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BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt1
Bruce Mayer, PE Chabot College Mathematics
Bruce Mayer, PELicensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Chabot Mathematics
§3.1 2-Var§3.1 2-VarLinear Linear
SystemsSystems
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt2
Bruce Mayer, PE Chabot College Mathematics
Review §Review §
Any QUESTIONS About• §’s2.4 → Point-Slope Eqn, Modeling
Any QUESTIONS About HomeWork• §’s2.4 → HW-07
2.4 MTH 55
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt3
Bruce Mayer, PE Chabot College Mathematics
Systems of EquationsSystems of Equations
System of Equations ≡ A group of two or more equations; e.g.,
5 (Equation 1)
3 4 8 (Equation 2)
x y
x y
Solution For A System Of Equations ≡ An ordered set of numbers that makes ALL equations in the system TRUE at the same time
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt4
Bruce Mayer, PE Chabot College Mathematics
Checking System SolutionChecking 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.
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt5
Bruce Mayer, PE Chabot College Mathematics
Example Example Chk System Soln Chk System Soln
Consider TheEquation System
7 (Equation 1)
3 2 (Equation 2)
x y
y x
Determine whether each ordered pair is a solution to the system of equations.
a. (−3, 2) b. (3, 4)
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt6
Bruce Mayer, PE Chabot College Mathematics
Example Example Chk System Soln Chk System Soln
SOLUTION →Chk True/False
7 (Equation 1)
3 2 (Equation 2)
x y
y x
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
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt7
Bruce Mayer, PE Chabot College Mathematics
Example Example Chk System Soln Chk System Soln
SOLUTION →Chk True/False
7 (Equation 1)
3 2 (Equation 2)
x y
y x
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
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt8
Bruce Mayer, PE Chabot College Mathematics
Example Example Chk System Soln Chk System Soln
SOLUTION →Chk True/False
7 (Equation 1)
3 2 (Equation 2)
x y
y x
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
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt9
Bruce Mayer, PE Chabot College Mathematics
Systems of Equations SolnSystems of Equations Soln A system-of-
equations 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
This Graph shows two lines which have one point in common
5xy1 xy
The common point is (–3,2) Satisfies BOTH Eqns
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt10
Bruce Mayer, PE Chabot College Mathematics
Solve Systems of Eqns by GraphingSolve 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
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt11
Bruce Mayer, PE Chabot College Mathematics
Solving by Graphing ProcedureSolving 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
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt12
Bruce Mayer, PE Chabot College Mathematics
Example Example Solve w/ Graphing 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) (−1, −2) The crossing
point provides the common solution
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt13
Bruce Mayer, PE Chabot College Mathematics
Example Example Solve w/ Graphing Solve w/ Graphing Chk (−1, −2) Soln:
• y = 3x + 1
• x − 2y = 3
21 ,
y = 3x + 1→• −2 = 3(−1) + 1
• −2 = −3 + 1
• −2 = −2
x − 2y = 3 →• (−1) − 2(−2) = 3
• −1+4 = 3
• 3 = 3
Thus (−1, −2) Chksas a Soln
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt14
Bruce Mayer, PE Chabot College Mathematics
Example Example Solve By Graphing Solve By Graphing Solve System:
SOLUTION: graph Both Equations
• As a check note that [4−2] = [6−4] is true.– The solution is (4, 2)
• 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).
xy
xy
6
2
(4, 2)
y = x 2
y = 6 x
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt15
Bruce Mayer, PE Chabot College Mathematics
The Substitution Soln MethodThe 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.
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt16
Bruce Mayer, PE Chabot College Mathematics
Substitution SummarizedSubstitution 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
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt17
Bruce Mayer, PE Chabot College Mathematics
Example Example Solve by Subbing Solve by Subbing
Solve theSystem
252
1423
xy
yx
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
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt18
Bruce Mayer, PE Chabot College Mathematics
Example Example Solve by Subbing Solve by Subbing
The Algebrato Solve
252
1423
xy
yx
423 yx Equation (1)
45223 xx Substitute: y = −2x + 5
41043 xx Distributive Property
4107 x Combine Like Terms
147 x Add 10 to Both Sides
2x Divide Both Sides by 7 to Find x
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt19
Bruce Mayer, PE Chabot College Mathematics
Example Example Solve by Subbing 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):
52 xy Equation (2)
522 y Substitute: x = 2
54 y Simplifying
1y When x = 2
The ordered pair (2, 1) appears to be the solution
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt20
Bruce Mayer, PE Chabot College Mathematics
Example Example Solve by Subbing Solve by Subbing
Check Tentative Solution (2,1) 3x − 2y = 4 y = −2x + 5
3(2) − 2(1) 4 1 −2(2) + 5
6 − 2 4 1 −4 + 5
4 = 4 True 1 = 1 True
Since (2, 1) checks in BOTH equations, it IS a solution.
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt21
Bruce Mayer, PE Chabot College Mathematics
Substitution Solution CAUTIONSubstitution Solution CAUTION
CautionCaution! 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.
