Download Chapter 7 - James Bac Dang

CHAPTER 7
Systems of Equations &
Inequalities
7.1 Systems of Linear Equations in
2 Variables
• Objectives
– Decide whether an ordered pair is a solutions
of a linear system.
– Solve linear systems by substitution
– Solve linear systems by addition.
– Identify systems that do not have exactly one
ordered-pair solution.
– Solve problems using systems of linear
equations.
What is a solution of a system of
linear equations?
• Linear equations have graphs that are lines and
are stated in terms of 2 variables, x & y.
• A system of equations indicates there are 2 or
more equations, thus 2 or more lines.
• A solution to a system of equations would be
the ordered pair that makes both equations true
simultaneously.
• Graphically, this would indicate the point of
intersection of the lines. (0 points if parallel, 1
point if non-parallel & distinct, or infinitely many if
the 2 lines are on top of each other)
Solving a system of 2 equations
• Substitution: isolate a variable (x or y) in
one equation & substitute the expression it
is equal to into the other equation
• Example:
– Solve: x – 3y = -11 and x + y = 5
– Add 3y to both sides of the 1st equation,
resulting in: x = 3y – 11
– Now replace x in the 2nd equation with this:
(3y – 11) + y = 5 OR 4y = 16 OR y = 4
NOW solve for x in EITHER equation (let y = 4)
x + 4 = 5 OR x = 1 SOLUTION: (1,4)
CHECK the solution in both equations…is it true?
Addition to solve a system of
equations
• Add (or subtract) the 2 equations to eliminate one variable,
then solve for the other
• Add the 2 equations together:
• Example: Solve: 2x – y = 2, 3x – y = 5
2x  y  2
3 x  y  13
5 x  15,
x3
• Let x = 3 in either equation to solve for y:
• 2(3) – y = 2, y = 4
• Solution: (3,4) (CHECK!)
Sometimes, you must multiply an
equation by a constant to get a
term to drop out when adding
• Example:
• Solve: 2x – 3y = 7, 4x + 5y = 25
• Note: If the 2 equations were added,
nothing would drop out.
• Multiply 1st equation (both sides) by (-2)
• -4x + 6y = -14 (NOW add to 2nd eq.)
• Result: 11y = 11, thus y = 1
• Solve for x: 2x – 3 = 7, x = 5
• Solution: (5,1) CHECK!!
Systems with NO solution
• Since the equations are linear, no
solutions would mean the lines are
parallel.
• How would we know this w/o graphing?
• When solving (either w/substitution or
graphing) we would encounter a FALSE
statement. (example, next slide)
System with NO solution
•
•
•
•
•
Solve : 3x + y = 5, -6x – 2y = 12
Substitute: let y = 5 – 3x
-6x – 2(5 – 3x) = 12
-6x – 10 + 6x = 12
-10 = 12 FALSE!!! No matter what
values you use for (x,y), this statement will
never be true:
• NO SOLUTION (lines are parallel)
Systems with infinitely many
solutions
•
•
•
•
•
•
Lines would lie on top of each other
Example: 2x + 4y = 6, x + 2y = 3
Multiply 2nd equation by (-2): -2x – 4y = -6
Add the 2 equations together:
0 + 0 = 0 When is this true??? ALWAYS
There are infinitely many solutions. They are all
ordered pairs that fall on that line.
• Possible pairs: (3,0), (5,-1), (-1,2), etc
• General form of solution:  3  x 
 x,

