M12 Systems of Equations.pdf

(1)

Systems of Equations

At the end of this lecture, a student must be able to:

Recognize a system of equations

Interpret solutions of systems as points of intersection

Algebraically solve for systems involving two linear equations in two variables

Solve a system involving three linear equations in three variables

(2)

System of Equations

Definition

•

A

system of equations

is a set of two or

more equations in several variables.

•

A

solution

of a system is a solution

common to all equations in the system.

•

The

solution set

of a system is the set of

(3)

Two Linear Equations in Two Variables

Definition

A

system of two linear equations in two

variables

has the form

a

1

x

+

b

1

y

=

c

1

a

2

x

+

b

2

y

=

c

2

(4)

Graphical Interpretation of Linear Systems

Recall: A solution to an equation in two variables is a point on the graph of the equation.

For systems of two linear equations in two variables, each equation represents a line.

⇒ A solution to each equation is a point on a line.

⇒ A solution to the system is a point on both lines.

⇒ A solution to the system gives a point of intersection of the lines.

Example: Solve the system

3x−4y = 1 2x+ 3y = 12

m

Solve for the points of intersection

of the lines with equations

3x−4y−1 = 0 and

(5)

Graphical Interpretation of Linear Systems

For systems of two linear equations in two variables,

each equation represents a line.

3x−4y = 1 2x+ 3y = 12

m

3x−4y−1 = 0 and

(6)

Graphical Interpretation of Linear Systems

3x−4y = 1 2x+ 3y = 12

m

3x−4y−1 = 0 and

(7)

Graphical Interpretation of Linear Systems

3x−4y = 1 2x+ 3y = 12

m

3x−4y−1 = 0 and

(8)

Graphical Interpretation of Linear Systems

3x−4y = 1 2x+ 3y = 12

m

3x−4y−1 = 0 and

(9)

Graphical Interpretation of Linear Systems

3x−4y = 1 2x+ 3y = 12

m

3x−4y−1 = 0 and

(10)

Algebraic Methods in Solving Linear Systems

Example: Solve for the points of intersection of the

lines with equations

3

x

−

4

y

−

1 = 0

and

2

x

+ 3

y

−

12 = 0

.

Two ways to solve such systems algebraically:

Elimination by Substitution Isolate one variable in one equation, and replace the variable in the other equation.

(11)

Algebraic Methods in Solving Linear Systems

Example: Solve for the points of intersection of the

lines with equations

3

x

−

4

y

−

1 = 0

and

2

x

+ 3

y

−

12 = 0

.

(12)

Algebraic Methods in Solving Linear Systems

Example: Solve for the points of intersection of the

lines with equations

3

x

−

4

y

−

1 = 0

and

2

x

+ 3

y

−

12 = 0

.

(13)

Two Linear Equations

Example:

Solve for the points of intersection of the lines with

equations

3

x

−

4

y

−

1 = 0

and

2

x

+ 3

y

−

12 = 0

.

Solution:

To eliminate x by addition,

`1 :

2·(

3x−4y

)

=

(

1

) ·2 (Make coefficients of

`2 :

(−3)·(

2x+ 3y

)

=

(

12

) ·(−3) xadditive inverses)

6x−8y = 2 (Add LHS and RHS)

−6x−9y = −36

⇒ −17y = −34 (Solve for y)

(14)

Two Linear Equations

Example:

equations

3

x

−

4

y

−

1 = 0

and

2

x

+ 3

y

−

12 = 0

.

Solution: To eliminate x by addition,

`1 :

2·(

3x−4y

)

=

(

1

`2 :

(−3)·(

2x+ 3y

)

=

(

12

−6x−9y = −36

⇒ −17y = −34 (Solve for y)

(15)

Example:

equations

3

x

−

4

y

−

1 = 0

and

2

x

+ 3

y

−

12 = 0

.

`1 :

2·(

3x−4y

)

=

(

1

`2 :

(−3)·(

2x+ 3y

)

=

(

12

−6x−9y = −36

⇒ −17y = −34 (Solve for y)

(16)

Two Linear Equations

Example:

equations

3

x

−

4

y

−

1 = 0

and

2

x

+ 3

y

−

12 = 0

.

`1 :

2·(

3x−4y

)

=

(

1

) ·2

(Make coefficients of

`2 :

(−3)·(

2x+ 3y

)

=

(

12

) ·(−3)

xadditive inverses)

−6x−9y = −36

⇒ −17y = −34 (Solve for y)

(17)

Example:

equations

3

x

−

4

y

−

1 = 0

and

2

x

+ 3

y

−

12 = 0

.

`1 : 2·(3x−4y) = (1) ·2 (Make coefficients of `2 : (−3)·(2x+ 3y) = (12) ·(−3) xadditive inverses)

−6x−9y = −36

⇒ −17y = −34 (Solve for y)

(18)

Two Linear Equations

Example:

equations

3

x

−

4

y

−

1 = 0

and

2

x

+ 3

y

−

12 = 0

.

6x−8y = 2

(Add LHS and RHS)

−6x−9y = −36

⇒ −17y = −34 (Solve for y)

(19)

Example:

equations

3

x

−

4

y

−

1 = 0

and

2

x

+ 3

y

−

12 = 0

.