2252
1423
xxy
yx
1,2,252
1423
yxxy
yx
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt22
Bruce Mayer, PE Chabot College Mathematics
Example Example Solve by Subbing Solve by Subbing
Solve theSystem
2535
13
yx
yx
SOLUTION: Sub 3 − y for x in Eqn (2)535 yx Equation (2)
5335 yy Substitute: x = 3−y
53515 yy Distributive Property
0210 y Combine Terms, Subtract 5 from Both sides
5102 yy Solve for y = 5
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt23
Bruce Mayer, PE Chabot College Mathematics
Solve by SubbingSolve by Subbing Find x for y = 5
• Use Eqn (1) yx 3 Eqn (1)
53 x Sub y = 5
2x Solve for x
Chk Soln pair (−2,5)
x = 3 − y 5x + 3y = 5
−2 = 3 − 5 5(−2)+3(5) = 5
−2 = −2 −10 + 15 = 5
5 = 5
Thus (−2,5) is the Soln• The graph below is
another check.
(−2, 5)
x = 3 y
5x + 3y = 5
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt24
Bruce Mayer, PE Chabot College Mathematics
Solving for the Variable FirstSolving for the Variable First Sometimes neither
equation has a 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:
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.
2825
16
yx
yx
36
16
yx
yx
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt25
Bruce Mayer, PE Chabot College Mathematics
Example Example Solve Solve substitute x = 6−y for x in equation (2)
of the original pair and solve for y
2825
16
yx
yx
825 yx Equation (2)
8265 yy Substitute: x = 6−y
82530 yy Distributive Property
8330 y Combine Like Terms
y322 Add −8, Add 3y to Both Sides
322y Divide Both Sides by 3 to Find y
Use Parens or Brackets when Subbing
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt26
Bruce Mayer, PE Chabot College Mathematics
Example Example Solve Solve 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):
2825
16
yx
yx
3
4
3
22
3
18
3
2266
yx
A Check of this Ordered Pair Shows that it is a Solution:
3
22,
3
4
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt27
Bruce Mayer, PE Chabot College Mathematics
Substitution Solution ProcedureSubstitution Solution Procedure1. 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.
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt28
Bruce Mayer, PE Chabot College Mathematics
Solving by Addition/EliminationSolving by Addition/Elimination
The Addition/Elimination method for solving systems of equations makes use of the addition principle
For Example;Solve System
273
1834
yx
yx
According to equation (2), x−3y and 7 are the samesame thingthing. Thus we can add 4x + 3y to the left side of the equation(1) and 7 to the right side
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt29
Bruce Mayer, PE Chabot College Mathematics
Elimination ExampleElimination Example
AddEquations
2
1
1505
73
834
yx
yx
yx
The resulting equation has just one variable: 5x = 15 • Dividing both sides by 5, find that x = 3
Next Sub 3 for x inEither Eqn to Findthe y-value• Using Eqn-1
8334 y
8312 y43 y34y
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt30
Bruce Mayer, PE Chabot College Mathematics
Elimination ExampleElimination Example
Check Tentative Solution (3, −4/3)
• Since (3, −4/3) checks in both equations, it is the solution– Graph confirms
4x + 3y = 8 x − 3y = 7 4(3) + 3(−4/3) 8 3 − 3(−4/3) 7 12 − 4 3 + 4 8 = 8 7 = 7
True True
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt31
Bruce Mayer, PE Chabot College Mathematics
Example Example Solve Solve 21332
11734
yx
yx
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 of equation (2) by −1 to find an equivalent eqn and then add:
2
1
4 2
1332
17 34
x
yx
yx
2x
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt32
Bruce Mayer, PE Chabot College Mathematics
Example Example Solve Solve 21332
11734
yx
yx
To Find y substitute 2 for x in either of the original eqns:
The graph shown below also checks.
11734 yx
17324 y
1738 y393 yy
We can check the ordered pair (2, 3) in Both Eqns
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt33
Bruce Mayer, PE Chabot College Mathematics
Which Variable to EliminateWhich 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;Solve System
216103
1152
yx
yx
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt34
Bruce Mayer, PE Chabot College Mathematics
Example Example Solve Solve
SOLUTION: No terms are opposites, but if both sides of equation (1) are multiplied by 2, the coefficients of y will be opposites.
2
1
41 7
16 103
2104
x
yx
yx
216103
1152
yx
yx
Mult Both Sides of Eqn-1 by 2
Add Eqns
2x Solve for x
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt35
Bruce Mayer, PE Chabot College Mathematics
Example Example Solve Solve
SOLUTION: Find y by Subbing x=2 into Either Eqn; Choosing Eqn-1
216103
1152
yx
yx
Sub 2 for x
Simplify
1522 y
Solve for y
154 y
155 yy
Student Exercise: confirm that (2, −1) checks and is the solution.
