Download Mod 1 Ch 4-5

Mth 95 Notes
Module 1
Spring 2014
Section 4.1- Solving Systems of Linear Equations in Two Variables
by Graphing , Substitution, and Elimination
A Solution to a System of Two (or more) Linear Equations is the common point(s) of
intersection and produces true math sentences when substituted into both equations
Determine by substitution whether (-3, 3) and (-6, -6) are solutions to the system.
3x  y   12
x  y 0
Number of Possible Solutions to a system of Linear Equations
y
B
A
y
y
C
x
______solution(s)
Slopes are ______________
Y-intercepts are __________
x
x
______ solution(s)
________________ of solution(s)
Slopes are the ________
Slopes are the __________
Y-intercepts are __________ Y-intercepts are the _______
A consistent system has at least one solution. An inconsistent system has no solution.
Which of the systems above are consistent?
Inconsistent?
Independent equations are different. Dependent equations are the same (slope and yintercept). Which equations above are independent?
Dependent?
To tell how many solutions a system of equations has, transform both equations into
slope-intercept form and examine their m-values and b-values.
Tell how many solutions each system has.
y  2x  3
2x  y 1
6 x  3 y  9
4x  y  4
Chapters 4 and 5
y  2 x  3
2 y  4 x
1
Mth 95 Notes
Module 1
Spring 2014
Methods of Solving a System of Equations
Method 1: Solving by Graphing: good for approximation, a good visual of the
situation, also a good check for your symbolic solutions.
Steps: 1) Graph both equations on the same coordinate plane.
2) Name the point(s) of intersection
3) Check your solution algebraically by substituting the point of intersection into
both equations or transforming both equations into slope intercept form and
examining m and b for infinitely many solutions and no solution.
Solve each system by graphing.
3x  y 1
x  2y   3
y
y  2x 1
y  2 x  1
6x  3y  3
3 y  6 x
y
x
Solution ____________
y
x
Solution ____________
x
Solution ____________
Show an algebraic check for your solution to the first system above. Check the other
systems by graphing them on your calculator.
Method 2: Solving by Substitution (works well when the coefficient of x or y is one.)
Steps: 1) Choose one of the equations and solve it for one of the variables.
2) Substitute what you get into the other equation and solve.
3) Check your solution.
Use substitution to solve the following systems of linear equations
3x  y 1
y  2x 1
y  2 x  1
x  2y   3
6x  3y  3
3 y  6 x
Solution ____________
Solution ____________
Chapters 4 and 5
Solution ____________
2
Mth 95 Notes
Module 1
Applications: Use substitution to solve the following:
Spring 2014
(Break Even problem)
Find the number of units, x, that must be sold to break even given the Cost (C) and
Revenue(R) functions for a business to be:
C ( x) 12 x 15,000
R( x)  32 x
Method 3: Elimination
Steps: 1) Get both equations in standard form. Be sure the same variables line up under each
other and the equal signs are under each other.
2) Multiply or divide one or both the equations to get the same coefficient but opposite
signs on one of the variables.
3) Combine the like variables and constants in a vertical manner. One of the variables
should be eliminated.
4) Solve for the remaining variable.
5) Check your solution.
Use the elimination method to solve the following:
3x  y  7
2x  y   2
3x  4 y 10
5 x 10 y  15
Solution ____________
Solution ____________
Chapters 4 and 5
3 x  4y  7
y  2x   3
x  3y  6
2x  6y  5
Solution ____________
Solution ____________
3
Mth 95 Notes
1
y  x4
3
3 y  x  12
Module 1
Spring 2014
Solution ____________
What does it mean when both my variables disappear in the process?
Section 4.3 – Systems of Linear Equations and Problem Solving
Applications: Begin by defining your variables. Then write a system of equations and
use the elimination or substitution method to solve each problem.
(Mixture Problem): One number is two less than a second number. Twice the first is 4
more than 3 times the second. Find the numbers.
(Mixture Problem): A grocer mixes two grades of coffee that sell for $6 and $8 a
pound. He has a bin that holds 50 pounds of coffee. How much of each must he buy to
make a mixture that sells for $7.20 a pound?
Chapters 4 and 5
4
Mth 95 Notes
Module 1
Spring 2014
(Mixture problem with investments): A sum of $1000 is invested for one year. Some
of the money is invested at 6% and the remainder at 12%. How much was invested at
each percent if the total interest income from the two investments is $108?
(Another mixture problem with percents): A pharmacist need 500 milliliters of a 20%
Phenobarbital solution but has only 5% and 25% Phenobarbital solutions available. Find
how many milliliters of each he should mix to get the desired solution.
(A motion problem where current affects the rate) A boat travels 18 miles down a
river in 1 hour. A return trip against the current takes 1.5 hours. Find the average speed
of the boat in still water and the rate of the current. (distance = rate x time)
Chapters 4 and 5
5
Mth 95 Notes
Module 1
Spring 2014
Section 5.1 – Integer Exponents
In the expression 43 , 4 is the ___________and 3 is the ____________. 43 means 4
is repeated as a ________________ 3 times, so 43 = _________________________
Properties of Exponents:
1) Product Rule
bm  bn  bmn
When you multiply two numbers with the same _____________
keep the ______________ and add the ____________________
x3  x5
(5 x 2 y 3 )(2 xy 2 )
32  35
(4 x5 y 2 )(2 x3 yz )( z 4 )
2) Zero Exponents
For any nonzero number a, a 0 = 1, so 4 0 =_______and 100 = __________ ,
but 00 is ____________________ .
80 
(8)0 
 80 
(2 y  3)0 
3( x  2)0 
3) Quotient Rule for Exponents
bm
 bmn
bn
When you divide two numbers with the same ___________, subtract
the exponents ( top exponent – bottom exponent).
x5

