- TODAY WE'LL DISCUSS THE THIRD METHOD
FOR SOLVING SYSTEMS OF EQUATIONS,
AND THAT METHOD WILL BE THE ELIMINATION OR ADDITION METHOD.
LET'S TAKE A LOOK AT HOW THIS IS GOING TO WORK.
ESSENTIALLY WHAT WE'RE GOING TO DO
IS ADD THESE EQUATIONS TOGETHER.
THE REASON WE'RE ALLOWED TO DO THAT IS BECAUSE
IF WE TAKE A LOOK AT THE SECOND EQUATION,
SINCE THIS EXPRESSION IS EQUAL TO 7,
WE COULD ADD THIS EXPRESSION TO THE LEFT SIDE OF THIS EQUATION
AND ADD 7 TO THE RIGHT SIDE OF THIS EQUATION
SINCE THEY ARE EQUAL.
NOW LET'S SEE WHAT HAPPENS WHEN WE DO THIS.
IF WE ADD THE X TERMS TOGETHER WE'D HAVE 5X.
NOW, WHEN WE ADD THE Y TERMS TOGETHER SINCE THEY'RE OPPOSITES
WE GET 0, SO THESE TERMS ARE ELIMINATED.
AND ON THE RIGHT, 8 + 7 WOULD GIVE US 15.
WE HAVE AN EQUATION WITH ONE VARIABLE.
WE CAN SOLVE THAT DIVIDING BOTH SIDES BY 5.
WE HAVE X = 3.
REMEMBER, ONE SOLUTION CONSISTS OF AN X AND A Y VALUE.
WE KNOW THE X VALUE IS 3, BUT WE STILL HAVE TO FIND THE Y-VALUE.
WHAT WE DO NOW IS WE REPLACE X WITH 3 IN EITHER EQUATION
AND SOLVE FOR Y.
I WILL GO AHEAD AND USE THE FIRST EQUATION FOR THIS PURPOSE.
SO WE KNOW THAT X IS 3, SO WE'D HAVE 4 x 3 + 3Y = 8.
SO WE'LL SIMPLIFY AND SOLVE.
- 12 ON BOTH SIDES.
3Y = -4.
DIVIDE BOTH SIDES BY 3.
Y = -4/3.
WE HAVE ONE SOLUTION FOR THIS SYSTEM.
THEREFORE, THE SYSTEM IS CONSISTENT AND INDEPENDENT.
YOU CAN SEE THIS METHOD WORKS VERY NICELY.
IT'S A LOT FASTER THAN TRYING TO GRAPH THESE TWO LINEAR EQUATIONS
AND ESTIMATE A POINT OF INTERSECTION.
LET'S TAKE A LOOK AT IN GENERAL HOW WE'RE GOING TO SOLVE
A SYSTEM USING THE ELIMINATION METHOD.
STEP ONE, WE'LL MULTIPLY ONE OR BOTH OF THE EQUATIONS
BY THE APPROPRIATE NUMBERS
SO THAT ONE OF THE VARIABLES WILL HAVE THE SAME COEFFICIENT
WITH OPPOSITE SIGNS.
THEN WE'LL ADD THE TWO EQUATIONS TOGETHER
BECAUSE ONE OF THE VARIABLES HAD THE SAME COEFFICIENT
WITH THE OPPOSITE SIGN AND WILL BE ELIMINATED
WHEN ADDED TO THE OTHER EQUATION.
THEN WE'LL SOLVE THE RESULTING EQUATION
AND THEN SUBSTITUTE THIS ANSWER BACK
INTO ONE OF THE ORIGINAL EQUATIONS
TO FIND THE REMAINING VARIABLE.
SO THE KEY HERE, IF WE WANT TO SOLVE THIS USING ELIMINATION,
EITHER THE X TERMS OR THE Y TERMS MUST BE OPPOSITES.
SINCE THE X TERMS ARE ALREADY OPPOSITE SIGNS,
I'M GOING TO GO AHEAD AND TRY TO ELIMINATE THE X TERMS.
NOW, THE LEAST COMMON MULTIPLE OF 2 AND 3 WOULD BE 6,
SO I'M GOING TO CONVERT THIS TO A 6X AND THIS TO A -6X.
THE WAY I'M GOING TO DO THAT
IS I'M GOING TO TAKE THIS FIRST EQUATION AND MULTIPLY BY A 2,
AND I'LL MULTIPLY THE SECOND EQUATION BY 3.
WHEN I DO THIS, I DO HAVE TO MULTIPLY THE ENTIRE EQUATION
BY 2 TO MAINTAIN THE EQUALITY, SO WE'D HAVE 6X + 10Y = 8.
HERE WE WOULD HAVE -6X + 9Y = 30.
NOW WE WILL ADD THE 2 EQUATIONS TOGETHER.
THAT'S SOMETIMES WHY IT IS CALLED
THE ADDITION METHOD.
OF COURSE, WHEN WE ADD A 6 AND A -6 THAT WOULD BE 0.
SO THE RESULT WOULD BE 19Y IS EQUAL TO 38.