6x−8y = 2

(Add LHS and RHS)

−6x−9y = −36

⇒ −17y = −34 (Solve for y)

(20)

Two Linear Equations

Example:

equations

3

x

−

4

y

−

1 = 0

and

2

x

+ 3

y

−

12 = 0

.

−6x−9y = −36

⇒ −17y = −34 (Solve for y)

(21)

Example:

equations

3

x

−

4

y

−

1 = 0

and

2

x

+ 3

y

−

12 = 0

.

−6x−9y = −36

⇒ −17y = −34

(Solve for y)

(22)

Two Linear Equations

Example:

equations

3

x

−

4

y

−

1 = 0

and

2

x

+ 3

y

−

12 = 0

.

−6x−9y = −36

⇒ −17y = −34 (Solve for y)

(23)

Example:

equations

3

x

−

4

y

−

1 = 0

and

2

x

+ 3

y

−

12 = 0

.

−6x−9y = −36

⇒ −17y = −34 (Solve for y)

(24)

Two Linear Equations

Example:

equations

3

x

−

4

y

−

1 = 0

and

2

x

+ 3

y

−

12 = 0

.

Solution (cont.): If y= 2,

then using either`1 or `2, solve for x `1

`2

3x−4(2)−1 = 0

2x+ 3(2)−12 = 0 3x−8−1 = 0 2x+ 6−12 = 0

3x = 9 2x = 6

x = 3 x = 3

(25)

Example:

equations

3

x

−

4

y

−

1 = 0

and

2

x

+ 3

y

−

12 = 0

.

Solution (cont.):

If y= 2, then using either `1 or `2, solve for x

`1

`2

3x−4(2)−1 = 0

2x+ 3(2)−12 = 0 3x−8−1 = 0 2x+ 6−12 = 0

3x = 9 2x = 6

x = 3 x = 3

(26)

Two Linear Equations

Example:

equations

3

x

−

4

y

−

1 = 0

and

2

x

+ 3

y

−

12 = 0

.

Solution (cont.):

If y= 2, then using either `1 or `2, solve for x `1

`2

3x−4(2)−1 = 0

2x+ 3(2)−12 = 0 3x−8−1 = 0 2x+ 6−12 = 0

3x = 9 2x = 6

x = 3 x = 3

(27)

Example:

equations

3

x

−

4

y

−

1 = 0

and

2

x

+ 3

y

−

12 = 0

.

Solution (cont.):

`2

3x−4(2)−1 = 0

2x+ 3(2)−12 = 0

3x−8−1 = 0

2x+ 6−12 = 0

3x = 9 2x = 6

x = 3 x = 3

(28)

Two Linear Equations

Example:

equations

3

x

−

4

y

−

1 = 0

and

2

x

+ 3

y

−

12 = 0

.

Solution (cont.):

`2

3x−4(2)−1 = 0

2x+ 3(2)−12 = 0

3x−8−1 = 0

2x+ 6−12 = 0

3x = 9

2x = 6

x = 3 x = 3

(29)

Example:

equations

3

x

−

4

y

−

1 = 0

and

2

x

+ 3

y

−

12 = 0

.

Solution (cont.):

`2

3x−4(2)−1 = 0

2x+ 3(2)−12 = 0

3x−8−1 = 0

2x+ 6−12 = 0

3x = 9

2x = 6

x = 3

(30)

Two Linear Equations

Example:

equations

3

x

−

4

y

−

1 = 0

and

2

x

+ 3

y

−

12 = 0

.

Solution (cont.):

If y= 2, then using either `1 or `2, solve for x `1 `2

3x−4(2)−1 = 0 2x+ 3(2)−12 = 0 3x−8−1 = 0 2x+ 6−12 = 0

3x = 9 2x = 6

x = 3 x = 3

(31)

Example:

equations

3

x

−

4

y

−

1 = 0

and

2

x

+ 3

y

−

12 = 0

.

Solution (cont.):

If y= 2, then using either `1 or `2, solve for x `1 `2

3x−4(2)−1 = 0 2x+ 3(2)−12 = 0 3x−8−1 = 0 2x+ 6−12 = 0

3x = 9 2x = 6

x = 3 x = 3

(32)

Two Linear Equations

Example:

Solve for the points of intersection of the lines with equations

3x−4y−1 = 0 and 2x+ 3y−12 = 0.

−1 1 2 3

−1 1 2 3 4

0

3x-4y=1

2x+3y=12

(33)

Example: Solve for the intersection of the lines

`

1

:

y

=

3

4

x

−

1

4

and

`

2

:

y

=

3

2

x

+

3

4

Solution:

By Substitution Method,

4·

3 4x−

1 4 = 3 2x+

3 4

·4 (Isolate y in `1

and replacey in `2) ⇒ 3x−1 = 6x+ 3 (Solve forx)

⇒ −4 = 3x

⇒ x = −4

(34)

`

1

:

y

=

3

4

x

−

1

4

and

`

2

:

y

=

3

2

x

+

3

4

Solution: By Substitution Method,

4·

3 4x−

1 4 = 3 2x+

3 4

·4 (Isolate y in `1

⇒ −4 = 3x

⇒ x = −4

(35)