x3
55

53
24c5 d 4

4cd 3
8 x3 y 6

40 x3 y 4
4) Negative Exponents (doesn’t change the sign of the coefficient)
1
bn
x 3 
52 
y 3  y 1 
1
 bn
b n
1

x 7
1

2 3
y5

x 3
bn 
Chapters 4 and 5
6
Mth 95 Notes
Module 1
Spring 2014
Use properties of exponents to evaluate each expression. Give all answers with positive
exponents.
32 
 3
2
y3

y 1
 3x y  5 x
24 x 2 y13

2 x5 y 2
15r 2t 5

25rt 2
2
5
7

y3 

(5)3 
6 x 5 y 3

12 xy 7
x 3

x5
r wt
7 5

7 3
2.4 x3 y 0 (3x 1 z ) 
2
3 1

Review Section 5.1
Product Rule for Exponents
(3x 4 )(5 x 2 ) 
Quotient Rule for Exponents
20c 3

5c
Zero Exponents
50 
Negative Exponents
n 3 
(4n)0 
3x 0 
00 
1

m 5
Section 5.2 - More Work with Properties of Exponents
5) Raising Powers to Powers - Multiply the exponents.
b 
m
n
 b m n
(32 )3 
Chapters 4 and 5
x 
3
5

(32 )3 
y 
3
2

(51 ) 3 
7
Mth 95 Notes
Module 1
Spring 2014
6) Raising Products to Powers - Take each factor to that power.
(ab)n  a nbn
 2x 
2
3
 

m n 
( xy )5 
3 a 2b
4
3 2
 5x 
2

3

 3x 
7
4

2

7) Raising Quotients to Powers - Take the numerator and the denominator to that power.



n
an
a

 
bn
b
5
3
 2m 2 

 
 n 
x
 
y
2
2
2
 y 3 
 5  
y 
 2x2 

 
7


 3
  
 4
Simplify the following expressions completely. Write answers using positive exponents
only.
3
1
 3 x 2 

 
 4 xy 
 2 x 3 y 4 

 
5


36c5 d 2

9cd 2
9 y5

1 6
y
3
 4 x  2 x  
 2x y 
x y 
62 x 3 y 0

33 x 5 y 3
 2c d  3c 
 5r 5t 2 
 3 6  
r t 
Chapters 4 and 5
5
2
3
4
2
3
2
4
3
2

2
2

8