Solve Linear Systems by Elimination

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Do Now 1/13/10
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Text p. 447, #6-28 evens
Text p. 454, #10-26 evens, #37 & #39
Quiz Sections 7.1 - 7.4 Friday
In your journal, list the possible ways to solve a
linear system. Then solve the following systems.
5x + 6y = 50
-x + 6y = 26
(4, 5)
-8y + 6x = 36
6x – y = 15
(2, -3)
Homework
Text p. 447, #6-28 evens
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6) (12,6)
8) (-1,2)
10) (-2,2)
12) (0,-10)
14) (-3,1)
16) (-15,-62)
18) (-6,-1)
20) (5,3)
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22) B
 24) Did not change the
sign for -3x when
rearranging the
equations. y = 9
 26) (4,-9)
 28) (5,8)
Objective
 SWBAT
to solve linear systems using
elimination (adding, subtracting, and
multiplying).
“How Do You Solve a Linear System???”
(1) Solve Linear Systems by Graphing (7.1)
(2) Solve Linear Systems by Substitution (7.2)
(3) Solve Linear Systems by ELIMINATION!!! (7.3)
Adding or Subtracting
(4) Solve Linear Systems by Multiplying First (7.4)
Then eliminate.
Section 7.3 “Solve Linear Systems
by Adding or Subtracting”
 ELIMINATIONadding or subtracting equations to obtain a
new equation in one variable.
Solving Linear Systems Using Elimination
(1) Add or Subtract the equations to eliminate one
variable.
(2) Solve the resulting equation for the other variable.
(3) Substitute in either original equation to find the value
of the eliminated variable.
“Solve Linear Systems by Elimination”
Equation 1
Equation 2
+
2x + 3y = 11
-2x + 5y = 13
8y = 24
y=3
2x + 3y = 11 Equation 1
2x + 3(3) = 11
2x + 9 = 11
x=1
2(1) + 3(3) = 11
11 = 11
Substitute value for
y into either of the
original equations
The solution is the point (1,3).
Substitute (1,3) into both
equations to check.
-2(1) + 5(3) = 13
13 = 13
Section 7.4 “Solve Linear Systems
by Multiplying First”
 ELIMINATIONadding or subtracting equations to obtain a
new equation in one variable.
Solving Linear Systems Using Elimination
(1) Multiply the whole equation by a constant in order to
be able to eliminate a variable.
(2) Add or Subtract the equations to eliminate one
variable.
(3) Solve the resulting equation for the other variable.
(4) Substitute in either original equation to find the value
of the eliminated variable.
“Solve Linear Systems by Elimination
Multiplying First!!”
Equation 1
6x + 5y = 19
Equation 2
2x + 3y = 5
x (-3)
+
6x + 5y = 19
-6x – 9y = -15
-4y = 4
y = -1
2x + 3y = 5 Equation 2
2x + 3(-1) = 5
2x - 3 = 5
x=4
6(4) + 5(-1) = 19
19 = 19
Substitute value for
y into either of the
original equations
The solution is the point (4,-1).
Substitute (4,-1) into both
equations to check.
2(4) + 3(-1) = 5
5=5
“Solve Linear Systems by Elimination
Multiplying First!!”
Equation 1
2x + 5y = 3
Equation 2
3x + 10y = -3
x (-2)
+
-4x - 10y = -6
3x + 10y = -3
-x
2x + 5y = 3 Equation 1
2(9) + 5y = 3
18 + 5y = 3
y = -3
2(9) + 5(-3) = 3
3=3
= -9
x=9
Substitute value for
x into either of the
original equations
The solution is the point (9,-3).
Substitute (9,-3) into both
equations to check.
3(9) + 10(-3) = -3
-3 = -3
“Solve Linear Systems by Elimination
Multiplying First!!”
Equation 1
4x + 5y = 35
x (2)
Equation 2
-3x + 2y = -9
x (-5)
+
8x + 10y = 70
15x - 10y = 45
23x
4x + 5y = 35 Equation 1
4(5) + 5y = 35
20 + 5y = 35
y=3
4(5) + 5(3) = 35
35 = 35
= 115
x=5
Substitute value for
x into either of the
original equations
The solution is the point (5,3).
Substitute (5,3) into both
equations to check.
-3(5) + 2(3) = -9
-9 = -9
“Solve Linear Systems by Elimination
Multiplying First!!”
Equation 1
Equation 2
9x + 2y = 39
x (2)
6x + 13y = -9 x (-3)
+
18x + 4y = 78
-18x - 39y = 27
-35y = 105
y = -3
9x + 2y = 39 Equation 1
9x + 2(-3) = 39
9x - 6 = 39
x=5
Substitute value for
y into either of the
original equations
9(5) + 2(-3) = 39 The solution is the point (5,-3).
Substitute (5,-3) into both
39 = 39
equations to check.
6(5) + 13(-3) = -9
-9 = -9
Guided Practice
x+y=2
2x + 7y = 9
6x – 2y = 1
-2x + 3y = -5
(1,1)
(-0.5, -2)
3x - 7y = 5
9y = 5x + 5
(-10,-5)
Communicators
 Solve
the linear systems using elimination
on your communicators. When you are
finished raise your board!
NJASK7 prep
Homework
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Text p. 454, #10-26 evens, #37 & #39
Quiz Sections 7.1 - 7.4 Friday.
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