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Chabot Mathematics
§3.1 2-Var
Linear Systems
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
Chabot College Mathematics
1
Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Review §
2.4
MTH 55
 Any QUESTIONS About
• §’s2.4 → Point-Slope Eqn, Modeling
 Any QUESTIONS About HomeWork
• §’s2.4 → HW-07
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Systems of Equations
 System of Equations ≡ A group of two
or more equations; e.g.,
 x y 5

3x  4 y  8
(Equation 1)
(Equation 2)
 Solution For A System Of Equations
≡ An ordered set of numbers that makes
ALL equations in the system TRUE at
the same time
Chabot College Mathematics
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Bruce Mayer, PE
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Checking System Solution
 To verify or check 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.
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Chk System Soln
x  y  7
 Consider The

Equation System  y  3x  2
(Equation 1)
(Equation 2)
 Determine whether each ordered pair is
a solution to the system of equations.
a. (−3, 2)
b. (3, 4)
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Chk System Soln
 SOLUTION →
Chk True/False
x  y  7

 y  3x  2
(Equation 1)
(Equation 2)
 a. (−3, 2) → Sub: −3 for x, & 2 for y
x+y=7
y = 3x − 2
−3 + 2 = 7
2 = 3(−3) − 2
−1 = 7
2 = −11
False
False
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Chk System Soln
 SOLUTION →
Chk True/False
x  y  7

 y  3x  2
(Equation 1)
(Equation 2)
 b. (3, 4) → Sub: 3 for x, & 4 for y
x+y=7
y = 3x − 2
3+4=7
4 = 3(3) − 2
7=7
4=7
True
False
Chabot College Mathematics
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Bruce Mayer, PE
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Example  Chk System Soln
 SOLUTION →
Chk True/False
x  y  7

