slide1 n.
Download
Skip this Video
Loading SlideShow in 5 Seconds..
METHODS: GRAPHICAL BY HAND WITH CALCULATOR SUBSTITUTION ELIMINATION PowerPoint Presentation
Download Presentation
METHODS: GRAPHICAL BY HAND WITH CALCULATOR SUBSTITUTION ELIMINATION

Loading in 2 Seconds...

play fullscreen
1 / 17

METHODS: GRAPHICAL BY HAND WITH CALCULATOR SUBSTITUTION ELIMINATION - PowerPoint PPT Presentation


  • 89 Views
  • Uploaded on

ARAPAHOE COMMUNITY COLLEGE MAT 121 COLLEGE ALGEBRA SOLVING SYSTEMS OF EQUATIONS IN TWO VARIABLES CHAPTER 6 SECTION 6.1 DIANA HUNT. METHODS: GRAPHICAL BY HAND WITH CALCULATOR SUBSTITUTION ELIMINATION THE SOLUTION TO THE SYSTEM OF EQUATIONS IS AN ORDERED PAIR, (X,Y):

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about 'METHODS: GRAPHICAL BY HAND WITH CALCULATOR SUBSTITUTION ELIMINATION' - maida


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
slide1

ARAPAHOE COMMUNITY COLLEGE MAT 121 COLLEGE ALGEBRASOLVING SYSTEMS OF EQUATIONS IN TWO VARIABLESCHAPTER 6 SECTION 6.1DIANA HUNT

slide2

METHODS:

GRAPHICAL

BY HAND

WITH CALCULATOR

SUBSTITUTION

ELIMINATION

THE SOLUTION TO THE SYSTEM OF EQUATIONS IS AN ORDERED PAIR, (X,Y):

THE POINT(S) WHICH LIE(S) ON BOTH LINES OR THE POINT OF INTERSECTION

THIS IS KNOWN AS AN ORDERED PAIR USUALLY ALPHABETICAL: (x,y) or (w,x)

slide3
GRAPHICAL: BY HAND

GIVEN 2 EQUATIONS: 2y – 1 = x (EQUATION 1)

-3y + 9 = 6x (EQUATION 2)

FIRST: REWRITE EQUATIONS IN y = mx + b FORMAT SO WE CAN EASILY GRAPH:

y = ½ x + ½ (y intercept of ½ and slope of ½) (EQN 1)

y = -2x + 3 (y intercept of 3 and slope of -2) (EQN 2)

slide4

y = ½ x + ½

y = -2x + 3

Solution: (1,1)

Point of Intersection

slide5

GRAPHICAL: BY CALCULATOR

IT IS ASSUMED THAT YOU ARE FAMILIAR WITH YOUR CALCULATOR SO

THESE DIRECTIONS ARE FAIRLY BRIEF. PLEASE SEE NEXT PAGE FOR

FURTHER CLARIFICATION.

TURN CALCULATOR ON

PRESS “Y=“ BUTTON (UPPER LEFT HAND CORNER)

ENTER IN “½ X + ½” FOR Y1

ENTER IN “-2X + 3” FOR Y2

PRESS GRAPH (UPPER RIGHT HAND CORNER)

YOU SHOULD SEE TWO LINES INTERSECTING IN QUADRANT I

PRESS 2ND TRACE (TOP ROW – RIGHT OF MIDDLE) – PUTS YOU IN CALC MODE

CHOOSE INTERSECT OPTION –USE DOWN ARROW KEY THEN PRESS ENTER

MOVE FLASHING CURSOR WITH THE RIGHT OR LEFT ARROW BUTTON

NEAR THE INTERSECTION – NOT TOO CLOSE!

PRESS ENTER

IF YOUR CURSOR HAS NOT GONE TO THE OTHER LINE:

PRESS UP OR DOWN ARROW (CURSOR SHOULD NOW BE ON OTHER LINE)

MOVE FLASHING CURSOR WITH THE RIGHT OR LEFT ARROW BUTTON

NEAR THE INTERSECTION – NOT TOO CLOSE!

PRESS ENTER TWICE

YOUR SCREEN SHOULD SAY “INTERSECTION AND X = 1 Y = 1”

slide6

CIRCLED IN WHITE ARE THE

NECESSARY KEYS FOR THIS

EXERCISE.

