SECTION 6.1

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SECTION 6.1 - PowerPoint PPT Presentation

SECTION 6.1. SYSTEMS OF LINEAR EQUATIONS: SUBSTITUTION AND ELIMINATION. MOVIE THEATER TICKET SALES. SEE EXAMPLE 1 SEE EXAMPLE 2. EQUIVALENT SYSTEMS OF EQUATIONS. Linear System More than one linear equation considered at a time.

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SECTION 6.1

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SECTION 6.1
• SYSTEMS OF LINEAR EQUATIONS:
• SUBSTITUTION AND ELIMINATION
MOVIE THEATER TICKET SALES

SEE EXAMPLE 1

SEE EXAMPLE 2

EQUIVALENT SYSTEMS OF EQUATIONS

Linear System

More than one linear equation considered at a time.

Solution - ordered pair (or triple) that satisfies both (or all) equations simultaneously.

CONSISTENT VS. INCONSISTENT

When a system of equations has at least one solution, it is said to be consistent; otherwise, it is called inconsistent.

THREE POSSIBILITIES FOR A LINEAR SYSTEM

x - y = 1

x - y = 3

x - y = 1

2x - y = 4

x - y = 1

2x - 2y = 2

No Solution

One Solution

Infinitely Many Solutions

SOLVING A SYSTEM BY SUBSTITUTION

2x + y = 5

- 4x + 6y = 12

RULES FOR OBTAINING AN EQUIVALENT SYSTEM

1. Interchange any two equations.

2. Multiply (or divide) each side of an equation by a nonzero constant.

3. Replace any equation in the system by the sum (or difference) of that equation and a nonzero multiple of any other equation in the system.

SOLVING A SYSTEM BY ELIMINATION
• 2x + 3y = 1
• x + y = - 3
• Multiply equation 2 by 2
• Replace equation 2 with the sum of equations 1 and 2.
AN INCONSISTENT SYSTEM

2x + y = 5

4x + 2y = 8

AN DEPENDENT SYSTEM

2x + y = 4

- 6x - 3y = - 12

3 EQUATIONS, 3 UNKNOWNS

2 x + 4y - 2 z = - 10

- 3x + 4y - 2 z = 5

5x + 6y + 3 z = 3

x + 2 y - z = - 5

- 3x + 4y - 2 z = 5

5x + 6y + 3 z = 3

Dividing 1st equation by 2 to make leading coefficient equal to 1.

EXAMPLE

x + 2 y - z = - 5

- 3x + 4y - 2 z = 5

3x + 6 y - 3 z = - 15

- 3x + 4y – 2z = 5

10y – 5z = -10

Mult. 1st eqn by 3 and add to 2nd

EXAMPLE

x + 2 y - z = - 5

5x + 6y + 3 z = 3

-5x - 10 y + 5 z = 25

5x + 6y + 3z = 3

- 4y + 8z = 28

Mult. 1st eqn by -5 and add to 3rd

EXAMPLE

Now we have 2 equations in only y & z:

10y - 5 z = -10

- 4y + 8 z = 28

Divide 1st equation by 5

Divide 2nd equation by 4

EXAMPLE

2y - z = - 2

- y + 2 z = 7

2y – z = -2

-2y + 4z = 14

3z = 12

z = 4

Multiply 2nd equation by 2 & add to 1st

EXAMPLE

2y – z = -2

2y – 4 = -2

2y = 2

y = 1

x + 2y – z = - 5

x + 2 (1) - 4 = - 5

x - 2 = - 5

x = - 3

(-3, 1, 4)

EXAMPLE

DO EXAMPLES 9, 10, 11

CONCLUSION OF
• SECTION 6.1