Linear and Quadratic Equation System

8/3/2019 Linear and Quadratic Equation System

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LINEAR AND QUADRATICEQUATION SYSTEMS

The First Semester of 10 Level

SMA 1 SLAWI


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STANDARD COMPETENCE :

3.1 Solving linear equation systems andlinear-quadratic equation systems in twovariable

3.2 Compose mathematics model from the

problems which have relation with linearequation systems

3.3 Solving mathematics model from theproblems which have relation with linearequation systems and its interpretation


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This chapter is about finding the solutionof linear and quadratic equation systems.

When you have completed it, you should be able to

*Find the solution set of linear equation systems in two variables*Find the solution set of linear equation systems in threevariables

*Find the solution set of simultaneous equations, one linear-onequadratic in two variable

*Find the solution set of simultaneous equations, two quadraticsin two variable

*Identify the problems which have relation with linear equationsystems

*Compose mathematics model from the problems which haverelation with linear equation systems

*Solve mathematics model from the problems whichhave relation with linear equation systems

*Interpret result the solution of problems which have

relation with linear equation systems


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A. SYSTEM OF LINEAR EQUATIONS IN TWO VARIABLES

The general form of linear equation systems intwo variables is :

,,

222

111

cybxa

cybxa

22 ,ba

212121,,,,, ccbbaa

Solution set of linear equation systems in two variablescan be found by :

1. Graphs2. Elimination3. Substitution4. Combination of elimination and substitution

11,ba 0

0

Are real number


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1. Solution by Graphs

To solve a system of linear equations in variable x and y by graphs,we must draw the two equations on the same coordinate system.Then find the point of their intersections. This point is calledthe solution of linear equation systems in two variablesExamples :

1. By graph, find the solution set of 2x + y = 4x y = - 1

Solution :Step 1. Draw the two equations on the same coordinate system

2x + y = 4 x y = - 1

x 0 2 x 0 -1

y 4 0 y 1 0


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The graph :

4

22x + y = 4

1

-1

x y = -1

1

2

Y

X

A(1,2)

Step 2. The point of intersections is A(1,2)

Step 3. The solution set is {(1,2)}

Conclusion :

If2

1

2

1

2

1

c

c

b

b

a

a , then system of linear equations in two variables

has only one solution and the graphs intersect inone and only one point


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2. By the graph, find the solution set of x + 2y = 4

2x + 4y = 12

Solution :

Step 1 : x + 2y = 4 2x + 4y = 12

x 0 4 x 0 6

y 2 0 y 3 0

The graph :Y

X

2

4

3

6

2x + 4y = 12

x + 2y = 4


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Conclusion :

If2

1

2

1

2

1

c

c

b

b

a

a

, the system of linear equations in twovariables has no solution. And the graphsare distinct parallel lines

3. By the graph, find the solution set of x y = 2

2x 2y = 4

Solution :

Step 1. x y = 2 2x 2y = 4

x 0 2 x 0 2

y -2 0 y -2 0

The graph :Y

X

-2

2

x - y = 0 2x 2y = 4


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Conclusion :

If2

1

2

1

2

1

c

c

b

b

a

a , then system of linear equations in two

variables has infinitely many solutions and

the graphs are in the same line

2. Solution by Elimination

There are two steps :

a. eliminate one variableb. eliminate other variable

Example :

By elimination, find the solution set of 2x y = 4

3x + 2y = 13

Solution :

2x y = 4..(1)

3x + 2y = 13(2)


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Elimination of y :

4x 2y = 8.(3)(multiply (1) by 2)3x + 2y =13

+

7x = 21

x = 3

(Addition of (3) and (2) so that eliminate thevariable y)

Elimination of x :

6x 3y = 12.(4)(multiply (1) by 3)

6x + 4y = 26.(5)(multiply (2) by 2)

- 7y = - 14

y = 2

(Subtraction (4) and (5) so that eliminatethe variable x)

The solution set is {(3,2)}


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3. Solution by Substitution

There are two steps :

a. Take one equation and express one variable with other variable

b. Substitute to other equation

Example :

By substitution, find the solution set of x + y = 4

4x + 3y = 13

Solution :

x + y = 4..(1) y = 4 x..(3)(express y with x)

4x + 3y = 13(2)

Substitute (3) to (2)

4x + 3y = 134x + 3(4 x) = 134x + 12 3x = 13

x = 1

Substitute x = 1 to (3)

y = 4 - x

= 4 - 1= 3



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4. Solution by combination of elimination and substitution

Example : By combination of elimination and substitution, find

the solution set of :

4x + 3y = 10

2x + y = 4

Solution :

4x + 3y = 10 x1 4x + 3y = 10

2x + y = 4 x3 6x + 3y = 12

-2x = -2

x = 1Substitute x = 1 to equation which is easy, for example to2x + y = 4 2.1 + y = 4

y = 2



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B. SYSTEM OF LINEAR EQUATIONS IN THREE VARIABLES

The general form is :

