# SECTION 6.1 SYSTEMS OF LINEAR EQUATIONS: SYSTEMS OF LINEAR EQUATIONS: SUBSTITUTION AND ELIMINATION SUBSTITUTION AND ELIMINATION.

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SECTION 6.1 SYSTEMS OF LINEAR EQUATIONS: SYSTEMS OF LINEAR EQUATIONS: SUBSTITUTION AND ELIMINATION 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 No Solution x - y = 1 2x - y = 4 One Solution x - y = 1 2x - 2y = 2 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.

MOVIE THEATER TICKET SALES DO EXAMPLE 5

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 Dividing 1st equation by 2 to make leading coefficient equal to 1. x + 2 y - z = - 5 x + 2 y - z = - 5 - 3x + 4y - 2 z = 5 5x + 6y + 3 z = 3

EXAMPLE x + 2 y - z = - 5 x + 2 y - z = - 5 - 3x + 4y - 2 z = 5 Mult. 1st eqn by 3 and add to 2nd 3x + 6 y - 3 z = - 15 3x + 6 y - 3 z = - 15 - 3x + 4y – 2z = 5 - 3x + 4y – 2z = 5 10y – 5z = -10 10y – 5z = -10

EXAMPLE x + 2 y - z = - 5 x + 2 y - z = - 5 5x + 6y + 3 z = 3 Mult. 1st eqn by -5 and add to 3rd -5x - 10 y + 5 z = 25 -5x - 10 y + 5 z = 25 5x + 6y + 3z = 3 5x + 6y + 3z = 3 - 4y + 8z = 28 - 4y + 8z = 28

EXAMPLE Now we have 2 equations in only y & z: 10y - 5 z = -10 - 4y + 8 z = 28 Divide 1 st equation by 5 Divide 2nd equation by 4

EXAMPLE 2y - z = - 2 - y + 2 z = 7 Multiply 2 nd equation by 2 & add to 1st 2y – z = -2 -2y + 4z = 14 3z = 12 3z = 12 z = 4 z = 4

EXAMPLE 2y – z = -2 2y – z = -2 2y – 4 = -2 2y – 4 = -2 2y = 2 2y = 2 y = 1 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 DO EXAMPLES 9, 10, 11

CONCLUSION OF CONCLUSION OF SECTION 6.1 SECTION 6.1

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