2 

7.2 Systems of Linear Equations in
3 Variables
• Objectives:
– Verify the solution of a system of linear
equations in 3 variables
– Solve systems of linear equations in 3
variables
– Solve problems using systems in 3 variables
What is a system of equations of 3
variables?
• Linear equations that exist in space (not
simply in one plane)
• 3 variables: (x,y,z)
• 3 lines could intersect at one point, no
points, or infinitely many points (fall on top
of each other)
• How do we solve a system of 3 equations?
Very much like we do a system of 2
equations! (next slide)
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•
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•
•
Use substitution or elimination to solve:
3x + y – z = -2
x + 2y + z = 4
2x – y + 2z = -2
ADD the 1st 2 eq: 4x + 3y = 2
Multiply the 1st eq. by (2): 6x + 2y – 2z = -4
NOW add this eq. to 3rd eq: 8x + y = -6
We NOW have 2 eq. with 2 variables:
Solve this new system by multiplying 1st by (-2)
& then adding the 2nd.
Result: (-8x – 6y = -4) + (8x + y = -6):
-5y = -10, y=2
Now solve for x: 8x + 2 = -6, x = -1
Now solve for z: -1 + 4 + z = 4, z = 1
Solution: (-1, 4, 1) (CHECK!!)
7.3 Partial fractions
• Objectives:
– Decompose P/Q where Q has only distinct
linear factors
– Decompose P/Q, where Q has repeated
linear factors
– Decompose P/Q, where Q has a nonrepeated
prime quadratic factor
– Decompose P/Q, where Q has a prime,
repeated quadratic factor
What is decomposition of partial
fractions?
• Writing a more complex fraction as the sum or
difference of simpler fractions.
3a  4b a b
• Examples:


12
4 3
5x  3 5 3
  2
2
x
x x
• Why would you ever want to do this? It’s
EXTREMELY helpful in calculus!
Distinct linear factors
• Factor the denominator into linear terms
• Each linear term will be the denominator of
a separate term (i.e. if there are 3 factors,
there will be 3 separate fractions added
together)
• Example next slide
12 x  27 x  15
Decompose
2
( x  2)( x  1)
2
• Factor denominator: (x+2)(x+1)(x-1)
• Now the original expression will be written as the sum of 3
fractions:
• A/(x+2) + B/(x+1) + C/(x-1)
• To find A,B,C, rewrite this sum as 1 rational expression
(get common denominators)
A( x  1)( x  1)
B( x  2)( x  1)
C ( x  2)( x  1)


( x  2)( x  1)( x  1) ( x  1)( x  2)( x  1) ( x  1)( x  2)( x  1)
A( x 2  1)  B( x 2  x  2)  C ( x 2  3x  2)
( x  2)( x  1)( x  1)
Ax 2  A  Bx 2  Bx  2 B  Cx 2  3Cx  2C
( x  2)( x  1)( x  1)
(example continued)
• Since denominators are
equivalent, so must be
numerators. Equate like x 2 ( A  B  C )  12 x 2 , A  B  C  12
parts.
x( B  3C )  27 x, B  3C  27
 A  2 B  2C  27
• Use substitition & addition
to solve (add 1st & 3rd
equations, then take the
result & add to 2nd eq)
• Substitute back in other
eq. to solve for all 3
variables
 B  3C  27(1st  3rd )
6C  54(2nd ), C  9
B  0, C  9, A  3
Check your solutions.
• Are these solutions correct?
3
9
12 x 2  27 x  15


2
x  2 x  1 ( x  2)( x  1)
• At first glance, no, but then with closer
examination, we realize the x+1 term in the
denominator would cancel with an x+1 term in the
numerator, (recall B=0), thus it is correct!
• If we had first simplified the numerator to
(x+1)(12x+15) and cancelled the x+1 terms, it
may have been easier!
What if one on the denominators is
a linear term squared?
• This is accounted for by having the nonsquared term
as one denominator and having the squared term as
another denominator.
• What is one denominator is a linear term cubed?
There would be 3 denominators in the
decomposition:
A
B
C


2
(term) (term) (term)3
What if denominator has a
quadratic factor (not reduced to
product of linear factors)?
• A quadratic denominator in decomposition would
have a linear numerator:
5x  2 x  13
A
Bx  C

 2
2
( x  3)(2 x  1) x  3 2 x  1
2
• To decompose the fraction, you proceed precisely
as was done with linear denominators.
7.4 Systems of Nonlinear
Equations in 2 Variables
• Objectives:
– Recognize systems of nonlinear equations in
2 variables
– Solve nonlinear systems by substitution
– Solve nonlinear systems by addition
– Solve problems using systems of nonlinear
equations
Decompose
2
3
1)

x3 x3
2
3
2)

