[email protected] mth55_lec-10_sec_3-1_2var_linsys_ppt 1 bruce mayer, pe chabot college...

[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt1

Bruce Mayer, PE Chabot College Mathematics

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

§3.1 2-Var§3.1 2-VarLinear Linear

SystemsSystems



Review §Review §

Any QUESTIONS About• §’s2.4 → Point-Slope Eqn, Modeling

Any QUESTIONS About HomeWork• §’s2.4 → HW-07

2.4 MTH 55



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



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.



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)




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





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





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



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



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



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



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



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



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



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.



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



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




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




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




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.



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




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



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



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



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



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



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.



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



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



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



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



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



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.


216103

1152

yx

yx



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




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.



Multiple MultiplicationMultiple Multiplication

Sometimes BOTH equations must be multiplied to find the Least Common Multiple (LCM) of two coefficients


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




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




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



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



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



Case-1 Consistent SystemsCase-1 Consistent Systems

In this case, the graphs of the two lines intersect at exactly one point.



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



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.



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



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



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



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



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



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




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




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



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



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



WhiteBoard WorkWhiteBoard Work

Problems From §3.1 Exercise Set• 92, 94, 96

InConsistentEquations

y x

y x

3

54

3

5

x

y



All Done for TodayAll Done for Today

WomenIn The

Professions



Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

AppendiAppendixx

–

srsrsr 22

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