33333333333

22222222222

11111111111

,,;,,,;

,,;,,,;

,,;,,,;

cbaRdcbadzcybxa

cbaRdcbadzcybxa

cbaRdcbadzcybxa

not triplet 0

not triplet 0

not triplet 0

Solution set of linear equation systems in three variables can befound by :

a. Substitutionb. Combination of elimination and substitution


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1. Solution by substitution

Example :

Find the solution set of

2x + y + 3z = 7x + 2y + z = 1

3x y + 2z = 8

Solution :

2x + y + 3z = 7(1)x + 2y + z = 1.(2)3x y + 2z = 8(3)

Equation (1) is made become y = 7 2x 3z(4)

Substitute equation (4) to (2) so :x + 2(7 2x 3z) + z = 1x + 14 4x 6z + z = 1-3x 5z = - 13

3x + 5z = 13(5)Substitute (4) to (3) so :3x (7 2x 3z) + 2z = 83x 7 + 2x + 3z + 2z = 85x + 5z = 15z = 3 x.(6)


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Substitute (6) to (5) so :3x + 5(3 x) = 133x + 15 5x = 13-2x = - 2

x = 2

Substitute x = 2 to (6) so z = 3 2 = 1

Substitute x = 2, z = 1 to (4) so y = 7 2.2 3.1= 7 4 3= 0

Solution set is {(2,0,1)}

2. Solution set by combination of substitution and elimination

Example :

Find solution set of2x + y z = 5x + 2y + z = 13x 3y + 2z = -4


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Solution :

2x + y z = 5(1)x + 2y + z = 13(2)x 3y + 2z = -2..(3)

Elimination z from (1) and (2)2x + y z = 5x + 2y + z = 13

x y = -8..(4)

Elimination z from (1) and (3)

2x + y z = 5 x2 4x + 2y 2z = 10x 3y + 2z = -2 x1 x 3y + 2z = -2

+

5x y = 8.(5)

Elimination y from (4) and (5)x y = -8

5x y = 8

-4x = -16x = 4

Substitute x = 4 to (4)4 y = -8

y = 12

Substitute x = 4 and y = 12 to (1)2x + y z = 52.4 + 12 z = 5

z = 25

The solution set is {(4,12,25)}

C


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C. SIMULTANEOUS EQUATIONS, ONE LINEAR-ONE QUADRATIC,AND SIMULTANEOUS EQUATIONS TWO QUADRATICS1. Solution set by substitution

Example :

a) Find the solution set ofy = x + 3y = x2 + 1

Solution :

y = x + 3.(1)y = x2+ 1(2)

Substitute (1) to (2) :x2 + 1 = x + 3x2 x 2 = 0

(x 2)(x + 1) = 0x = 2 or x = -1

For x = 2, then y = 5 and the intercept point is (2,5)For x = -1, then y = 2 and the intercept point is (-1,2)

Solution set is {(2,5);(-1,2)}


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b. Find the solution set ofy= 2x2 3x + 1y = x2 + x + 6

Solution :

2x2

3x + 1 = x2

+ x + 6x2 4x 5 = 0(x + 1)(x 5) = 0x = -1 or x = 5

For x = -1, then y = 6For x = 5, then y = 36

Solution set is : {(-1,6);(5,36)}

2. Solution by factoring of the quadratic form

Example :

Find the solution set ofy x = 3x2 + 2xy + y2 25 = 0


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Solution :

y x = 3..(1)x2 + 2xy + y225 = 0 (2)

Factorize the equation (2)x2 + 2xy + y2 25 = 0(x + y + 5)(x + y 5) = 0x + y + 5 = 0 or x + y 5 = 0

For x + y + 5 = 0 then y = -x 5

Substitute y = -x 5 to (1) :(-x 5) x = 3

- 2x = 8x = -4 and y = -1, so the intercept point is (-4,-1)

For x + y 5 = 0 then y = 5 xSubstitute y = 5 x to (1) :

(5 x) x = 3- 2x = -2

x = 1 and y = 4, so the intercept point is (1,4)

Solution set is {(-4,-1);(1,4)}


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SO THAT YOU CAPABLE TO SOLVING LINEAR QUADRATIC

EQUATION SYSTEMS AND COMPOSE MATHEMATICS MODEL FROMPROBLEMS WHICH HAVE RELATION WITH LINEAR EQUATION

SYSTEMS, SO TO SOLVING IT

NEXT

PLEASE YOU DO THE EXERCISES IN YOUR MATHEMATICS BOOK

THANK YOU FOR YOUR ATTENTION

AFGRIZ PRASETIYAWATI, S.Pd Guru SMA Negeri 1 SLAWI

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Linear and Quadratic Equation System