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt36
Bruce Mayer, PE Chabot College Mathematics
Multiple MultiplicationMultiple Multiplication
Sometimes BOTH equations must be multiplied to find the Least Common Multiple (LCM) of two coefficients
For Example;Solve System
2453
1232
yx
yx
SOLUTION: It is often helpful to write both equations in Standard form before attempting to eliminate a variable:
4453
3232
yx
yx
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt37
Bruce Mayer, PE Chabot College Mathematics
Example Example Solve Solve
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
4453
3232
yx
yx
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt38
Bruce Mayer, PE Chabot College Mathematics
Example Example Solve Solve
The Algebra
Solve for x using y = −2 in Eqn-3
4453
3232
yx
yx
6
5
2
8 106
6 9 6
y
yx
yx Mult Both Sides of Eqn-3 by 3
Add Eqns
Mult Both Sides of Eqn-4 by −2
2232 x262 x
42 x2x
Students to Verify that solution (−2, −2) checks
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt39
Bruce Mayer, PE Chabot College Mathematics
Solving Systems by EliminationSolving Systems by Elimination1. Write the equations in standard form (Ax + By = C).2. Use the multiplication principle to clear fractions or
decimals.3. Multiply one or both equations by a number (or numbers)
so that they have a pair of terms that are additive inverses.
4. Add the equations. The result should be an equation in terms of one variable.
5. Solve the equation from step 4 for the value of that variable.
6. 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.
7. Check your solution in the original equations
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt40
Bruce Mayer, PE Chabot College Mathematics
Types of Systems of EquationsTypes 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
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt41
Bruce Mayer, PE Chabot College Mathematics
Case-1 Consistent SystemsCase-1 Consistent Systems
In this case, the graphs of the two lines intersect at exactly one point.
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt42
Bruce Mayer, PE Chabot College Mathematics
Case-2 Inconsistent SystemsCase-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
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt43
Bruce Mayer, PE Chabot College Mathematics
Case-3 Dependent SystemsCase-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.
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt44
Bruce Mayer, PE Chabot College Mathematics
Testing for System CaseTesting for System Case
Given System of Two Eqns
222111 & bxmybxmy 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
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt45
Bruce Mayer, PE Chabot College Mathematics
Example Example Solve by Graphing Solve by Graphing Solve System:
• Thus the system is INCONSISTENT and has NO solution.
34
3&2
4
3 xyxy
SOLUTION: Graph Eqns• equations are in slope-
intercept 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.
y = 3x/4 + 2
y = 3x/4 3
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt46
Bruce Mayer, PE Chabot College Mathematics
Example Example Solve System by Sub Solve System by Sub
SolveSystem
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
244
312
4
3 xyxy
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt47
Bruce Mayer, PE Chabot College Mathematics
Example Example Solve System by Sub Solve System by Sub
Solve by Algebra
24
3 xy Equation (1)
24
34
4
3
xx Substitute: y = (3/4)x–4
24 Subtract (3/4)x from Both Sides
We arrive at contradiction; the system is inconsistent and thus has NO solution
244
312
4
3 xyxy
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt48
Bruce Mayer, PE Chabot College Mathematics
Example Example Solve By Graphing Solve By Graphing Solve System:
842
1684
yx
yx
SOLUTION: graph Both Equations
equation is a solution of the OTHER equation as well.
• Both equations represent the SAME line.
• Because the equations are EQUIVALENT, any solution of ONE
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt49
Bruce Mayer, PE Chabot College Mathematics
Example Example Solve By Subbing Solve By Subbing
SolveSystem
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
2842
11684
yx
yx
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt50
Bruce Mayer, PE Chabot College Mathematics
Example Example Solve By Subbing Solve By Subbing
Solve by Algebra
1684 yx Equation (1)
162
1284
xx Use Eqn-2 to Substitute for y
164164 xx Simplifying
2842
11684
yx
yx
164164 xx ReArranging
???00161644 xx Final ReArrangement
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt51
Bruce Mayer, PE Chabot College Mathematics
Example Example Solve By Subbing Solve By Subbing
ExamineSoln to
2842
11684
yx
yx
164164 xx 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
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt52
Bruce Mayer, PE Chabot College Mathematics
Example Example Elim/Addition Soln Elim/Addition Soln Example – Solve:
232
112
yx
yx
SOLUTION: To eliminate y multiply equation (2) by −1. Then add
400
32
1 2
yx
yx
?¿
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
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt53
Bruce Mayer, PE Chabot College Mathematics
Example Example Elim/Addition Soln Elim/Addition Soln Example – Solve:
215129
1543
yx
yx
SOLUTION: To eliminate x, we multiply both sides of eqn (1) by −3, and then add the two eqns
000
15129
1512 9
yx
yx
?¿
Again, we have eliminated both variables. The resulting equation, 0 = 0, is always true, indicating that the equations are dependentdependent
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt54
Bruce Mayer, PE Chabot College Mathematics
WhiteBoard WorkWhiteBoard Work
Problems From §3.1 Exercise Set• 92, 94, 96
InConsistentEquations
y x
y x
3
54
3
5
x
y
BMayer@ChabotCollege.edu • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt55
Bruce Mayer, PE Chabot College Mathematics
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