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):
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)
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=½x+½
(y intercept of ½ and slope of ½) (EQN 1)
y = -2x + 3
(y intercept of 3 and slope of -2) (EQN 2)
y=½x+½
y = -2x + 3
5
Solution: (1,1)
Point of Intersection
4
3
2
1
-5
-4 -3
-2
0
-1-1 0
-2
-3
-4
-5
y = 0.5x+0.5
1
2
3
4
5
y = -2x+3
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”
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)
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
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)
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.
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 = 6 TRUE STATEMENT
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
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
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)
3(6x + 5y = -34)
5x – 15y = 10
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
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 = -1
MULTIPLY ALL TERMS BY 6
THEN APPLY ANY OF THE THREE METHODS TO SOLVE THE SYSTEM.
THE SOLUTION IS (-12, -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)
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
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