TRY THE FOLLOWING ON

YOUR OWN:

Y = -2X + 5

Y = 3X - 2

THE ANSWER IS (1.4, 2.2)

slide7

TERMINOLOGY, SOLUTIONS,

PARALLEL LINES AND THE SAME LINE:

PARALLEL LINES

(LINES HAVING THE SAME SLOPE BUT DIFFERENT Y INTERCEPTS)

IF YOU GRAPH YOUR LINES AND THEY ARE PARALLEL, THERE IS NO

INTERSECTION AND THEREFORE NO SOLUTION BECAUSE THERE ARE NO

POINTS THAT LIE ON BOTH LINES. THIS SYSTEM IS CALLED INCONSISTENT

AND INDEPENDENT

SAME LINE

IF YOU GRAPH YOUR SYSTEM AND IT IS THE SAME LINE (YOU’LL SEE THIS

WHEN YOU WRITE THE EQUATIONS IN Y = MX + B FORM)

THERE ARE MANY COMMON POINTS SO THE SOLUTION IS INFINITELY MANY

POINTS AND THE SYSTEM IS CALLED CONSISTENT AND DEPENDENT

INTERSECTING LINES

INTERSECTING LINES HAVE ONE SOLUTION AND THE SYSTEM IS CALLED

CONSISTENT AND INDEPENDENT

slide8

SUBSTITUTION METHOD:

GIVEN A SYSTEM OF EQUATONS: 2y = x + 3 (EQUATION 1)

x = 5y – 1 (EQUATION 2)

SOLVE EITHER EQUATION FOR ONE OF THE TWO VARIABLES

(IN THIS CASE EQUATION 2 IS SOLVED FOR X)

SUBSTITUTE IN FOR X IN EQUATION 1:

2y = (5y - 1) + 3 (EQUATION 1)

SOLVE THIS EQUATION FOR y:

2y – 5y = -1 + 3

-3y = 2

y = -2/3

SUBSTITUTE THE VALUE YOU GOT FOR y INTO EITHER ORIGINAL EQUATION

2(-2/3) = x + 3 (EQUATION 1)

-4/3 = x + 3

-4/3 – 3 = x

-4/3 – 9/3 = x

-13/3 = x

THE SOLUTION IS (-13/3, -2/3)

slide9

YOUR TURN: SOLVE THIS SYSTEM OF EQUATIONS USING THE

SUBSTITUTION METHOD.

x + y = 9

2x – 3y = -2

THE SOLUTION IS: (5, 4)

YOU CAN CHECK YOUR WORK BY GRAPHING OR ALGEBRAICALLY

TO CHECK YOUR WORK ALGEBRAICALLY:

FIND X AND Y

PLUG X AND Y INTO BOTH EQUATIONS AND YOU SHOULD

GET TWO TRUE STATEMENTS. IF NOT, YOU DO NOT HAVE

THE CORRECT SOLUTION OR YOU MADE AN ARITHMETIC ERROR

WARNING: IT IS VERY IMPORTANT TO CHECK BOTH EQUATIONS

BECAUSE YOU COULD HAVE MADE A MISTAKE AND ONE

EQUATION WOULD BE TRUE AND THE OTHER FALSE.

slide10

LET’S CHECK OUR FIRST PROBLEM ALGEBRAICALLY:

2y – 1 = x (EQUATION 1)

-3y + 9 = 6x (EQUATION 2)

SOLUTION WAS FOUND TO BE (1,1)

2(1) – 1 = 1 (EQUATION 1)

-3(1) + 9 = 6(1) (EQUATION 2)

SIMPLIFY:

1 = 1 (EQUATION 1)

-3 + 9 = 6 (EQUATION 2)

ONE MORE STEP:

1 = 1 TRUE STATEMENT

6 = 6TRUE STATEMENT

slide11

SOLUTIONS:

PARALLEL LINES

DURING THE SUBSTITUTION PROCESS YOU END UP WITH A FALSE

STATEMENT SUCH AS 4 = 2. IN THIS CASE YOU HAVE NO SOLUTION.

SAME LINE

DURING THE SUBSTITUTION PROCESS YOU END UP WITH A TRUE

STATEMENT SUCH AS 3 = 3. IN THIS CASE YOU HAVE INFINITELY MANY

SOLUTIONS.

INTERSECTING LINES

DURING THE SUBSTITUTION PROCESS YOU END UP WITH ONE VALUE

FOR X AND ONE VALUE FOR Y AND THAT IS YOUR SOLUTION.