Example: Solve for the intersection of the lines

`

1

:

y

=

3

4

x

−

1

4

and

`

2

:

y

=

3

2

x

+

3

4

4·

3 4x−

1 4 = 3 2x+

3 4

·4

(Isolate y in`1

and replacey in `2)

⇒ 3x−1 = 6x+ 3 (Solve forx)

⇒ −4 = 3x

⇒ x = −4

(36)

`

1

:

y

=

3

4

x

−

1

4

and

`

2

:

y

=

3

2

x

+

3

4

4·

3 4x−

1 4 = 3 2x+

3 4

·4

(Isolate y in`1

and replacey in `2)

⇒ −4 = 3x

⇒ x = −4

(37)

`

1

:

y

=

3

4

x

−

1

4

and

`

2

:

y

=

3

2

x

+

3

4

4·

3 4x−

1 4 = 3 2x+

3 4

·4

(Isolate y in`1

and replacey in `2)

⇒ −4 = 3x

⇒ x = −4

(38)

Example: Solve for the intersection of the lines

`

1

:

y

=

3

4

x

−

1

4

and

`

2

:

y

=

3

2

x

+

3

4

4·

3 4x−

1 4 = 3 2x+

3 4

·4

(Isolate y in`1

and replacey in `2)

⇒ 3x−1 = 6x+ 3

(Solve forx)

⇒ −4 = 3x

⇒ x = −4

(39)

`

1

:

y

=

3

4

x

−

1

4

and

`

2

:

y

=

3

2

x

+

3

4

4·

3 4x−

1 4 = 3 2x+

3 4

·4 (Isolate y in`1

and replacey in `2)

⇒ 3x−1 = 6x+ 3

(Solve forx)

⇒ −4 = 3x

⇒ x = −4

(40)

`

1

:

y

=

3

4

x

−

1

4

and

`

2

:

y

=

3

2

x

+

3

4

4·

3 4x−

1 4 = 3 2x+

3 4

·4 (Isolate y in`1

⇒ −4 = 3x

⇒ x = −4

(41)

Example: Solve for the intersection of the lines

`

1

:

y

=

3

4

x

−

1

4

and

`

2

:

y

=

3

2

x

+

3

4

4·

3 4x−

1 4 = 3 2x+

3 4

·4 (Isolate y in`1

⇒ −4 = 3x

⇒ x = −4

(42)

`

1

:

y

=

3

4

x

−

1

4

and

`

2

:

y

=

3

2

x

+

3

4

4·

3 4x−

1 4 = 3 2x+

3 4

·4 (Isolate y in`1

⇒ −4 = 3x

⇒ x = −4

(43)

`

1

:

y

=

3

4

x

−

1

4

and

`

2

:

y

=

3

2

x

+

3

4

Solution (cont):

y

=

3

4

−

4

3

−

1

4

y

=

−1

−

1

4

y

=

−

5

4

Therefore,

`

1

∩

`

2

=

−

4

3

,

−

5

4

(44)

`

1

:

y

=

3

4

x

−

1

4

and

`

2

:

y

=

3

2

x

+

3

4

Solution (cont):

y

=

3

4

−

4

3

−

1

4

y

=

−1

−

1

4

y

=

−

5

4

Therefore,

`

1

∩

`

2

=

−

4

3

,

−

5

4

(45)

`

1

:

y

=

3

4

x

−

1

4

and

`

2

:

y

=

3

2

x

+

3

4

Solution (cont):

y

=

3

4

−

4

3

−

1

4

y

=

−1

−

1

4

y

=

−

5

4

Therefore,

`

1

∩

`

2

=

−

4

3

,

−

5

4

(46)

`

1

:

y

=

3

4

x

−

1

4

and

`

2

:

y

=

3

2

x

+

3

4

Solution (cont):

y

=

3

4

−

4

3

−

1

4

y

=

−1

−

1

4

y

=

−

5

4

Therefore,

`

1

∩

`

2

=

−

4

3

,

−

5

4

(47)

`

1

:

y

=

3

4

x

−

1

4

and

`

2

:

y

=

3

2

x

+

3

4

Illustration:

−4 −2 2

−2 0

y

=

3₄

x

−

1₄

y

=

3₂

x

+

3₄

(48)

Two Lines

In a system of two linear equations in two variables,

if the lines representing the equations..

•

..intersect at exactly one point, the system is

said to be

consistent

and

independent.

Previous example:

(49)

Two Lines

if the lines representing the equations..

•

..intersect at exactly one point, the system is

said to be

consistent

and

independent.

Previous example:

(50)

Two Lines

•

..are parallel lines, then the system has no

solution and is said to be

inconsistent.

Example:

`1 : 4x+ 6y= 6

⇔ y =−2

3x+ 1

`2 : 2x+ 3y= 5

⇔ y=−2

3x+ 5 3

    

   

(51)

Two Lines

if the lines representing the equations..

•

inconsistent.

Example:

`1 : 4x+ 6y= 6

⇔ y =−2

3x+ 1

`2 : 2x+ 3y= 5

⇔ y=−2

3x+ 5 3

    

   

(52)

Two Lines

•

inconsistent.

