Download 1 - Lamar County School District

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2.
2.
Describe each transformation of f to g:
f(x) = |x| and g(x) = -3|x + 2| - 1.
Write a function g if f(x) = x2 has a vertical shrink of 1/3 followed by
a translation up 2.
The data shows the humerus lengths ( in centimeters) and heights
(in centimeters( of several females.
Use the graphing calculator to find a line of best fit for the data.
Estimate the height of a female whose humerus is 40 centimeters long.
Estimate the humerus length of a female with a height of 130 cm.
Algebra II
1
Systems of Equations
with Two Variables
Algebra II

two or more linear equations.

ì 1st equation
Looks like í
î2nd equation

A solution is an ordered pair that makes all
equations true.
Algebra II
3
3x – 2y = 2
x + 2y = 6
a) (0, -1)
no
b) (2,2)
yes
Algebra II
4
 Graphing
 Substitution
 Elimination
Algebra II
5
To find the solution of a system of two linear
equations: (steps)
1. Graph each equation
2. Identify the intersection
3. This is the solution to the system
because it is the point that satisfies both
equations.
**Remember that a graph is just a picture
of the solutions.
Algebra II
6
Graph
Number of
Solutions
intersecting lines
one solution
Two lines intersect at one point.
parallel lines
no solutions
Parallel lines
Lines coincide
Algebra II
coincident lines
(same line)
infinitely many solutions
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Solve the system of equations
by graphing.
ìï 2x - 2y = -8
í
ïî 2x + 2y = 4
First, graph 2x – 2y = -8.
Second, graph 2x + 2y = 4.
The lines intersect at (-1, 3)
The solution is (-1, 3)
Algebra II
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Solve the system of equations
by graphing.
ì -x + 3y = 6
í
î 3x - 9y = 9
(3, 3)
(0, 2)
(-3, 1)
First, graph -x + 3y = 6.
Second, graph 3x – 9y = 9.
(-3, -2)
(3, 0)
(0, -1)
The lines are parallel.
No solution
Algebra II
9
Solve the system of equations
by graphing.
ì 2x - y = 6
í
î x + 3y = 10
First, graph 2x – y = 6.
Second, graph x + 3y = 10.
The lines intersect at (4, 2)
The solution is (4, 2)
Algebra II
10
Solve the system of equations
by graphing.
ì x = 3y -1
í
î2x - 6y = -2
First, graph x = 3y – 1.
Second, graph 2x – 6y = -2.
(-1, 0)
(2, 1)
(-4, -1)
The lines are identical.
Infinitely many
solutions
Algebra II
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Steps for Substitution:
1.
Solve one of the equations for one variable (try to
solve for the variable with a coefficient of one)
2.
Substitute the expression into the other equation
and solve the new equation.
3.
Substitute the value from step 2 into one of your
original equations to complete the ordered pair
Algebra II
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1. 3x – y = 6
-4x + 2y = –8
-4x + 6x – 12 = -8
2x = 4
x=2
Step 1:
Step 3:
3x – y = 6
-y = -3x + 6
y = 3x – 6
y = 3x – 6
y = 3(2) – 6
y=0
(2,0)
Step
Algebra
II
2:
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2. x – 3y = 4
6x – 2y = 4
18y + 24 – 2y = 4
16y = -20
y = -5/4
Step 1:
x – 3y = 4
x = 3y +4
Step 3:
x = 3y + 4
x = 3(-5/4) + 4
x = 1/4
(1/4, -5/4)
Step 2:
6x – 2y = 4
Algebra II
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3. y = 2x – 5
8x – 4y = 20
Step 1:
Y = 2x – 5
(already done)
8x – 4(2x – 5) = 20
8x – 8x + 20 = 20
20 = 20
0=0
True Statement!
Infinitely Many Solutions
Step 2:
Algebra II
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4. -4x + y = 6
-5x – y = 21
-5x – 4x – 6 = 21
-9x = 27
x = -3
Step 1:
-4x + y = 6
y = 4x + 6
Step 3:
y = 4x + 6
y = 4(-3) + 6
y = -6
(-3,-6)
Step 2:
-5x – y = 21
Algebra II
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
Steps for elimination:
1. Make one of the variables have opposite coefficients
(multiply by a constant if necessary)
2. Add the equations together and solve for the
remaining variable
3. Substitute the value from step 3 into one of the
original equations to complete the ordered pair
Algebra II
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2 30x – 15y = -15
12x + 15y = -27
6x – 3y = –3
42x + 0 = -42
4x + 5y = –9
42x = -42
42
42
1 5(6x – 3y = –3)
3(4x + 5y = –9)
x = -1
Solve the following
system by elimination
Algebra II
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Use x = -1 to find y
3
2nd equation:
4x + 5y = -9
4(-1) + 5y = -9
-4 + 5y = -9
+4
+4
5y = -5
5
5
y = -1
Algebra II
(-1, -1)
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Solve the following
system by elimination
3x – y = 4
6x – 2y = 4
1
-2(3x – y = 4)
(6x – 2y = 4)
2
-6x + 2y = -8
6x – 2y = 4
0 + 0 = -4
0 ≠= -4
False!
No Solution
Algebra II
20
Solve the following
system by elimination
2
3x + 5y = -6
2x – 2y = -8
1
Algebra II
2(3x + 5y = -6)
-3(2x – 2y = -8)
6x + 10y = -12
-6x + 6y = 24
0 + 16y = 12
16y = 12
16
16
y = 3/4
21
Use y = 3/4 to find x
3
1st equation:
3x + 5y = -6
3x+ 5(3/4) = -6
3x + 15/4 = -6
-15/4 -15/4
3x = -39/4
3
3
y = -13/4
Algebra II
(-13/4, 3/4)
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Solve the following
system by elimination
-2x + y = -5
8x – 4y = 20
1
Algebra II
4(-2x + y = -5)
(8x – 4y = 20)
2
-8x + 4y = -20
8x – 4y = 20
0+0=0
0 =
= 0
True!
Infinitely Many
Solutions
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1. 4x – 3y = 10
2x + 2y = 7
4. 3x + 2y = 8
2y + 4x = -2
2. Y = 3x – 5
2x + 3y = 8
5. 2x + 7y = 10
x + 4y = 9
3. X – 3y = 10
4x + 3y = 21
Algebra II
6. x – 3y = -6
x = 2y
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1. 4x – 3y = 10
8x – 6y = 5
2. 3x + 3y = 10
2x – 2y = 15
M = 4/3, b= -10/3
M = -1, b = 10/3
M = 4/3 b = -5/6
M = 1, b = -15/2
No solution
Algebra II
One solution
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3.
Algebra II
y = 2x + 8
2x – y = -8
4.
1/2x + 3y = 6
1/3x – 5y = -3
M = 2, b= 8
M = -1/6, b = 2
M = 2, b = 8
M = 1/15, b = 3/5
Infinitely
many
One
solution
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1. Your family is planning a 7 day trip to Florida.
You estimate that it will cost $275 per day in Tampa
and $400 per day in Orlando. Your total budget for
the 7 day is $2300. How many days should you
spend in each location?
 X = # of days in Tampa
 Y = # of days in Orlando


