Intermediate Algebra Chapter 5

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Intermediate Algebra
Chapter 4
• Systems
• of
• Linear Equations
Objective
• Determine if an ordered
pair is a solution for a
system of equations.
System of Equations
• Two or more equations
considered simultaneously form
a system of equations.
a1 x  b1 y  c1
a2 x  b2 y  c2
Checking a solution to a system
of equations
• 1. Replace each variable
in each equation with its
corresponding value.
• 2. Verify that each
equation is true.
Graphing Procedure
• 1. Graph both equations in the
same coordinate system.
• 2. Determine the point of
intersection of the two graphs.
• 3. This point represents the
estimated solution of the system
of equations.
Graphing observations
• Solution is an estimate
• Lines appearing parallel have to
be checked algebraically.
• Lines appearing to be the same
have to be checked
algebraically.
Classifying Systems
• Meet in Point – Consistent –
independent
• Parallel – Inconsistent –
Independent
• Same – Consistent - Dependent
Def: Dependent Equations
• Equations with
identical graphs
Independent Equations
• Equations with
different graphs.
Algebraic Check
• Same Line
a1 x  b1 y  c1
a2 x  b2 y  c2
a1
b1
c1


a2
b2
c2
Algebraic Check
• Parallel Lines
a1 x  b1 y  c1
a2 x  b2 y  c2
a1
b1
c1


a2
b2
c2

Algebraic Check
• Meet in a point {(x,y)}
a1 x  b1 y  c1
a2 x  b2 y  c2
a1
b1
c1


a2
b2
c2
Calculator Method for Systems
• Solve each equation for y
• Input each equation into Y=
• Graph
• Set Window
• Use CalIntersect
Calculator Problem
y  2 x  3
x  2y  4
 2, 1
Calculator Problem 2
3x  4 y  8
3
y   x3
4

Calculator Problem 3
6x  2 y  4
y  3x  2
 x, y  | y  3x  2
Objective
• Solve a System of
Equations using the
Substitution Method.
Substitution Method
• 1. Solve one equation for one variable
• 2. In other equation, substitute the
expression found in step 1 for that
variable.
• 3. Solve this new equation (1 variable)
• 4. Substitute solution in either original
equation
• 5. Check solution in original equation.
Althea Gibson – tennis player
• “No matter what
accomplishments you
make, someone helped
you.”
Intermediate Algebra
•The
• Elimination
•Method
Notes on elimination method
• Sometimes called addition
method
• Goal is to eliminate on of the
variables in a system of equations
by adding the two equations, with
the result being a linear equation in
one variable.
Write both equations in ax +
by = c form
• 2. If necessary, multiply one or
both of the equations by
appropriate numbers so that the
coefficients of one of the
variables are opposites.
• 1.
Procedure for addition method cont.
• 3. Add the equations to eliminate a
variable.
• 4. Solve the resulting equation
• 5. Substitute that value in either of
the original equations and solve for
the other variable.
• 6. Check the solution.
Procedure for addition method cont.
• Solution could be ordered pair.
• If a false statement results i.e. 1 =
0, then lines are parallel and
solution set is empty set.
(inconsistent)
• If a true statement results i.e. 0 = 0,
then lines are same and solution set
is the line itself. (dependent)
Practice Problem
x y 6
2 x  5 y  16
• Answer {(2,4)}
Practice Problem Hint: eliminate
x first
4 x  3 y  2
6 x  7 y  7
• Answer {(-7/2,-4)}
Practice Problem
2x  y  1
2 x  y  3
Practice problem
3x  4 y  5
9 x  12 y  15
Special Note on Addition Method
• Having solved for one variable,
one can eliminate the other
variable rather than substitute.
• Useful with fractions as
answers.
Practice Problem – eliminate one
variable and than the other
15
3
3
y
x
4
4
4
5
4
y3
x
3
3
• Answer: {(8/3,1/3)}
Confucius
• “It is better to light one
small candle than to
curse the darkness.”
Intermediate Algebra 4.2
• Systems
• Of
• Equations
• In
• Three Variables
Objective
• To use algebraic
methods to solve linear
equations in three
variables.
Def: linear equation in 3
variables
• is any equation that can be
written in the standard form
ax + by +cz =d where a,b,c,d
are real numbers and a,b,c
are not all zero.
Def: Solution of linear equation
in three variables
• is an ordered triple
(x,y,z) of numbers that
satisfies the equation.
Procedure for 3 equations, 3
unknowns
• 1. Write each equation in the form
ax +by +cz=d
• Check each equation is written
correctly.
• Write so each term is in line with a
corresponding term
• Number each equation
Procedure continued:
• 2. Eliminate one variable from
one pair of equations using the
elimination method.
• 3. Eliminate the same variable
from another pair of
equations.
• Number these equations
Procedure continued
• 4. Use the two new equations
to eliminate a variable and solve
the system.
• 5. Obtain third variable by back
substitution in one of original
equations
Procedure continued
• Check the ordered
triple in all three of the
original equations.
Sample problem 3 equations
(1) x  y  z  2
(2) 2 x  y  2 z  1
(3) 3x  2 y  z  1
Answer to 3 eqs-3unknowns
•{(-2,3,1)}
Bertrand Russell – mathematician
(1872-1970)
• “Mathematics takes us still
further from what is human,
into the region of absolute
necessity, to which not only the
actual world, but every possible
world, must conform.”
Cramer’s Rule
• Objective: Evaluate
determinants of 2 x 2 matrices
• Objective: Solve systems of
equations using Cramer’s Rule
Determinant
a b 
If A  
then
det[
A
]


