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THIS LECTURE IS ON SOLVING A SYSTEM OF LINEAR EQUATIONS
USING THE SUBSTITUTION METHOD.
SO LET'S WRITE DOWN THE METHOD
AND HOPEFULLY YOU WON'T FORGET IT.
SO THE STEPS INVOLVED
IS FIRST YOU NEED TO ISOLATE A VARIABLE
FROM ONE OF THE EQUATIONS.
SO CHOOSE ONE OF THE EQUATIONS
AND ISOLATE ONE VARIABLE. OKAY.
SO MOST OF THE TIME THE VARIABLES ARE X AND Y
SO BASICALLY WHAT THAT MEANS
IS YOU HAVE TO EITHER GET X ALONE
OR YOU HAVE TO GET Y ALONE.
OKAY. SO YOU HAVE TO ISOLATE ONE VARIABLE.
AND THE NEXT STEP
WHAT YOU NEED TO DO IS TO SUBSTITUTE THE RESULT
INTO THE OTHER EQUATION.
SO THAT'S CRUCIAL.
MAKE SURE YOU PUT IT INTO THE OTHER EQUATION. OKAY.
SO YOU CANNOT PUT BACK INTO THE SAME EQUATION YOU GOT IT FROM.
AND THEN WHAT YOU NEED TO DO IS TO SOLVE THE EQUATION
AND WHEN YOU'RE DONE,
STEP THREE IS ASKING YOU TO CONTINUE
TO FIND THE SOLUTION TO THE SYSTEM.
BUT REMEMBER WHAT THAT MEANS.
THE SOLUTION IS ALWAYS MADE UP OF AN ORDERED PAIR
AND YOU WILL NEED TO GIVE ME TWO NUMBERS. OKAY.
SO AN ORDERED PAIR IS MADE UP OF TWO NUMBERS IN OUR CASE
AND WE NEED THEM NEED AS COORDINATES.
OKAY. SO LET'S LOOK AT THE FIRST EXAMPLE.
SUPPOSEDLY YOU ARE ASKED TO SOLVE
AND SAY THE SYSTEM IS Y - 1.2X = 0.4
AND THE OTHER EQUATION IS 6Y - 5X = 9.
SO IN FRONT OF YOU THERE'S TWO EQUATIONS.
THE TRICK IS TO LOOK FOR A VARIABLE
THAT IS BY ITSELF ALREADY.
LIKE THIS.
SO FROM THE FIRST EQUATION
I'M GOING TO ISOLATE THE VARIABLE.
SO IT'S GOING TO READ Y = 1.2X + 0.4.
OKAY. SO THAT'S MY RESULT THERE.
AND WHAT I WANT TO DO
IS PUT THE RESULT INTO THE OTHER EQUATION
AND IN PARTICULAR I AM PUTTING IT INTO THE PLACE OF Y.
SO WHERE Y IS SUBSTITUTE THAT IN. OKAY.
SO IT'S GOING TO READ 6 x OF IN PLACE OF Y
YOU WILL REPLACE IT BY 1.2X + 0.4 - 5X = 9.
SO ALL WE'RE REPLACING IS JUST THE Y PORTION RIGHT THERE. OKAY.
SO THE Y HAS BEEN REPLACED TO THE 1.2X + 0.4.
LIKE THAT. OKAY.
NOW WE HAVE THIS NEW EQUATION
THAT IS ONLY MADE UP OF ONE VARIABLE.
AND ALL YOU HAVE TO DO NOW IS TO DISTRIBUTE.
SO 1.2 x 6,
IT COMES OUT TO BE 7.2X + 6 x 0.4 IS 2.4 - 5X = 9.
IF YOU COMBINE LIKE TERMS, 7.2X -5X IS 2.2X, + 2.4 = 9.
SO 2.2X IS = 9 - 2.4 WHICH MEANS 2.2X = 6.6.
SO IT'D BE THAT IF YOU DIVIDE BOTH SIDES BY 2.2,
WHAT'S GOING TO HAPPEN IS X = 3.
BUT REMEMBER,
FOR YOU TO SAY IT'S A SOLUTION YOU NEED TO HAVE TWO NUMBERS.
IT'S THE X FOLLOWED BY THE Y. OKAY.
SO WHAT WE FOUND IS ONLY X
SO PUT THE NUMBER IN THE FIRST SPOT.
AND TO FIND THE SECOND NUMBER RIGHT THERE
YOU HAVE TO USE THE EQUATION.