Example:

`1 : 4x+ 6y= 6 ⇔ y =−

2 3x+ 1

`2 : 2x+ 3y= 5 ⇔ y=−

2 3x+

5 3

    

   

(53)

Two Lines

•

inconsistent.

Example:

`1 : 4x+ 6y= 6 ⇔ y =−

2 3x+ 1

`2 : 2x+ 3y= 5 ⇔ y=−

2 3x+

5 3

    

   

(54)

Two Lines

if the lines representing the equations..

•

..coincide, then there are infinitely many

solutions to the system. The system is said to

be

dependent.

Example:

`1 : 4x+ 6y= 6

⇔ y=−2

3x+ 1

`2 : 2x+ 3y= 3

⇔ y=−2

3x+ 1

    

   

Same line

(55)

Two Lines

•

solutions to the system. The system is said to

be

dependent.

Example:

`1 : 4x+ 6y= 6

⇔ y=−2

3x+ 1

`2 : 2x+ 3y= 3

⇔ y=−2

3x+ 1

    

   

Same line

(56)

Two Lines

•

be

dependent.

Example:

`1 : 4x+ 6y= 6 ⇔ y=−

2 3x+ 1

`2 : 2x+ 3y= 3 ⇔ y=−

2 3x+ 1

    

   

Same line

(57)

Two Lines

In a system of two linear equations in two variables,

•

be

dependent.

Example:

`1 : 4x+ 6y= 6 ⇔ y=−

2 3x+ 1

`2 : 2x+ 3y= 3 ⇔ y=−

2 3x+ 1

    

   

Same line

(58)

Two Lines

•

solutions to the system. The system is said to

be

dependent.

Example:

`1 : 4x+ 6y= 6 ⇔ y=−

2 3x+ 1

`2 : 2x+ 3y= 3 ⇔ y=−

2 3x+ 1

    

   

Same line

(59)

Linear Equations in Three Variables

Definition

•

An equation of the form

ax

+

by

+

cz

+

d

= 0

,

where

a, b, c, d

∈

R

and

a, b, c

not all zero, is

called a

linear equation in

the

three variables

x, y,

and

z

.

•

An ordered triple

(

r, s, t

)

of real numbers is a

solution

of an equation in three variables

x, y, z

if the equation is satisfied when

r, s,

and

t

are

substituted into

x, y,

and

z

respectively.

Example: One of the solutions ofx+y−z−4 = 0 is

(60)

Definition

•

ax

+

by

+

cz

+

d

= 0

,

where

a, b, c, d

∈

R

and

a, b, c

not all zero, is

called a

linear equation in

the

three variables

x, y,

and

z

.

•

An ordered triple

(

r, s, t

)

solution

x, y, z

if the equation is satisfied when

r, s,

and

t

are

substituted into

x, y,

and

z

respectively.

(61)

Definition

•

ax

+

by

+

cz

+

d

= 0

,

where

a, b, c, d

∈

R

and

a, b, c

not all zero, is

called a

linear equation in

the

three variables

x, y,

and

z

.

•

An ordered triple

(

r, s, t

)

solution

x, y, z

if the equation is satisfied when

r, s,

and

t

are

substituted into

x, y,

and

z

respectively.

(62)

Definition

A

system of three linear equations in three variables

has the form







a

1

x

+

b

1

y

+

c

1

z

=

d

1

a

2

x

+

b

2

y

+

c

2

z

=

d

2

a

3

x

+

b

3

y

+

c

3

z

=

d

3

where

a

1

, a

2

, a

3

, b

1

, b

2

, b

3

, c

1

, c

2

, c

3

, d

1

, d

2

, d

3

∈

R

.

Example:











(63)

System of Three Linear Equations in Three

Variables

To solve such systems:

1.

Choose any two pairs of equations and

eliminate the same variable

.

2.

Solve the resulting system of two linear

equations.

(64)

Variables

1.

.

2.

equations.

(65)

Variables

1.

eliminate the same variable

.

2.

equations.

(66)

Variables

1.

.

2.

equations.

(67)

Example: Solve for x, y and z if

  

 

2x+ 3y−z =−3 (1)

3x+ 2y+ 2z = 5 (2)

4x−4y−3z = 2 (3)

Solution: Eliminatez first.

From eq. 1 and eq. 2,

4x+ 6y−2z =−6 3x+ 2y+ 2z = 5

7x+ 8y=−1 (4)

−6x−9y+ 3z = 9 4x−4y−3z = 2

(68)

  

 

2x+ 3y−z =−3 (1)

3x+ 2y+ 2z = 5 (2)

4x−4y−3z = 2 (3)

4x+ 6y−2z =−6 3x+ 2y+ 2z = 5

7x+ 8y=−1 (4)

−6x−9y+ 3z = 9 4x−4y−3z = 2

(69)

  

 

2x+ 3y−z =−3 (1)

3x+ 2y+ 2z = 5 (2)

4x−4y−3z = 2 (3)

4x+ 6y−2z =−6 3x+ 2y+ 2z = 5

7x+ 8y=−1 (4)

−6x−9y+ 3z = 9 4x−4y−3z = 2

(70)

