Transcript solving the system

Section 3.2
Solving Systems of Equations Algebraically
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Use algebraic computation to get precise solutions
The Substitution Method (One Equation into Another)
The Elimination Method (Adding Equations)
How to identify
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Consistent Systems (one solution – lines cross)
Inconsistent Systems (no solution – parallel lines)
Dependent Systems (infinitely many solutions – same line)
Comparing the Methods
Definition
Simultaneous Linear Equations
Consider the pair of equations together
4x + y = 10
-2x + 3y = -12
Each line has infinitely many pairs (x, y) that satisfy it.
But taken together, only one pair (3, -2) satisfies both.
Finding this pair is called solving the system.
In 3.1, you learned to solve a system of two equations in
two variables by graphing (approximation).
In this section you will learn two ways to solve linear
systems algebraically (precision).
Solving Systems of Linear Equations
Using the Substitution Method
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if a=b and b=c then a=c
Substitution Method - Example
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You can pick either variable to start –
you will get the same (x,y) solution. It
may take some work to isolate a variable:
Solve for (A)’s y
or
( A) 2 x  y  6
( B) 3x  4 y  4
( A1 ) y  6  2 x
( B ) 3 x  4(6  2 x)  4
( B )  5 x  24  4
( B )  5 x  20
( B ) x  4 1 2 a solution
( A1 ) y  6  2(4)
( A1 ) y  2
Solution : (4,2)
( A)  2 x  y  6
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( B)  3x  4 y  4
Solve for (A)’s x
( A) 2 x  y  6
( B) 3x  4 y  4
( A1 ) 2 x  6  y
( A1 ) x  3  12 y
( B ) 3(3  12 y )  4 y  4
( B ) 9  32 y  4 y  4
( B ) 52 y  5
( B ) y  2 1 2 a solution
( A) 2 x  (2)  6
( A) x  4
Solution : (4,2)
if a=b and c=d then a+c=b+d
Using the Elimination (Addition) Method
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if a=b then ca=cb
Elimination Method – Even Multiples
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When one variable term is an even
multiple of its matching variable term,
you only have to multiply one equation.
Let’s eliminate y: Multiply (A) by -2
( A) 5 x  4 y  22
( B)  3 x  8 y  18
 2( A)  10 x  8 y  44
( B)  3 x  8 y  18
(C )  13x
 26
(C ) x  2 1 2 a solution
( A) 5(2)  4 y  22
( A) 10  4 y  22
( A)
4 y  12
( A) y  3
Solution : (2,3)
( A)  5 x  4 y  22
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( B)  3 x  8 y  18
Elimination Method – No Multiples
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When no variable terms are even
( A)  2 x  3 y  17
multiples, you have to multiply both
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equations by different numbers.
( B )  5 x  7 y  29
Let’s eliminate y: Multiply (A) by 7 and (B) by -3
( A) 2 x  3 y  17
( B ) 5 x  7 y  29
7( A) 14 x  21 y  119
 3( B )  15 x  21 y  87
(C )  x
 32
(C ) x  32 1 2 a solution
( A) 2(32)  3 y  17
( A)  64  3 y  17
( A)
3 y  81
( A) y  27
Solution : (32,27)
Clearing Fractions or Decimals
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Systems without a Single Point Solution
0 = 4 untrue
Inconsistent Systems - how can you tell?
An inconsistent system
has no solutions.
(parallel lines)
Substitution Technique
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( A)
( B)
y  3 x  5
y  3 x  2
 3 x  5  3 x  2
 3x
 3x
5
2
2
2
7  0 inconsiste nt
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Elimination Technique
( A)
( B)
y  3 x  5
y  3 x  2
 1( A)  y  3 x  5
( B ) y  3 x  2
0  7 inconsiste nt
0 = 0 or n = n
Dependent Systems – how can you tell?
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A dependent system has
infinitely many solutions.
(it’s the same line!)
Substitution Technique
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Elimination Technique
( A)
3y  2x  6
( B)  12 y  8 x  24
( A)
3y  2x  6
( B )  12 y  8 x  24
( B) 8 x  12 y  24
x  32 y  3
( A) 3 y  2( 32 y  3)  6
3y  3y  6  6
6  6 dependent
4( A) 12 y  8 x  24
( B )  12 y  8 x  24
00
dependent
Application Problem
TEMPORARY HELP
A law firm had to hire several
workers to help finish a large
project. From the billing records
shown in Illustration 3,
determine the daily fee charged
by the employment agency for a
clerk-typist and for a computer
programmer.
The daily fees:
$105 for a clerk-typist
$185 for a programmer