AND I'M GOING TO CHOOSE MY RESULTING EQUATION
WHICH SAYS Y = 1.2X + 0.4 AND SINCE X IS 3,
THAT MEANS I NEED TO MULTIPLY 1.2 x 3
WHICH COMES TO 3.6 + 0.4.
AND THAT MAKES IT 4.
OKAY. SO THAT'S HOW WE SAY THE SOLUTION IS 3,4.
SOMETIMES THE EQUATIONS,
ALL THE VARIABLES ARE SQUEEZED TO ONE SIDE.
FOR EXAMPLE, IN EXAMPLE TWO,
X + Y - 7 = O AND 5X + 12Y - 7 = 0.
SO WHAT YOU HAVE TO DO IS PICK A VARIABLE TO ISOLATE
AND SINCE X IS ALONE, IT'S ALWAYS Y
SO LET'S TRY X, FOR EXAMPLE.
SO X = -Y + 7.
SO WHAT I JUST DID WAS TO MOVE ALL THE OTHER TERMS
TO THE OTHER SIDE OF THE EQUAL SIGN,
LEAVING X ALONE.
OKAY. SO THIS BECOMES MY RESULT.
AND WHAT I'M DOING IS I'M GOING TO PUT THE RESULT
INTO THE OTHER EQUATION AND IN PARTICULAR
I AM SUBSTITUTING INTO WHERE X USED TO BE.
SO IT'S GOING TO BE READING 5 x OF.
IN PLACE OF X IT'S NOW GOING TO READ -Y + 7. OKAY.
BUT THERE'S OTHER TERMS SO CONTINUE WRITING,
IT'S GOING TO BE + 12Y - 7 = 0.
SO ALL THAT HAS CHANGED
IS THE X IS NOW BEEN REPLACED BY -Y + 7. OKAY.
SO NOW THAT I HAVE AN EQUATION IN ONE VARIABLE
I'M GOING TO DISTRIBUTE
WHICH GIVES IT -5Y + 35 + 12Y - 7 = 0.
IF I COMBINE LIKE TERMS,
-5Y + 12Y = 7Y AND 35 - 7 MAKES IT 28 POSITIVE.
SO BRING THE NUMBER TO THE OTHER SIDE OF THE EQUAL SIGN,
MAKES IT -28 AND WHEN I DIVIDE BY 7 BOTH SIDES,
TELLS ME Y = -4.
SO THIS IS JUST TO REMIND ME
THAT THE SOLUTION IS MADE UP OF TWO NUMBERS
AND THIS TIME BECAUSE THE NUMBER WE FOUND WAS FOR Y,
IT SITS IN THE SECOND POSITION.
AND SO FOR ME TO FIND THE FIRST POSITION
I NEED TO USE ONE OF THE EQUATIONS.
I'M GOING TO CHOOSE MY RESULTING EQUATION WHICH SAYS X = -Y + 7.
OKAY, THE ONE IN THE BOX.
AND I'M REPLACING THE Y BY -4.
WATCH OUT.
ALL YOU'RE REPLACING IS JUST THE Y.
SO THIS MINUS ACTUALLY STAYS. OKAY.
SO IT'S GOING TO BE -4 + 7 WHICH MAKES IT 4 + 7
AND THAT MAKES IT 11.
SO THE SOLUTION IS GOING TO BE 11,-4.
SO IF YOU RECALL,
A SYSTEM OF EQUATIONS BASICALLY
IS LOOKING FOR THE POINT WHERE THEY MEET, OKAY,
AND SO THIS COORDINATES,
11,-4 IS WHERE THE TWO LINES MEET ON A GRAPH.
NOW LET'S CONSIDER EXAMPLE THREE.
AGAIN, WE ARE ASKING YOU TO SOLVE
AND THE SYSTEM GIVEN TO YOU IS X = 4 - Y AND 4X + 4Y = 16.
NOW IN THIS QUESTION YOU NOTICE THAT THE FIRST EQUATION
IS ALREADY NICELY ISOLATED FOR YOU.
SO THAT STRAIGHTAWAY BECOMES MY RESULT
AND I'M GOING TO PUT THIS RESULTING
INTO THE SECOND EQUATION WHERE THE X USED TO BE.
SO IT'S GOING TO READ 4 x OF.
IN PLACE OF X IT'S GOING TO READ 4 - Y + 4Y = 16.
NOW, SAME THING. I'M GOING TO DISTRIBUTE.
IT'S GOING TO READ 16 - 4Y + 4Y = 16.
THE STRANGE THING ABOUT THIS IS YOU NOTICE
THE VARIABLES CANCEL. OKAY.