THE KEY IS AS LONG AS YOU HAVE A TRUE STATEMENT YOU HAVE A

SOLUTION, IT’S JUST A MATTER OF WHAT KIND OF SOLUTION

THE SYSTEM HAS THE SAME TERMINOLOGY AS STATED ON SLIDE 7

slide12

ELIMINATION METHOD:

WRITE THE EQUATIONS SO THAT THE VARIABLES AND THE CONSTANTS

LINE UP IN COLUMNS:

x – 3y = 2

6x + 5y = -34

NOTICE THE X’S, THE Y’S, THE =‘S AND THE NUMBERS ARE LINED UP IN

COLUMNS

FIRST TRY ADDING THE TWO EQUATIONS TOGETHER TO SEE IF ONE

VARIABLE IS ELIMINATED

SECOND TRY SUBTRACTING ONE EQUATION FROM THE OTHER TO SEE IF

ONE VARIABLE IS ELIMINATED

THIRD: DECIDE ON A VARIABLE TO ELIMINATE – IT’S YOUR CHOICE, IT

DOES NOT MATTER

slide13

x – 3y = 2 (EQUATION 1)

6x + 5y = -34 (EQUATION 2)

I’VE DECIDED TO ELIMINATE Y. TO DO THIS WE NEED TO MULTIPLY EQUATION 1 BY 5 AND EQUATION 2 BY 3. NOTICE THAT ONE OF THE COEFFICIENTS OF Y IS ALREADY NEGATIVE. THIS IS VERY HELPFUL.

5(x – 3y = 2) 5x – 15y = 10

3(6x + 5y = -34) 18x + 15y = -102

BE SURE TO MULTIPLY EVERY TERM IN THE EQUATION

NOW LET’S TRY ADDING:

5x – 15y = 10

18x – 15y = -102

23x = -92

x = -4

NOW SUBSTITUTE -4 IN FOR x IN EITHER OF THE ORIGINAL EQUATIONS

-4 – 3y = 2 (EQUATION 1)

- 3y = 6

y = -2

THE SOLUTION IS (-4, -2) BE SURE TO PUT THE SOLUTION IN (x, y) FORMAT

slide14

THIS IS AN EXAMPLE OF SIMPLY ADDING THE TWO EQUATIONS TO

ELIMINATE A VARIABLE:

3x + 4y = -2

-3x – 5y = 1

NOTICE THE X GETS ELIMINATED, THE SOLUTION IS (-2,1)

THIS IS AN EXAMPLE OF SUBTRACTING THE TWO EQUATIONS:

x + 2y = 7

x – 2y = -5

NOTICE THE X GETS ELIMINATED, THE SOLUTION IS (1, 3)

SOMETIMES THERE ARE FRACTIONS OR DECIMALS IN THE EQUATION

IF THERE ARE FRACTIONS MULTIPLY EACH EQUATION BY ITS LOWEST

COMMON DENOMINATOR, SUCH AS:

2/3 x + 3/5 y = -17 MULTIPLY ALL TERMS BY 15

½ x – 1/3 y = -1MULTIPLY ALL TERMS BY 6

THEN APPLY ANY OF THE THREE METHODS TO SOLVE THE SYSTEM.

THE SOLUTION IS (-12, -15)

slide15

IF THERE ARE DECIMALS IN THE EQUATIONS MULTIPLY EACH EQUATION BY AN APPROPRIATE FACTOR OF 10 TO ELIMINATE ALL DECIMALS. FOR EXAMPLE:

0.3x – 0.02y = -0.9 MULTIPLY ALL TERMS BY 100

0.2x – 3y = -0.6 MULTIPLY ALL TERMS BY 10

THE SOLUTION IS (-3, 0)

slide16

SOLUTIONS:

PARALLEL LINES

DURING THE ELIMINATION PROCESS YOU END UP WITH A FALSE

STATEMENT SUCH AS 4 = 2. IN THIS CASE YOU HAVE NO SOLUTION.

SAME LINE

DURING THE ELIMINATION PROCESS YOU END UP WITH A TRUE

STATEMENT SUCH AS 3 = 3. IN THIS CASE YOU HAVE INFINITELY MANY

SOLUTIONS.

INTERSECTING LINES

DURING THE ELIMINATION PROCESS YOU END UP WITH ONE VALUE

FOR X AND ONE VALUE FOR Y AND THAT IS YOUR SOLUTION.

THE KEY IS AS LONG AS YOU HAVE A TRUE STATEMENT YOU HAVE A

SOLUTION, IT’S JUST A MATTER OF WHAT KIND OF SOLUTION

THE SYSTEM HAS THE SAME TERMINOLOGY AS STATED ON SLIDE 7

slide17

SUMMARY

THREE METHODS FOR SOLVING A SYSTEM OF EQUATIONS IN TWO VARIABLES

GRAPHING

SUBSTITUTION

ELIMINATION

CHECK YOUR ANSWERS

ALGEBRAICALLY

GRAPHICALLY

WHEN TO USE A GIVEN METHOD

GRAPHING – WHEN y IS EASILY SOLVED FOR

SUBSTITUTION – WHEN ONE OF THE VARIABLES IS SOLVED FOR

ELIMINATION – ANYTIME

TIME TO PRACTICE IN MYMATHLAB –

USE THE “SHOW ME HOW” IF YOU GET STUCK