2
x  3 ( x  3)
2
9
3)

2
x  3 ( x  3)
2
6
4)

x  3 ( x  3) 2
2x  9
2
( x  3)
Non-linear systems: what are
they?
• If a linear systems involves finding the
intersection of lines, consider what the
graph of nonlinear equations might be.
• Graphs could be that of circles, parabolas,
ellipses.
• Where might they intersect? A circle and
parabola might intersect once, twice, three
times, or not at all!
• A line and an ellipse may not intersect, of
may have one or 2 points of intersection.
When solving these systems, be
aware of ALL possible solutions.
Don’t overlook some!
• Graphing the system will provide a clue as
to what the solutions may look like (Be
careful when entering some functions.
They may need to be entered as
piecewise functions.).
Solve: 9 x  4 y  36, x  y  4
2
2
2
2
• Solve using addition (multiply 2nd equations by -9
before adding together)
9 x  4 y  36
2
2
 9 x  9 y  36
 5 y 2  0, y  0, x  2
solution : {( 2,0), (2,0)}
2
2
• Note: this is a circle that sits inside an ellipse &
they share the 2 points of intersection.
7.5 Systems of Inequalities
• Objectives:
– Graph a linear inequality in two variables
– Graph a nonlinear inequality in two variables
– Graph a system of inequalities
– Solve applied problems involving systems of
inequalities
Linear inequality in two variables
• A linear inequality divides the xy-plane into 2
parts. Either the points on one side of the line
make the inequality true or points on the other
side do.
• Once the line is determined, select any point
on either side to test in the original inequality
to determine if that point is a solution or not.
• If the point makes the inequality true, all points
on that side are also in the solution set. If the
point makes the inequality false, all points on
the other side of the line are in the solution
set.
System of linear inequalities
• If the 2 lines intersect at one point, the
plane is divided into 4 areas. The solution
could be found in one of these areas.
• Often graphing and looking for overlapping
areas is easier than looking at points in
each region.
Nonlinear inequalities
• Consider the graph of a circle. The plan is
divided into the area inside the circle, and
that outside the circle. Solve it as you
would a linear system.
• Consider the graph of other nonlinear
inequalities (parabolas, ellipses,
hyperbolas). Again, the graph would show
that the points that make the inequality
true would be found inside OR outside of
the graph.
7.6 Linear Programming
• Objectives
– Write an objective function describing a
quantity that must be maximized or minimized
– Use inequalities to describe limitations in a
situation
– Use linear programming to solve problems
Find a linear equation that
represents a specific situation,
taking into account restrictions
(linear inequalities).
• Each restriction creates a linear inequality
• Plot the equation and the inequalities and
examine the vertices (points of intersection
of lines)
• The original function is maximized and
minimized at one or more corner points.
You are taking a test with a multiple-choice
section & a full response section. Each M-C
question is worth 3 pts. & full-response is 7 pts.
You have 50 min. to complete the test. M-C
requires 2 min/prob & full-response,
5min/prob. If you answer at most 13 questions
& all your answers are correct, how many of
each should you answer for the highest score?
What’s the score?
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Example continued
3x + 7y = score (objective eq.)
2x+5y < 50, x >0,y>0, x +y < 13
Intersection pt.=(5,8)
Other vertices: (0,10), (0,0),(13,0)
(1st quad)
Put all 4 pairs into objectives eq:
3(0)+7(10)=70
3(0)+7(0)=0
3(13)+7(0)=39
3(5)+7(8)=71
5 M-C & 8 full-response for 71 pts.
State the objective equation:
Student tickets will be sold for $4 each and adults
for $6 each for a play. There are 200 seats in the
auditorium. Every adult will bring at most 3
children. How many of each type of ticket should
be sold to raise the most money?
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1) Total = 4x + 6y
2) x + y < 200
3) y  3x
4) x>0, y>0