AND IN ITS PLACE I END UP WITH 16 = 16.
SO WHAT I WAS TELLING YOU IN CLASS
IS WHEN THE NUMBERS MATCH,
IT MEANS THE TWO LINES ARE COINCIDENT LINES.
THAT MEANS THEY ARE ACTUALLY ON TOP OF EACH OTHER.
SO THERE ARE INFINITELY MANY SOLUTIONS.
SO THE TYPE OF LINES THAT YOU WILL SEE
IS WHAT YOU CALL GRACING THE LINES,
MEANING IF YOU WERE TO DRAW THEM ON THE GRAPH
THEY ARE RIGHT ON TOP OF EACH OTHER
AND BEING RIGHT ON TOP OF EACH OTHER
THAT MEANS THEY PRETTY MUCH MEET EVERYWHERE ON THE LINE.
AND IF THEY MEET EVERYWHERE ON THE LINE.
THERE IS INFINITELY MANY OF THEM.
OKAY. LET'S LOOK AT EXAMPLE FOUR.
SUPPOSEDLY YOU'RE ASKED THIS TIME TO SOLVE
1.5X - Y = 1, 6X = 4Y + 7.
SO I'M STILL LOOKING AT THE FIRST EQUATION
BECAUSE I SEE THE Y BY ITSELF AND I NEED TO GET Y ALONE.
SO IT'S GOING TO READ -Y = -1.5X + 1.
BUT I CANNOT HAVE -Y
SO I'M GOING TO DIVIDE EVERYBODY BY -1
WHICH MAKES IT Y = 1.5X - 1.
SO THAT BECOMES MY RESULT.
AND I'M GOING TO USE THAT RESULT INTO THE OTHER EQUATION
WHERE Y USED TO BE AND SO IT'S GOING TO READ 6X = 4 x OF.
IN PLACE OF Y I'M NOW GOING TO PUT IN 1.5X - 1 + 7. OKAY.
SO WHAT'S GOING TO HAPPEN NOW
WHEN I DISTRIBUTE IS 4 x 1.5X = 6X
AND 4 x -1 = -4.
SO WHEN I BRING THE 6X TO THE SAME SIDE
WHAT'S GOING TO HAPPEN IS
I END UP WITH 0 ON ONE SIDE AND 3 ON THE OTHER SIDE.
SO THIS TIME, UNLIKE THE OTHER EXAMPLE,
THE NUMBERS DON'T MATCH.
SO WHEN NUMBERS DON'T MATCH,
OKAY, THIS MEANS THAT THE TWO LINES ARE PARALLEL LINES.
SO IF YOU LOOKING FOR POINT OF INTERSECTION
BUT THE LINES ARE PARALLEL,
THAT'S WHEN WE SAY THAT THE SYSTEM HAS NO SOLUTION
BECAUSE THERE IS NO POINT OF MEETING.
IN SUMMARY,
YOU WILL NOTICE THERE ARE THREE TYPES OF SYSTEMS.
FIRST TYPE IS WHERE WE SAY THAT THEY ARE CONSISTENT
AND BASICALLY THEY ARE MADE UP OF INTERSECTING LINES.
AND YOU NOTICE BECAUSE THE LINES ARE INTERSECTING
YOU WILL NOTICE THAT YOU END UP WITH ONLY ONE SOLUTION POINT.
AND THE SOLUTION ALWAYS ENDS UP WITH AN ORDERED PAIR,
SOMETHING LIKE THAT. OKAY.
NOW, THE OTHER TYPE THAT WE SAW AS IN EXAMPLE THREE
WAS WHERE WE ACTUALLY SAW THE DEPENDENT SYSTEM
AND WHAT DEPENDENT SYSTEM IS MADE UP OF
IS THEY ARE MADE UP OF COINCIDENT LINES.
SO THEY ARE BASICALLY THE SAME LINES.
SO WHEN THEY ARE THE SAME LINES
YOU WILL NOTICE THAT THERE IS INFINITELY MANY SOLUTIONS
BECAUSE THEY MEET EVERYWHERE ON THE TWO LINES.
SO THIS WAS SEEN IN THE EXAMPLE ONE AND TWO
AND THIS WAS SEEN IN EXAMPLE THREE.
AND IN EXAMPLE FOUR,
I WAS SHOWING YOU THE FOURTH TYPE
WHICH WE CALL INCONSISTENT.
AND THIS IS MADE UP OF PARALLEL LINES.
AND FOR THIS KIND THERE ARE NO SOLUTIONS. �