X+y=7
275x + 400 y = 2300
Algebra II
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2. You plan to work 200 hours this summer mowing
lawns or babysitting. You need to make a total of
$1300. Babysitting pays $6 per hour and lawn
mowing pays $8 per hour. How many hours should
you work at each job?
 X = # of hours babysitting
 Y = # of hours of mowing


X + y = 200
6x + 8y = 1300
Algebra II
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3. You make small wreaths and large wreaths to sell
at a craft fair. Small wreaths sell for $8 and large
wreaths sell for $12. You think you can sell 40 wreaths
all together and want to make $400. How many of each
type of wreath should you bring to the fair?
X = # small wreaths
Y = # large wreaths
X + y = 40
8x + 12y =
Algebra II
400
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4. You are buying lotions or soaps for 12 of your
friends. You spent $100. Soaps cost $5 a piece
and lotions are $8. How many of each did you
buy?
x = # of soaps
y = # of lotions
x + y = 12
5x + 8y = 100
Algebra II
30
5. Becky has 52 coins in nickels and dimes.
She has a total of $4.65. How many of each
coin does she have?
x = # of nickels
y = # of dimes
x + y = 52
.05x + .10y = 4.65
Algebra II
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6. There were twice as many students as adults
at the ball game. There were 2500 people at the
game. How many students and how many
parents were at the game?
x = # of students
y = # of parents
x = 2y
x + y = 2500
Algebra II
32
1. Using substitution, solve the system:
{
3x + 4y = -4
x + 2y = 2
(-8, 5)
2. Using elimination, solve the system:
{
-3x + y = 11
5x – 2y = -16
Algebra II
(-6, -7)
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