c d 
a b
 ad  bc
c d
Cramer’s rule intuitive
• Each denominator, D is the
determinant of a matrix containing
only the coefficients in the system. To
find D with respect to x, we replace the
column of s-coefficients in the
coefficient matrix with the constants
form the system. To find D with
respect to y, replace the column of ycoefficients in the coefficient matrix sit
the constant terms.
Sample Problem: Evaluate:
3 2
2 4
• Answer = 16
Sample Cramer’s Rule problem
• Solve by Cramer’s Rule
2 x  3 y  5
3x  y  9
Cramer’s Rule Answer
D  11
Dx  22
Dy  33
Dx
22
x

2
D
11
Dy
33
y

 3
D
11
Senecca
• “It is not because things
are difficult that we do
not dare, it is because we
do not dare that they are
difficult.”
Intermediate Algebra 5.5
• Applications
• Objective: Solve application
problems using 2 x 2 and 3 x 3
systems.
Mixture Problems
• ****Use table or chart
• Include all units
• Look back to test
reasonableness of answer.
Sample Problem
• How many milliliters of a
10% HCl solution and
30% HCl solution must
be mixed together to
make 200 milliliters of
15% HCl solution?
Mixture problem equations
x  y  200
0.10 x  0.30 y  30
Mixture problem answers
• 150 mill of 10% sol
• 50 mill of 30% sol
• Gives 200 mill of 15% sol
Distance Problems
• Include Chart and/or picture
• Note distance, rate, and time in
chart
• D = RT and T = D/R and R=D/T
• Include units
• Check reasonableness of answer.
Sample Problem
• To gain strength, a rowing crew
practices in a stream with a fairly
quick current. When rowing
against the stream, the team takes
15 minutes to row 1 mile, whereas
with the stream, they row the same
mile in 6 minutes. Find the team’s
speed in miles per hour in still
water and how much the current
changes its speed.
Distance problem equations
0.25( x  y )  1
0.1( x  y )  1
 x y 4
 x  y  10
Answer
• Team row 7 miles per hour
in still water
• Current changes speed by
3 miles per hour
Joe Paterno – college football
coach
• “The will to win is
important but the will
to prepare is vital.”
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