Equations 1/7

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

7y – 5z = -29

7(-2) – 5z = -29

y = -2

Elimination


Notes solving systems of equations 1 723
Notes: Solving Systems of Equations 1/7

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

7y – 5z = -29

7(-2) – 5z = -29

-14 – 5z = -29

y = -2

Elimination


Notes solving systems of equations 1 724
Notes: Solving Systems of Equations 1/7

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

7y – 5z = -29

7(-2) – 5z = -29

-14 – 5z = -29

- 5z = -15

y = -2

Elimination


Notes solving systems of equations 1 725
Notes: Solving Systems of Equations 1/7

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

7y – 5z = -29

7(-2) – 5z = -29

-14 – 5z = -29

- 5z = -15

z = 3

y = -2, z = 3

Elimination


Notes solving systems of equations 1 726
Notes: Solving Systems of Equations 1/7

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

x – 2y + z = 15

y = -2, z = 3

Elimination


Notes solving systems of equations 1 727
Notes: Solving Systems of Equations 1/7

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

x – 2y + z = 15

x – 2(-2) + 3 = 15

y = -2, z = 3

Elimination


Notes solving systems of equations 1 728
Notes: Solving Systems of Equations 1/7

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

x – 2y + z = 15

x – 2(-2) + 3 = 15

x + 4 + 3 = 15

y = -2, z = 3

Elimination


Notes solving systems of equations 1 729
Notes: Solving Systems of Equations 1/7

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

x – 2y + z = 15

x – 2(-2) + 3 = 15

x + 4 + 3 = 15

x + 7 = 15

y = -2, z = 3

Elimination


Notes solving systems of equations 1 730
Notes: Solving Systems of Equations 1/7

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

x – 2y + z = 15

x – 2(-2) + 3 = 15

x + 4 + 3 = 15

x + 7 = 15

x = 8

y = -2, z = 3

Elimination


Notes solving systems of equations 1 731
Notes: Solving Systems of Equations 1/7

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

x – 2y + z = 15

x – 2(-2) + 3 = 15

x + 4 + 3 = 15

x + 7 = 15

x = 8

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

Elimination


Notes solving systems of equations 1 732
Notes: Solving Systems of Equations 1/7

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

Final Answer  (8, -2, 3)

(Yay!! We’re done with 1 problem!!)

Elimination


Notes solving systems of equations 1 733
Notes: Solving Systems of Equations 1/7

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

Final Answer  (4,5,-2)

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

Substitution


Notes solving systems of equations 1 734
Notes: Solving Systems of Equations 1/7

Solve for an easy variable

4x + 8z = 0  this is easy

3x – 2y + z = 0

-2x + y – z = -1 ignore

these two

Substitution


Notes solving systems of equations 1 735
Notes: Solving Systems of Equations 1/7

Solve for an easy variable

4x + 8z = 0  this is easy

Substitution


Notes solving systems of equations 1 736
Notes: Solving Systems of Equations 1/7

Solve for an easy variable

4x + 8z = 0  this is easy

4x = -8z

Substitution


Notes solving systems of equations 1 737
Notes: Solving Systems of Equations 1/7

Solve for an easy variable

4x + 8z = 0  this is easy

4x = -8z

x = -2z

Substitution


Notes solving systems of equations 1 738
Notes: Solving Systems of Equations 1/7

Replace every x with -2z in the original equations

3x– 2y + z = 0

-2x + y – z = -1

Substitution


Notes solving systems of equations 1 739
Notes: Solving Systems of Equations 1/7

Replace every x with -2z in the original equations

3(-2z) – 2y + z = 0

-2(-2z) + y – z = -1

Substitution


Notes solving systems of equations 1 740
Notes: Solving Systems of Equations 1/7

Simplify both equations

3(-2z) – 2y + z = 0  -6z – 2y + z = 0

-2(-2z) + y – z = -1

Substitution


Notes solving systems of equations 1 741
Notes: Solving Systems of Equations 1/7

Simplify both equations

3(-2z) – 2y + z = 0  -6z – 2y + z = 0

- 2y – 5z = 0

-2(-2z) + y – z = -1

Substitution


Notes solving systems of equations 1 742
Notes: Solving Systems of Equations 1/7

Simplify both equations

3(-2z) – 2y + z = 0  -6z – 2y + z = 0

- 2y – 5z = 0

-2(-2z) + y – z = -1  4z + y – z = -1

Substitution


Notes solving systems of equations 1 743
Notes: Solving Systems of Equations 1/7

Simplify both equations

3(-2z) – 2y + z = 0  -6z – 2y + z = 0

- 2y – 5z = 0

-2(-2z) + y – z = -1  4z + y – z = -1

y + 3 = -1

Substitution


Notes solving systems of equations 1 744
Notes: Solving Systems of Equations 1/7

Solve y + 3z = -1 for y

y = -3z – 1

Substitute y = -3z – 1 into -2y – 5z = 0

-2y – 5z = 0

-2(-3z – 1) – 5z = 0

6z + 2 – 5z = 0

z = -2

z = -2

Substitution


Notes solving systems of equations 1 745
Notes: Solving Systems of Equations 1/7

Substitute z = -2 into the following two equations

y = -3z – 1 x = -2z

y = -3(-2) – 1

z = -2

Substitution


Notes solving systems of equations 1 746
Notes: Solving Systems of Equations 1/7

Substitute z = -2 into the following two equations

y = -3z – 1 x = -2z

y = -3(-2) – 1

y = 6 – 1

z = -2

Substitution


Notes solving systems of equations 1 747
Notes: Solving Systems of Equations 1/7

Substitute z = -2 into the following two equations

y = -3z – 1 x = -2z

y = -3(-2) – 1

y = 6 – 1

y = 5

y = 5, z = -2

Substitution


Notes solving systems of equations 1 748
Notes: Solving Systems of Equations 1/7

Substitute z = -2 into the following two equations

y = -3z – 1 x = -2z

y = -3(-2) – 1 x = -2(-2)

y = 6 – 1

y = 5

y = 5, z = -2

Substitution


Notes solving systems of equations 1 749
Notes: Solving Systems of Equations 1/7

Substitute z = -2 into the following two equations

y = -3z – 1 x = -2z

y = -3(-2) – 1 x = -2(-2)

y = 6 – 1 x = 4

y = 5

y = 5, z = -2

Substitution


Notes solving systems of equations 1 750
Notes: Solving Systems of Equations 1/7

Substitute z = -2 into the following two equations

y = -3z – 1 x = -2z

y = -3(-2) – 1 x = -2(-2)

y = 6 – 1 x = 4

y = 5

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

Substitution