Objective - To solve a system of linear equations using linear combinations (Elimination).

Graphic Method Algebraic Approaches

solution

SubstitutionIf a = b, and b = c

then a = c

- Time consuming- Not always accurate

Elimination

then a = c.

If a = band c = d

then a + c = b + d.

Solve using elimination.2x + y = 63x − y = 4

⎧ ⎨ ⎩

2x + y = 63x − y = 4

5x = 10

2x + y = 62 2( )+ y = 6

5x =105 5x = 2

4 + y = 6−4 − 4

y = 2

Solution is (2,2) Check!

Solve using elimination.x + 3y = 5−x + y = 3

⎧ ⎨ ⎩

x + 3y = 5−x + y = 3

4y = 8

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

y4 4y = 2

x + 6 = 5−6 − 6x = −1

Solution is (−1,2) Check!

Solve using elimination.y = 3x + 42x − y = 5

⎧ ⎨ ⎩

y = 3x + 42x − y = 5

−3x + y = 42x − y = 5

−x = 9x 9−1• −x = 9 •−1

x = −9y = 3x + 4

y = 3 −9( )+ 4y = −27 + 4y = −23 Solution is (−9,−23) Check!

Solve using elimination.2x + 3y = −4x − 4y = 9

⎧ ⎨ ⎩

2x + 3y = −4x − 4y = 93x − y = 5y

−2( )

Solve using elimination.2x + 3y = −4x − 4y = 9

⎧ ⎨ ⎩

2x + 3y = −4x − 4y = 9

2x + 3y = −4−2x + 8y = −18

11y = −2211y 22

y = −2x − 4y = 9

x − 4(−2) = 9x + 8 = 9

x = 1 Solution is (1,−2) Check!

11 11

Lesson 6-3

Algebra Slide Show: Teaching Made Easy As Pi, by James Wenk © 2010

5( )

Solve using elimination.2x + 5y = 102x − y = 7

⎧ ⎨ ⎩

2x + 5y =102x − y = 7

2x + 5y =1010x − 5y = 35

12x = 452x y 7

x =4512

2x − y = 7

( )152 y 74 − =

712

− y = 7

y =12

( )3 1Solution is 3 ,4 2 Check!

12 12

=154

2( )5( )

Solve using elimination.−2x + 3y = 55x − 2y = 4

⎧ ⎨ ⎩

−2x + 3y = 55x − 2y = 4

−10x +15y = 2510x − 4y = 8

11y = 335x 2y 4

y = 3

5x − 2y = 45x − 2 3( ) = 4

5x − 6 = 4

5x =10Solution is 2,3( ) Check!

11 11

+6 + 6

x = 2

6

9

Joe bought 15 items for $135. If the 15 items consisted of notebooks for $4.50 each, and calculators for $12.00 each, how many of eachdid he buy?

Number x Unit Value = Total ValueNotebooksCalculators

xy

4.5012.00

4.5x12y

-12( )15 135

x + y = 154.5x + 12y = 135

6 notebooks9 calculators

-12x + -12y = -1804.5x + 12y = 135

-7.5x = -45-7.5 -7.5

x = 6

16

24

Merna raised $24 by selling 40 baked items for the Builders Club. She sold cookies for 50 centseach and brownies for 75 cents each. How many of each did she sell?

Number x Unit Value = Total Valuecookiesbrownies

xy

0.500.75

0.50x0.75y

40 24

24 cookies16 brownies

-2( )x + y = 40

0.5x + 0.75y = 24x + y = 40

-x - 1.5y = -48-0.5y = -8-0.5 -0.5

y = 16

105

90

A movie theater sells matinee tickets for $5 andregular admission tickets for $6.50. If 195 ticketswere sold for total revenues of $1,110, how manyof each type were sold?

Number x Unit Value = Total ValueMatineeRegular

xy

56.50

5x6.5y

-5( )

g y195

y1110

x + y = 1955x + 6.5y = 1110

90 regular105 matinee

-5x + -5y = -9755x + 6.5y = 1110

1.5y = 1351.5 1.5

y = 90

Number:Value:

Coin Problem #1Mike has nickels and dimes. If the total of his13 coins is one dollar, how much of each doeshe have? Let n = the # of nickels

Let d = the # of dimesNumber Equation

n + d = 13Value Equation

5n + 10d = 100or+ 7 13

= 7= 6

-5( )

or0.05n + 0.10d = 1.00

n + d = 135n + 10d = 100

-5n - 5d = -655n + 10d = 100

5d = 355 5d = 7

n + 7 = 13n = 6

Lesson 6-3 (cont.)

Algebra Slide Show: Teaching Made Easy As Pi, by James Wenk © 2010

Coin Problem #2Julie has 19 coins made up of dimes and quarters.Find the number of each coin if the value of thecoins is $3.85. Let d = the # of dimes

Let q = the # of quartersNumber Equation

d + q = 19Value Equation

10d + 25q = 385ord + 13 19

= 13= 6

-10( )

or0.10d + 0.25q = 3.85

d + q = 1910d + 25q = 385

-10d - 10q = -19010d + 25q = 385

15q = 19515 15

q = 13

d + 13 = 19d = 6

Lesson 6-3 (cont.)

Algebra Slide Show: Teaching Made Easy As Pi, by James Wenk © 2010