 y  3x  2
(Equation 1)
(Equation 2)
 Because (−3, 2) does NOT satisfy
EITHER equation, it is NOT a solution
for the system.
 Because (3, 4) satisfies ONLY ONE
equation, it is NOT a solution to the
system of equations
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Systems of Equations Soln
 A system-ofequations problem
involves finding the
solutions that satisfy
the conditions set
forth in two or more
Equations
 For Equations of
Lines, The System
Solution is the
CROSSING Point
Chabot College Mathematics
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 This Graph shows
two lines which have
one point in common
y  x5
y  x 1
 The common point
is (–3,2) Satisfies
BOTH Eqns
Bruce Mayer, PE
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Solve Systems of Eqns by Graphing
 Recall that a graph of an equation is a
set of points representing its solution set
 Each point on the graph corresponds
to an ordered pair that is a
solution of the equation
 By graphing two equations using one
set of axes, we can identify a solution
of both equations by looking for a
point of intersection
Chabot College Mathematics
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Bruce Mayer, PE
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Solving by Graphing Procedure
1. Write the equations of the lines in
slope-intercept form.
2. Use the slope and y-intercept of each
line to plot two points for each line on
the same graph.
3. Draw in each line on the graph.
4. Determine the point of intersection
(the Common Pt) and write this point
as an ordered pair for the Solution
Chabot College Mathematics
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Bruce Mayer, PE
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Example  Solve w/ Graphing
 Solve this system
• y = 3x + 1
• x – 2y = 3
 SOLUTION:
Graph Each Eqn
• y = 3x + 1
– Graph (0, 1) and
“count off” a slope of 3
• x – 2y = 3
– Graph using the
intercepts:
(0,–3/2) & (3, 0)
Chabot College Mathematics
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(−1, −2) The crossing
point provides the
common solution
Bruce Mayer, PE
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Example  Solve w/ Graphing
 Chk (−1, −2) Soln:
• y = 3x + 1
• x − 2y = 3
 y = 3x + 1→
• −2 = 3(−1) + 1
• −2 = −3 + 1
• −2 = −2 
1,2
 x − 2y = 3 →
• (−1) − 2(−2) = 3
• −1+4 = 3
• 3=3 
Chabot College Mathematics
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 Thus (−1, −2) Chks
as a Soln
Bruce Mayer, PE
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Example  Solve By Graphing
 Solve System:
y  x2
y=6x
y  6 x
 SOLUTION: graph
Both Equations
• The graphs intersect
at (4, 2), indicating
that for the x-value 4
both x−2 and 6−x
share the same
value (in this case 2).
(4, 2)
y=x2
• As a check note that
[4−2] = [6−4] is true.
– The solution is (4, 2)
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
The Substitution Soln Method
 Graphing can be an imprecise method
for solving systems of equations.
 We are now going to look at ways of
finding exact solutions using algebra
 One method for solving systems is
known as the substitution method. It
uses algebra instead of graphing and is
thus considered an algebraic method.
Chabot College Mathematics
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Bruce Mayer, PE
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Substitution Summarized
 The substitution method involves
isolating either variable in one
equation and substituting the
result for the same variable in the
second equation. The numerical
result is then back-substituted
into the first equation to find the
numerical result for the
second variable
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Solve by Subbing
4 1
 Solve the 3x  2 y 
System
y   2 x  5 2
 SOLUTION: The second equation
says that y and −2x + 5 represent
the same value.
 Thus, in the first equation we can
substitute −2x + 5 for y
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Solve by Subbing
4 1
 The Algebra 3x  2 y 
to Solve
y   2 x  5 2
3x  2 y  4
Equation (1)
3x  2 2 x  5  4
3x  4 x 10  4
Substitute: y = −2x + 5
Distributive Property
7 x 10  4 Combine Like Terms
7 x  14 Add 10 to Both Sides
x2
Chabot College Mathematics
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Divide Both Sides by 7 to Find x
Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Solve by Subbing
 We have found the x-value of the
solution. To find the y-value, we return
to the original pair of equations.
Substituting x=2 into either equation will
give us the y-value. Choose eqn (2):
y  2 x  5 Equation (2)
 The ordered
y  22  5
y  4  5
y 1
Chabot College Mathematics
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Substitute: x = 2
Simplifying
When x = 2
pair (2, 1)
appears to be
the solution
Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Solve by Subbing
 Check Tentative Solution (2,1)
3x − 2y = 4
3(2) − 2(1) 4
6−2 4
4 = 4 True
y = −2x + 5
1 −2(2) + 5
1 −4 + 5
1 = 1 True
 Since (2, 1) checks in BOTH equations,
it IS a solution.
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Substitution Solution CAUTION
 Caution! A solution of a system of
equations in two variables is an
ordered pair of numbers. Once you
have solved for one variable, do not
forget the other. A common mistake is
to solve for only one variable.
3x  2 y 
4 1
 x2
y   2 x  5 2
3x  2 y 
4 1
 x, y   2,1
y   2 x  5 2
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Solve by Subbing
 Solve the x  3  y
System 5 x  3 y  5
1
2
 SOLUTION: Sub 3 − y for x in Eqn (2)
5 x  3 y  5 Equation (2)
53  y 3 y  5
Substitute: x = 3−y
15  5 y  3 y  5
Distributive Property
10  2 y  0 Combine Terms, Subtract 5 from Both sides
2 y  10  y  5 Solve for y = 5
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Solve by Subbing
 Find x for y = 5
• Use Eqn (1)
x  3  y Eqn (1)
x  3  5
x  2 Solve for x
Sub y = 5
 Thus (−2,5) is the Soln
• The graph below is
another check.
5x + 3y = 5
(−2, 5)
 Chk Soln pair (−2,5)
x=3−y
−2 = 3 − 5
−2 = −2
5x + 3y = 5
5(−2)+3(5) = 5
−10 + 15 = 5
5=5
Chabot College Mathematics
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x=3y
Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Solving for the Variable First
 Sometimes neither
x  y  6 1
equation has a
5 x  2 y  8 2
variable alone on
one side. In that
case, we solve one
equation for one of
the variables and
then proceed as
before
 For Example; Solve:
Chabot College Mathematics
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 We can solve either
equation for either
variable. Since the
coefficient of x is
one in equation (1),
it is easier to solve
that equation for x.
x y 6
x  6 y
1
3
Bruce Mayer, PE
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x  y  6 1
Example  Solve 5 x  2 y  8 2
 substitute x = 6−y for x in equation (2) of
the original pair and solve for y
5 x  2 y  8 Equation (2)
56  y 2 y  8
Substitute: x = 6−y
Use Parens or Brackets when Subbing
30  5 y  2 y  8
Distributive Property
30  3 y  8 Combine Like Terms
22  3 y Add −8, Add 3y to Both Sides
y  22 3 Divide Both Sides by 3 to Find y
Chabot College Mathematics
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Bruce Mayer, PE
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x  y  6 1
Example  Solve 5 x  2 y  8 2
 To find x we substitute 22/3 for y in
equation (1), (2), or (3). Because it is
generally easier to use an equation that
has already been solved for a specific
variable, we decide to use equation (3):
22 18 22  4
x  6 y  6
 