DIVIDE BY 19.
WE HAVE A Y VALUE EQUAL TO 2.
WE'RE GOING TO TAKE THIS VALUE,
SUB IT INTO ONE OF THESE EQUATIONS AND THEN SOLVE FOR X.
I'LL GO AHEAD AND JUST USE THE FIRST EQUATION. WE HAVE 3X + 5Y,
BUT WE NOW KNOW Y IS 2, SO THAT WOULD BE A 10 = 4.
SUBTRACTING 10 ON BOTH SIDES, DIVIDING BY 3 WE HAVE X = -2.
THEREFORE, OUR SOLUTION TO THIS SYSTEM WOULD BE X = -2, Y = 2,
SO WE CAN CLASSIFY THIS SYSTEM AS CONSISTENT AND INDEPENDENT.
LET'S TAKE A LOOK AT ANOTHER.
THIS TIME I'LL TRY TO ELIMINATE THE Y TERMS.
SINCE WE HAVE A -3Y AND 6Y,
IF WE CAN MAKE THIS A -6Y THEY'D BE OPPOSITES.
SO IN THIS CASE WE ONLY HAVE TO MULTIPLY
THE FIRST EQUATION BY 2.
WE CAN LEAVE THE SECOND EQUATION EXACTLY THE SAME,
SO THE RESULT WOULD BE 4X - 6Y = -2.
SECOND EQUATION REMAINS THE SAME.
NOW, WHEN WE TRY TO ADD THESE TWO EQUATIONS TOGETHER,
NOTICE HOW THE X TERMS AND THE Y TERMS ARE OPPOSITES.
SO ON THE LEFT SIDE OF THIS EQUATION WE'D HAVE 0 = 3.
NOW, IF YOU RECALL FROM THE PREVIOUS VIDEO,
ALGEBRAICALLY IF THIS STATEMENT IS FALSE, WHICH IT IS,
THAT'S THE INDICATION THAT THIS SYSTEM HAS NO SOLUTION
WHICH ALSO MEANS THIS SYSTEM IS INCONSISTENT AND INDEPENDENT.
SO YOU CAN SEE THE ELIMINATION IS A NICE METHOD.
IT'S MUCH FASTER THAN TRYING TO GRAPH THESE LINEAR EQUATIONS.
LET'S GO AHEAD AND TAKE A LOOK AT AN APPLICATION.
SUNSET PHONE OFFERS 2 LONG DISTANCE PLANS.
PLAN A IS $5 PER MONTH AND 8 CENTS A MINUTE.
PLAN B HAS NO MONTHLY FEE BUT CHARGES 12 CENTS A MINUTE.
FOR WHAT NUMBER OF MINUTES WILL THE TWO PLANS COST THE SAME?
SO LET'S LET X = THE NUMBER OF MINUTES.
SO IF WE LET C = THE TOTAL COST FOR THE MONTH FOR PLAN A
WE HAVE A FIXED COST OF $5
AND A VARIABLE COST OF 8 CENTS PER MINUTE.
SO TO FIND THE TOTAL COST FOR THE MONTH
WE'D HAVE TO TAKE THE COST PER MINUTE,
MULTIPLY IT BY THE NUMBER OF MINUTES WHICH IS X
AND THEN ADD THE FIXED COST OF $5.
FOR PLAN B THE TOTAL COST FOR THE MONTH
IS A LITTLE BIT SIMPLER.
IT'S JUST 12 CENTS PER MINUTE
WHICH WOULD REPRESENT AS 0.12 x X.
AGAIN, THERE IS NO FIXED COSTS FOR PLAN B.
NOW THE QUESTION IS FOR WHAT NUMBER OF MINUTES
WILL THESE TWO PLANS BE THE SAME?
SO WE WANT TO KNOW WHEN THESE TWO C'S WILL BE EQUAL,
AND THAT WILL OCCUR WHEN 0.08X + 5 IS EQUAL TO 0.12X.
SO WHAT WE'RE REALLY DOING IS WE'RE PERFORMING SUBSTITUTION.
WE'LL TAKE THIS EXPRESSION AND SUB IT IN FOR THIS C
OR VICE VERSA.
THE RESULT WOULD BE 0.08X + 5.00 = 0.12X.
NOW IF WE DON'T LIKE TO WORK WITH DECIMALS
WE CAN MULTIPLY THIS ENTIRE EQUATION BY 100.
REMEMBER, MULTIPLYING BY 100
MOVES THE DECIMAL TO THE RIGHT 2 PLACES,
SO WE COULD SOLVE THE EQUATION 8X + 500 = 12X.
THEN WE WOULD SUBTRACT 8X ON BOTH SIDES TO GIVE US 4X.
DIVIDE BY 4.
X = 125.
THEREFORE THESE PLANS WOULD COST THE SAME
WHEN 125 MINUTES ARE USED FOR THE SAME MONTH.
I HOPE THAT HELPS EXPLAIN HOW TO USE THE ELIMINATION METHOD.
THANK YOU FOR WATCHING.