  

 

2x+ 3y−z =−3 (1)

3x+ 2y+ 2z = 5 (2)

4x−4y−3z = 2 (3)

4x+ 6y−2z =−6

3x+ 2y+ 2z = 5

7x+ 8y=−1 (4)

−6x−9y+ 3z = 9 4x−4y−3z = 2

(71)

  

 

2x+ 3y−z =−3 (1)

3x+ 2y+ 2z = 5 (2)

4x−4y−3z = 2 (3)

4x+ 6y−2z =−6 3x+ 2y+ 2z = 5

7x+ 8y=−1 (4)

−6x−9y+ 3z = 9 4x−4y−3z = 2

(72)

  

 

2x+ 3y−z =−3 (1)

3x+ 2y+ 2z = 5 (2)

4x−4y−3z = 2 (3)

4x+ 6y−2z =−6 3x+ 2y+ 2z = 5

7x+ 8y=−1

(4)

−6x−9y+ 3z = 9 4x−4y−3z = 2

(73)

  

 

2x+ 3y−z =−3 (1)

3x+ 2y+ 2z = 5 (2)

4x−4y−3z = 2 (3)

4x+ 6y−2z =−6 3x+ 2y+ 2z = 5

7x+ 8y=−1 (4)

−6x−9y+ 3z = 9 4x−4y−3z = 2

(74)

  

 

2x+ 3y−z =−3 (1)

3x+ 2y+ 2z = 5 (2)

4x−4y−3z = 2 (3)

4x+ 6y−2z =−6 3x+ 2y+ 2z = 5

7x+ 8y=−1 (4)

−6x−9y+ 3z = 9 4x−4y−3z = 2

(75)

  

 

2x+ 3y−z =−3 (1)

3x+ 2y+ 2z = 5 (2)

4x−4y−3z = 2 (3)

4x+ 6y−2z =−6 3x+ 2y+ 2z = 5

7x+ 8y=−1 (4)

−6x−9y+ 3z = 9

(76)

  

 

2x+ 3y−z =−3 (1)

3x+ 2y+ 2z = 5 (2)

4x−4y−3z = 2 (3)

4x+ 6y−2z =−6 3x+ 2y+ 2z = 5

7x+ 8y=−1 (4)

−6x−9y+ 3z = 9 4x−4y−3z = 2

(77)

  

 

2x+ 3y−z =−3 (1)

3x+ 2y+ 2z = 5 (2)

4x−4y−3z = 2 (3)

4x+ 6y−2z =−6 3x+ 2y+ 2z = 5

7x+ 8y=−1 (4)

−6x−9y+ 3z = 9 4x−4y−3z = 2 −2x−13y= 11

(78)

  

 

2x+ 3y−z =−3 (1)

3x+ 2y+ 2z = 5 (2)

4x−4y−3z = 2 (3)

4x+ 6y−2z =−6 3x+ 2y+ 2z = 5

7x+ 8y=−1 (4)

−6x−9y+ 3z = 9 4x−4y−3z = 2

(79)

  

 

2x+ 3y−z =−3 (1)

3x+ 2y+ 2z = 5 (2)

4x−4y−3z = 2 (3)

Solution (cont.):

System from eq. 4 and eq. 5,

(

7x+ 8y=−1 (4)

−2x−13y= 11 (5)

Solving the new system,

14x+ 16y =−2 −14x−91y = 77

−75y = 75

(80)

  

 

2x+ 3y−z =−3 (1)

3x+ 2y+ 2z = 5 (2)

4x−4y−3z = 2 (3)

Solution (cont.):

(

7x+ 8y=−1 (4)

−2x−13y= 11 (5)

14x+ 16y =−2 −14x−91y = 77

−75y = 75

(81)

  

 

2x+ 3y−z =−3 (1)

3x+ 2y+ 2z = 5 (2)

4x−4y−3z = 2 (3)

Solution (cont.):

(

7x+ 8y=−1 (4)

−2x−13y= 11 (5)

14x+ 16y =−2 −14x−91y = 77

−75y = 75

(82)

  

 

2x+ 3y−z =−3 (1)

3x+ 2y+ 2z = 5 (2)

4x−4y−3z = 2 (3)

Solution (cont.):

(

7x+ 8y=−1 (4)

−2x−13y= 11 (5)

14x+ 16y =−2

−14x−91y = 77 −75y = 75

(83)

  

 

2x+ 3y−z =−3 (1)

3x+ 2y+ 2z = 5 (2)

4x−4y−3z = 2 (3)

Solution (cont.):

(

7x+ 8y=−1 (4)

−2x−13y= 11 (5)

14x+ 16y =−2 −14x−91y = 77

−75y = 75

(84)

  

 

2x+ 3y−z =−3 (1)

3x+ 2y+ 2z = 5 (2)

4x−4y−3z = 2 (3)

Solution (cont.):

(

7x+ 8y=−1 (4)

−2x−13y= 11 (5)

14x+ 16y =−2 −14x−91y = 77

−75y = 75

(85)

  

 

2x+ 3y−z =−3 (1)

3x+ 2y+ 2z = 5 (2)

4x−4y−3z = 2 (3)