3
3 3
3
 A Check of this
 4 22 
Ordered Pair Shows   , 
3 3 

that it is a Solution:
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Substitution Solution Procedure
1. Solve for a variable in either one of the
equations if neither equation already
has a variable isolated.
2. Using the result of step (1), substitute in
the other equation for the variable
isolated in step (1).
3. Solve the equation from step (2).
4. Substitute the ½-solution from step (3) into
one of the other equations to solve for
the other variable.
5. Check that the ordered pair resulting from
steps (3) and (4) checks in both of the
original equations.
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Solving by Addition/Elimination
 The Addition/Elimination method for
solving systems of equations makes
use of the addition principle
 For Example; 4 x  3 y  8 1
Solve System x  3 y  7 2
 According to equation (2), x−3y and 7
are the same thing. Thus we can add
4x + 3y to the left side of the
equation(1) and 7 to the right side
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Elimination Example
4 x  3 y  8 1
 Add
x  3 y  7 2
Equations
5 x  0 y  15
 The resulting equation has just
one variable: 5x = 15
• Dividing both sides by 5, find that x = 3
 Next Sub 3 for x in
Either Eqn to Find
the y-value
• Using Eqn-1
Chabot College Mathematics
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43  3 y  8
12  3 y  8
3 y  4
y  4 3
Bruce Mayer, PE
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Elimination Example
 Check Tentative Solution (3, −4/3)
4x + 3y = 8
4(3) + 3(−4/3) 8
12 − 4
8=8
True
x − 3y = 7
3 − 3(−4/3) 7
3+4
7=7
True
• Since (3, −4/3)
checks in both
equations, it is
the solution
– Graph confirms
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Solve
4 x  3 y  17
2 x  3 y  13
1
2
 SOLUTION: Adding the two equations
as they appear will not eliminate a
variable.
 However, if the 3y were −3y in one
equation, we could eliminate y. We
multiply both sides
4 x  3 y  17 1
of equation (2) by
 2 x  3 y  13 2
−1 to find an
2x
 4
equivalent eqn
x2
and then add:
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Solve
 To Find y substitute
2 for x in either of
the original eqns:
4 x  3 y  17 1
42  3 y  17
8  3 y  17
3y  9  y  3
4 x  3 y  17
2 x  3 y  13
1
2
 The graph shown
below also checks.

