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METHODS: GRAPHICAL BY HAND WITH CALCULATOR SUBSTITUTION ELIMINATION

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):

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METHODS: GRAPHICAL BY HAND WITH CALCULATOR SUBSTITUTION ELIMINATION

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

  2. 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)

  3. 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)

  4. y = ½ x + ½ y = -2x + 3 Solution: (1,1) Point of Intersection

  5. 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”

  6. 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)

  7. 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

  8. 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)

  9. 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.

  10. 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

  11. 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

  12. 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

  13. 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

  14. 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)

  15. 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)

  16. 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

  17. 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

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