Transcript Do Now 1/29/07

Do Now 1/12/10
 Take
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Text p. 430, #4-20, 30-34 evens
Text p. 439, #4-24 even, #32, & #36
 Copy
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out HW from last night.
HW in your planner.
Text p. 441, #1-9 all
Text p. 447, #6-28 evens
Homework
Text p. 430, #4-20, 30-34 evens
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4) not a solution
6) B
8) (1,-3)
10) (3,2)
12) (1,2)
14) (-2,-2)
16) (1.8, -0.4)
18) (4,3)
20) (1, -2)
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30) (3,5), (1,2), & (5,1)
 32) B
 34a) 25 bike, 15 stairs
b) 25 elliptical trainer
and 5 stairs
Homework
Text p. 439, #4-24, 32 & 36 evens
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4) (1,1)
6) (13,6)
8) (3,22)
10) (8,-7)
12) (-1,11)
14) (4,3)
16) (-1,1)
18) A
20) (5,1)
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22) (10.5, 11.75)
 24) (14, 10)
 32) 22 tubes, 4 “cooler
tubes”
 36) 12 quarters
Section 7.1
“Solve Linear Systems by Graphing”
Linear System–
consists of two more linear equations.
x + 2y = 7
3x – 2y = 5
Equation 1
Equation 2
A solution to a linear system is an ordered
pair (a point) where the two linear equations
(lines) intersect (cross).
Section 7.2
“Solve Linear Systems by Substitution”
Solving a Linear System by Substitution
(1) Solve one of the equations for one of its variables. (When
possible, solve for a variable that has a coefficient of 1 or -1).
(2) Substitute the expression from step 1 into the other equation
and solve for the other variable.
(3) Substitute the value from step 2 into the revised equation from
step 1 and solve.
“Solve Linear Systems by Substituting”
Equation 1
x – 2y = -6
Equation 2
4x + 6y = 4
x = -6 + 2y
4(-6+ +6y2y)
4x
= 4+ 6y = 4
-24 + 8y + 6y = 4
-24 + 14y = 4
y =2
x – 2y = -6
Equation 1
x = -6 + 2(2)
x = -2
(-2) - 2(2) = -6
-6 = -6
Substitute
Substitute value for
x into the original
equation
The solution is the point (-2,2).
Substitute (-2,2) into both
equations to check.
4(-2) + 6(2) = 4
4=4
“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
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
“Solve Linear Systems by Elimination”
Equation 1
Equation 2
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+
4x + 3y = 2
-5x
5x + -3y
3y ==-2
2
-x
4x + 3y = 2 Equation 1
4(-4) + 3y = 2
-16 + 3y = 2
y=6
4(-4) + 3(6) = 2
2=2
= 4
x = -4
Substitute value for
x into either of the
original equations
The solution is the point (-4,6).
Substitute (-4,6) into both
equations to check.
5(-4) + 3(6) = -2
-2 = -2
“Solve Linear Systems by Elimination”
Equation 1
Equation 2
+
8x - 4y = -4
-3x
4y +
= 4y
3x =
+ 14
5x
8x - 4y = -4 Equation 1
8(2) - 4y = -4
16 - 4y = -4
y=5
8(2) - 4(5) = -2
-2 = -2
Arrange like terms
= 10
x= 2
Substitute value for
x into either of the
original equations
The solution is the point (2,5).
Substitute (2,5) into both
equations to check.
4(5) = 3(2) + 14
20 = 20
Guided Practice
4x – 3y = 5
-2x + 3y = -7
7x – 2y = 5
7x – 3y = 4
(-1, -3)
(1,1)
3x + 4y = -6
2y = 3x + 6
(-2,0)
NJASK7 prep
Homework
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Text p. 447, #6-28 evens