 We can check the
ordered pair (2, 3)
in Both Eqns
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Which Variable to Eliminate
 When deciding which variable to
eliminate, we inspect the coefficients in
both equations. If one coefficient is a
simple multiple of the coefficient of the
same variable in the other equation, that
one is the easiest variable to eliminate.
 For Example; 2 x  5 y  1 1
Solve System
3x  10 y  16
Chabot College Mathematics
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2
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Example  Solve
2 x  5 y  1
3x  10 y  16
1
2
 SOLUTION: No terms are opposites,
but if both sides of equation (1) are
multiplied by 2, the coefficients of y will
be opposites.
4 x  10 y  2
3x  10 y  16
7x
 14
x2
Chabot College Mathematics
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1 Mult Both Sides of Eqn-1 by 2
2
Add Eqns
Solve for x
Bruce Mayer, PE
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Example  Solve
2 x  5 y  1
3x  10 y  16
1
2
 SOLUTION: Find y by Subbing x=2 into
Either Eqn; Choosing Eqn-1
22  5 y  1
4  5 y  1
Sub 2 for x
Simplify
5 y  5  y  1 Solve for y
 Student Exercise: confirm that (2, −1)
checks and is the solution.
Chabot College Mathematics
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Bruce Mayer, PE
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Multiple Multiplication
 Sometimes BOTH equations must be
multiplied to find the Least Common
Multiple (LCM) of two coefficients
 For Example; 2 x  3 y  2 1
Solve System 3x  5 y  4 2
 SOLUTION: It is often helpful to write
both equations in Standard
form before attempting 2 x  3 y  2 3
to eliminate a variable: 3x  5 y  4 4
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Solve
2x  3 y  2
3x  5 y  4
3
4
 Since neither coefficient of x is a
multiple of the other and neither
coefficient of y is a multiple of the other,
we use the multiplication principle with
both equations.
 We can eliminate the x term by
multiplying both sides of equation (3) by
3 and both sides of equation (4) by −2
Chabot College Mathematics
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Bruce Mayer, PE
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Example  Solve
 The Algebra
6x  9 y  6
 6 x  10 y   8
y  2
2x  3 y  2
3x  5 y  4
3
4
5 Mult Both Sides of Eqn-3 by 3
6 Mult Both Sides of Eqn-4 by −2
Add Eqns
 Solve for x using y = −2 in Eqn-3
2 x  3 2  2
2x  6  2
2 x  4
x  2
Chabot College Mathematics
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 Students to Verify
that solution
(−2, −2) checks
Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Solving Systems by Elimination
1.
2.
Write the equations in standard form (Ax + By = C).
Use the multiplication principle to clear fractions or
decimals.
Multiply one or both equations by a number (or numbers)
so that they have a pair of terms that are additive
inverses.
Add the equations. The result should be an equation in
terms of one variable.
Solve the equation from step 4 for the value of that
variable.
Using an equation containing both variables, substitute
the value you found in step 5 for the corresponding
variable and solve for the value of the other variable.
Check your solution in the original equations
3.
4.
5.
6.
7.
Chabot College Mathematics
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Bruce Mayer, PE
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Types of Systems of Equations

