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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

Precalculus

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

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

Three Variables

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 1/7

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

- Elimination

- Substitution

Three Variables

Notes: Solving Systems of Equations 1/7

Solve using elimination

x – 2y + z = 15

2x + 3y – 3z = 1

4x + 10y – 5z = -3

Elimination

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 Elimination here

Elimination

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

-2(x – 2y + z = 15) =

Elimination

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

-2(x – 2y + z = 15) =

Elimination

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

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

Elimination

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

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

2x + 3y – 3z = 1__

Elimination

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

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

2x + 3y – 3z = 1__

7y – 5z = -29

Elimination

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

-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 1/7

x – 2y + z = 15

2x + 3y – 3z = 1

4x + 10y – 5z = -3

*HINT* the first equation has x by itself

-4(x – 2y + z = 15

Elimination

Use Elimination here

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

-4(x – 2y + z = 15) =

Elimination

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

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

Elimination

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

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

4x + 10y – 5z = -3

Elimination

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

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

4x + 10y – 5z = -3

18y – 9z = -63

Elimination

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

-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 1/7

Line up the two equations you already found

7y – 5z = -29

18y – 9z = -63

Elimination

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 1/7

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 1/7

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 1/7

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 1/7

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 1/7

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 1/7

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 1/7

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 1/7

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 1/7

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 1/7

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 1/7

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 1/7

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 1/7

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 1/7

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

Final Answer  (8, -2, 3)

(Yay!! We’re done with 1 problem!!)

Elimination

Notes: Solving Systems of Equations 1/7

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

Final Answer  (4,5,-2)

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

Substitution

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 1/7

Solve for an easy variable

4x + 8z = 0  this is easy

Substitution

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 1/7

Solve for an easy variable

4x + 8z = 0  this is easy

4x = -8z

x = -2z

Substitution

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 1/7

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 1/7

Simplify both equations

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

-2(-2z) + y – z = -1

Substitution

Notes: Solving Systems of Equations 1/7

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 1/7

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 1/7

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 1/7

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 1/7

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 1/7

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 1/7

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 1/7

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 1/7

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 1/7

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