Solution (cont.):

(

7x+ 8y=−1 (4)

−2x−13y= 11 (5)

14x+ 16y =−2 −14x−91y = 77

−75y = 75

(86)

Example: Solve for

x, y

and

z

if











2

x

+ 3

y

−

z

=

−3

(1)

3

x

+ 2

y

+ 2

z

= 5

(2)

4

x

−

4

y

−

3

z

= 2

(3)

Solution (cont.):

Usingy=−1and the system from eq. 4 and eq. 5,

(

7x+ 8y=−1 (4)

−2x−13y= 11 (5)

Using eq. 4,7x+ 8(−1) =−1 ⇒ 7x=−1 + 8 ⇒ x= 1

Using eq. 1,2(1) + 3(−1)−z =−3 ⇒z = 2

(87)

Example: Solve for

x, y

and

z

if











2

x

+ 3

y

−

z

=

−3

(1)

3

x

+ 2

y

+ 2

z

= 5

(2)

4

x

−

4

y

−

3

z

= 2

(3)

Solution (cont.):

(

7x+ 8y=−1 (4)

−2x−13y= 11 (5) Using eq. 4,7x+ 8(−1) =−1

⇒ 7x=−1 + 8 ⇒ x= 1

Using eq. 1,2(1) + 3(−1)−z =−3 ⇒z = 2

(88)

Example: Solve for

x, y

and

z

if











2

x

+ 3

y

−

z

=

−3

(1)

3

x

+ 2

y

+ 2

z

= 5

(2)

4

x

−

4

y

−

3

z

= 2

(3)

Solution (cont.):

(

7x+ 8y=−1 (4)

−2x−13y= 11 (5) Using eq. 4,7x+ 8(−1) =−1 ⇒ 7x=−1 + 8

⇒ x= 1

Using eq. 1,2(1) + 3(−1)−z =−3 ⇒z = 2

(89)

Example: Solve for

x, y

and

z

if











2

x

+ 3

y

−

z

=

−3

(1)

3

x

+ 2

y

+ 2

z

= 5

(2)

4

x

−

4

y

−

3

z

= 2

(3)

Solution (cont.):

(

7x+ 8y=−1 (4)

−2x−13y= 11 (5)

Using eq. 4,7x+ 8(−1) =−1 ⇒ 7x=−1 + 8 ⇒ x= 1

Using eq. 1,2(1) + 3(−1)−z =−3 ⇒z = 2

(90)

Example: Solve for

x, y

and

z

if











2

x

+ 3

y

−

z

=

−3

(1)

3

x

+ 2

y

+ 2

z

= 5

(2)

4

x

−

4

y

−

3

z

= 2

(3)

Solution (cont.):

(

7x+ 8y=−1 (4)

−2x−13y= 11 (5)

Using eq. 4,7x+ 8(−1) =−1 ⇒ 7x=−1 + 8 ⇒ x= 1

Using eq. 1,2(1) + 3(−1)−z =−3

⇒z = 2

(91)

Example: Solve for

x, y

and

z

if











2

x

+ 3

y

−

z

=

−3

(1)

3

x

+ 2

y

+ 2

z

= 5

(2)

4

x

−

4

y

−

3

z

= 2

(3)

Solution (cont.):

(

7x+ 8y=−1 (4)

−2x−13y= 11 (5)

Using eq. 4,7x+ 8(−1) =−1 ⇒ 7x=−1 + 8 ⇒ x= 1

Using eq. 1,2(1) + 3(−1)−z =−3 ⇒z = 2

(92)

Example: Solve for

x, y

and

z

if











2

x

+ 3

y

−

z

=

−3

(1)

3

x

+ 2

y

+ 2

z

= 5

(2)

4

x

−

4

y

−

3

z

= 2

(3)

Solution (cont.):

(

7x+ 8y=−1 (4)

−2x−13y= 11 (5)

Using eq. 4,7x+ 8(−1) =−1 ⇒ 7x=−1 + 8 ⇒ x= 1

Using eq. 1,2(1) + 3(−1)−z =−3 ⇒z = 2

(93)

Line and a Parabola

Example: Find the points of intersection of the

graphs of

p

:

y

=

x

2

−

4

x

+ 2

and

`

:

y

= 2

x

−

3

2x−3 = x2₋₄_x_{+ 2} _{(Take value of} _y _in _`

and replace y in p)

⇒ 0 = x2₋₆_x_{+ 5} _{(Solve for} _x₎

0 = (x−1)(x−5)

x= 1 or x= 5

(94)

Line and a Parabola

graphs of

p

:

y

=

x

2

−

4

x

+ 2

and

`

:

y

= 2

x

−

3

and replace y in p)

⇒ 0 = x2₋₆_x_{+ 5} _{(Solve for} _x₎

0 = (x−1)(x−5)

x= 1 or x= 5

(95)

Line and a Parabola

graphs of

p

:

y

=

x

2

−

4

x

+ 2

and

`

:

y

= 2

x

−

3

2x−3 = x2₋₄_x_{+ 2}

(Take value of y in `

and replace y in p)

⇒ 0 = x2₋₆_x_{+ 5} _{(Solve for} _x₎

0 = (x−1)(x−5)

x= 1 or x= 5

(96)

Line and a Parabola

graphs of

p

:

y

=

x

2

−

4

x

+ 2

and

`

:

y

= 2

x

−

3

2x−3

= x2₋₄_x_{+ 2} _{(Take value of} _y _in _`

and replace y in p)

⇒ 0 = x2₋₆_x_{+ 5} _{(Solve for} _x₎

0 = (x−1)(x−5)

x= 1 or x= 5

(97)

Line and a Parabola

graphs of

p

:

y

=

x

2

−

4

x

+ 2

and

`

:

y

= 2

x

−

3

and replace y in p)

⇒ 0 = x2₋₆_x_{+ 5} _{(Solve for} _x₎

0 = (x−1)(x−5)

x= 1 or x= 5

(98)

Line and a Parabola

graphs of

p

:

y

=

x

2

−

4

x

+ 2

and

`

:

y

= 2

x

−

3

and replace y in p)

⇒ 0 = x2₋₆_x_{+ 5}

(Solve for x)

0 = (x−1)(x−5)

x= 1 or x= 5

(99)

Line and a Parabola

graphs of

p

:

y

=

x

2

−

4

x

+ 2

and

`

:

y

= 2

x

−

3

and replace y in p)

⇒ 0 = x2₋₆_x_{+ 5} _{(Solve for} _x₎

0 = (x−1)(x−5)

x= 1 or x= 5

(100)

Line and a Parabola

graphs of

p

:

y

=

x

2

−

4

x

+ 2

and

`

:

y

= 2

x

−

3

and replace y in p)

⇒ 0 = x2₋₆_x_{+ 5} _{(Solve for} _x₎

0 = (x−1)(x−5)

x= 1 or x= 5

(101)

Line and a Parabola

graphs of

p

:

y

=

x

2

−

4

x

+ 2

and

`

:

y

= 2

x

−

3

and replace y in p)

⇒ 0 = x2₋₆_x_{+ 5} _{(Solve for} _x₎

0 = (x−1)(x−5)

x= 1 or x= 5

(102)

Line and a Parabola

graphs of

p

:

y

=

x

2

−

4

x

+ 2

and

`

:

y

= 2

x

−

3

and replace y in p)

⇒ 0 = x2₋₆_x_{+ 5} _{(Solve for} _x₎

0 = (x−1)(x−5)

x= 1 or x= 5

(103)

A Line and a Parabola

graphs of

p

:

y

=

x

2

−

4

x

+ 2

and

`

:

y

= 2

x

−

3

Solution (cont): Using the equation for

`

,

If

x

= 1

,

If

x

= 5

,

y

= 2(1)

−

3

y

= 2(5)

−

3

=

−1

= 7

(104)

graphs of

p

:

y

=

x

2

−

4

x

+ 2

and

`

:

y

= 2

x

−

3

`

,

If

x

= 1

,

If

x

= 5

,

y

= 2(1)

−

3

y

= 2(5)

−

3

=

−1

= 7

(105)

graphs of

p

:

y

=

x

2

−

4

x

+ 2

and

`

:

y

= 2

x

−

3

`

,

If

x

= 1

,

If

x

= 5

,

y

= 2(1)

−

3

y

= 2(5)

−

3

=

−1

= 7

(106)

graphs of

p

:

y

=

x

2

−

4

x

+ 2

and

`

:

y

= 2

x

−

3

Solution (cont): Using the equation for

`

,

If

x

= 1

,

If

x

= 5

,

y

= 2(1)

−

3

y

= 2(5)

−

3

=

−1

= 7

(107)

graphs of

p

:

y

=

x

2

−

4

x

+ 2

and

`

:

y

= 2

x

−

3

`

,

If

x

= 1

,

If

x

= 5

,

y

= 2(1)

−

3

y

= 2(5)

−

3

=

−1

= 7

(108)

graphs of

p

:

y

=

x

2

−

4

x

+ 2

and

`

:

y

= 2

x

−

3

`

,

If

x

= 1

,

If

x

= 5

,

y

= 2(1)

−

3

y

= 2(5)

−

3

=

−1

= 7

(109)

Example: Find the points of intersection of the

graphs of

p

:

y

=

x

2

−

4

x

+ 2

and

`

:

y

= 2

x

−

3

`

,

If

x

= 1

,

If

x

= 5

,

y

= 2(1)

−

3

y

= 2(5)

−

3

=

−1

= 7

(110)

`

∩

p

=

{(1

,

−1)

,

(5

,

7)}

`

:

y

= 2

x

−

3

y

−

int. :

−3

x

−

int. :

3

2

p

:

y

=

x

2

−

4

x

+ 2

y

−

int. :

2

x

−

int. :

2 +

√

2

,

2

−

√

2

vertex :

(2

,

−

2)

−2 2 4

−4

−2 2 4 6 8

(5,7)

(1,−1)

(111)

`

∩

p

=

{(1

,

−1)

,

(5

,

7)}

`

:

y

= 2

x

−

3

y

−

int. :

−3

x

−

int. :

3

2

p

:

y

=

x

2

−

4

x

+ 2

y

−

int. :

2

x

−

int. :

2 +

√

2

,

2

−

√

2

vertex :

(2

,

−

2)

−2 2 4

−4

−2 2 4 6 8

(5,7)

(1,−1)

(112)

`

∩

p

=

{(1

,

−1)

,

(5

,

7)}

`

:

y

= 2

x

−

3

y

−

int. :

−3

x

−

int. :

3

2

p

:

y

=

x

2

−

4

x

+ 2

y

−

int. :

2

x

−

int. :

2 +

√

2

,

2

−

√

2

vertex :

(2

,

−

2)

−2 2 4

−4

−2 2 4 6 8

(5,7)

(1,−1)

(113)

A Line and a Parabola

`

∩

p

=

{(1

,

−1)

,

(5

,

7)}

`

:

y

= 2

x

−

3

y

−

int. :

−3

x

−

int. :

3

2

p

:

y

=

x

2

−

4

x

+ 2

y

−

int. :

2

x

−

int. :

2 +

√

2

,

2

−

√

2

vertex :

(2

,

−

2)

−2 2 4

−4

−2 2 4 6 8

(5,7)

(1,−1)

(114)

`

∩

p

=

{(1

,

−1)

,

(5

,

7)}

`

:

y

= 2

x

−

3

y

−

int. :

−3

x

−

int. :

3

2

p

:

y

=

x

2

−

4

x

+ 2

y

−

int. :

2

x

−

int. :

2 +

√

2

,

2

−

√

2

vertex :

(2

,

−

2)

−2 2 4

−4

−2 2 4 6 8

(5,7)

(1,−1)

(115)

`

∩

p

=

{(1

,

−1)

,

(5

,

7)}

`

:

y

= 2

x

−

3

y

−

int. :

−3

x

−

int. :

3

2

p

:

y

=

x

2

−

4

x

+ 2

y

−

int. :

2

x

−

int. :

2 +

√

2

,

2

−

√

2

vertex :

(2

,

−

2)

−2 2 4

−4

−2 2 4 6 8

(5,7)

(1,−1)

(116)

`

∩

p

=

{(1

,

−1)

,

(5

,

7)}

`

:

y

= 2

x

−

3

y

−

int. :

−3

x

−

int. :

3

2

p

:

y

=

x

2

−

4

x

+ 2

y

−

int. :

2

x

−

int. :

2 +

√

2

,

2

−

√

2

vertex :

(2

,

−

2)

−2 2 4

−4

−2 2 4 6 8

(5,7)

(1,−1)

(117)

`

∩

p

=

{(1

,

−1)

,

(5

,

7)}

`

:

y

= 2

x

−

3

y

−

int. :

−3

x

−

int. :

3

2

p

:

y

=

x

2

−

4

x

+ 2

y

−

int. :

2

x

−

int. :

2 +

√

2

,

2

−

√

2

vertex :

(2

,

−

2)

−2 2 4

−4

−2 2 4 6 8

(5,7)

(1,−1)

(118)

`

∩

p

=

{(1

,

−1)

,

(5

,

7)}

`

:

y

= 2

x

−

3

y

−

int. :

−3

x

−

int. :

3

2

p

:

y

=

x

2

−

4

x

+ 2

y

−

int. :

2

x

−

int. :

2 +

√

2

,

2

−

√

2

vertex :

(2

,

−

2)

−2 2 4

−4

−2 2 4 6 8

(5,7)

(1,−1)

(119)

`

∩

p

=

{(1

,

−1)

,

(5

,

7)}

`

:

y

= 2

x

−

3

y

−

int. :

−3

x

−

int. :

3

2

p

:

y

=

x

2

−

4

x

+ 2

y

−

int. :

2

x

−

int. :

2 +

√

2

,

2

−

√

2

vertex :

(2

,

−

2)

−2 2 4

−4

−2 2 4 6 8

(5,7)

(1,−1)

(120)

`

∩

p

=

{(1

,

−1)

,

(5

,

7)}

`

:

y

= 2

x

−

3

y

−

int. :

−3

x

−

int. :

3

2

p

:

y

=

x

2

−

4

x

+ 2

y

−

int. :

2

x

−

int. :

2 +

√

2

,

2

−

√

2

vertex :

(2

,

−

2)

−2 2 4

−4

−2 2 4 6 8

(5,7)

(1,−1)

(121)

`

∩

p

=

{(1

,

−1)

,

(5

,

7)}

`

:

y

= 2

x

−

3

y

−

int. :

−3

x

−

int. :

3

2

p

:

y

=

x

2

−

4

x

+ 2

y

−

int. :

2

x

−

int. :

2 +

√

2

,

2

−

√

2

vertex :

(2

,

−

2)

−2 2 4

−4

−2 2 4 6 8

(5,7)

(1,−1)

(122)

`

∩

p

=

{(1

,

−1)

,

(5

,

7)}

`

:

y

= 2

x

−

3

y

−

int. :

−3

x

−

int. :

3

2

p

:

y

=

x

2

−

4

x

+ 2

y

−

int. :

2

x

−

int. :

2 +

√

2

,

2

−

√

2

vertex :

(2

,

−

2)

−2 2 4

−4

−2 2 4 6 8

(