When solving problems concerning
systems of two linear equations and
two variables there are
three possible outcomes.
1. Consistent Systems
2. INConsistent Systems
3. Dependent Systems
Chabot College Mathematics
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Bruce Mayer, PE
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Case-1 Consistent Systems
 In this case, the
graphs of the
two lines
intersect at
exactly one
point.
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Case-2 Inconsistent Systems
 In this case the
graphs of the two
lines show that
they are parallel.
 Since there is
NO Intersection,
there is NO
Solution to this
system
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Case-3 Dependent Systems
 In this case the
graphs of the two
lines indicate
that there are
infinite solutions
because they
are, in reality,
the same line.
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Testing for System Case
 Given System of Two Eqns
y1  m1 x  b1 & y2  m2 x  b2
 Case-1 Consistent → m1 ≠ m2
 Case-2 Inconsistent →
• m1 = m2
• b1 ≠ b2
 Case-3 Dependent →
• m1 = m2 AND b1 = b2;
– Or for a constant K: m1 = Km2 AND b1 = Kb2
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Bruce Mayer, PE
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Example  Solve by Graphing
 Solve System:
3
3
y  x  2 & y  x 3
4
4
 SOLUTION: Graph Eqns
y = 3x/4 + 2
• equations are in slopeintercept form so it is easy
to see that both lines have
the same slope. The
y-intercepts differ so the
lines are parallel. Because
the lines are parallel, there
is NO point of intersection.
Chabot College Mathematics
45
y = 3x/4  3
• Thus the system is
INCONSISTENT and
has NO solution.
Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Solve System by Sub
3
 Solve
y  x2
System
4
1
3
y  x4
4
 SOLUTION: This system is from The
previous Graphing example. The lines
are parallel and the system has no
solution. Let’s see what happens if we
try to solve this system by substitution
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46
Bruce Mayer, PE
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2
Example  Solve System by Sub
3
 Solve by y  3 x  2 1
y  x4
Algebra
4
4
3
y  x  2 Equation (1)
4
3
 3
 x  4  x  2 Substitute: y = (3/4)x–4
4
 4
 4  2 Subtract (3/4)x from Both Sides
2
 We arrive at contradiction; the system
is inconsistent and thus has NO solution
Chabot College Mathematics
47
Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Solve By Graphing
 Solve System:
4 x  8 y  16
2x  4 y  8
 SOLUTION: graph
Both Equations
• Both equations
represent the
SAME line.
• Because the
equations are
EQUIVALENT, any
solution of ONE
Chabot College Mathematics
48
equation is a solution of
the OTHER equation
as well.
Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Solve By Subbing
1
2
4 x  8 y  16
 Solve
System 2 x  4 y  8
 SOLUTION: Notice that Eqn-1 is simply
TWICE Eqn-2. Thus these Eqns are
DEPENDENT, and will Graph as
COINCIDENT Lines
• Recall that Dependent Equations have an
INFINITE number of Solutions
 Checking by Algebra
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49
Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Solve By Subbing
 Solve by Algebra 4 x  8 y  16 1
2 x  4 y  8 2
4 x  8 y  16
Equation (1)
 1 
4 x  82  x   16 Use Eqn-2 to Substitute for y
 2 
4 x  16  4 x  16 Simplifying
4 x  16  4 x  16
ReArranging
4 x  4 x  16  16  0  0 ??? Final ReArrangement
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50
Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Solve By Subbing
4 x  8 y  16 1
 Examine
Soln to
2 x  4 y  8 2
4 x  16  4 x  16
 This Solution equation is true for any
choice of x. When the solution leads to
an equation that is true for all real
numbers, we state that the system has
an infinite number of solutions.
• Simplification to 0=0 → infinite solns
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Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Elim/Addition Soln
 Example – Solve:
2 x  y  1 1
2 x  y  3 2
 SOLUTION: To
eliminate y multiply
equation (2) by −1.
Then add
2x  y  1
 2x  y  3
0  0  4 ?¿
Chabot College Mathematics
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 Note that in
eliminating y, we
eliminated x as well.
The resulting
equation 0 = 4,
is false (a
contradiction) for
any pair (x, y), so
there is no solution
• The Lines are
thus PARALLEL
Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Elim/Addition Soln
 Example – Solve:
 Again, we have
3x  4 y  5 1
eliminated both
variables. The
9 x  12 y  15 2
resulting equation,
 SOLUTION: To
0 = 0, is always true,
eliminate x, we
indicating that the
multiply both sides of
equations are
eqn (1) by −3, and
then add the two eqns dependent
 9 x  12 y  15
9 x  12 y  15
00  0
?¿
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Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
WhiteBoard Work
 Problems From §3.1 Exercise Set
• 92, 94, 96
y

InConsistent
Equations
3
y x4
5
3
y x
5
Chabot College Mathematics
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x
Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
All Done for Today
Women
In The
Professions
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Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Chabot Mathematics
Appendix
r  s  r  s r  s 
2
2
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
–
Chabot College Mathematics
56
Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt