- WE WANT TO SOLVE THE SYSTEM OF THREE EQUATIONS
WITH THREE UNKNOWNS USING THE ELIMINATION METHOD.
TO SOLVE A SYSTEM LIKE THIS, WE WANT TO FIND THE VALUES OF X, Y,
AND Z THAT WOULD SATISFY ALL THREE EQUATIONS,
AND WHEN SOLVING A SYSTEM LIKE THIS
THERE ARE ACTUALLY THREE POSSIBILITIES.
WE CAN HAVE ONE SOLUTION, NO SOLUTION,
OR AN INFINITE NUMBER OF SOLUTIONS.
SO BEFORE WE SOLVE THIS, TO GET A BETTER UNDERSTANDING OF THIS,
IF WE GRAPH THESE IN SPACE, EACH OF THESE WOULD BE A PLANE.
SO IF WE GRAPH THREE PLANES IN SPACE,
THERE ARE THREE POSSIBILITIES.
AS WE SEE HERE, THE PLANES COULD INTERSECT IN ONE POINT
WHICH MEANS A SYSTEM WOULD HAVE ONE SOLUTION.
SO WE CAN SAY THE SYSTEM IS CONSISTENT AND INDEPENDENT.
WE SAY CONSISTENT BECAUSE IT HAS A SOLUTION
AND INDEPENDENT BECAUSE IT HAS ONE SOLUTION.
IT'S ALSO POSSIBLE IF WE GRAPH THREE PLANES IN SPACE,
THERE AREN'T ANY POINTS THAT ARE ON ALL THREE PLANES.
THEREFORE THE SYSTEM WOULD HAVE NO SOLUTIONS,
AND WE CAN SAY THE SYSTEM IS INCONSISTENT.
THE THIRD POSSIBILITY IS THERE ARE AN INFINITE NUMBER OF POINTS
ON ALL THREE PLANES, AS WE SEE HERE.
NOTICE OUR THREE PLANES INTERSECT IN THIS YELLOW LINE,
AND THERE IS AN INFINITE NUMBER OF POINTS ON A LINE.
THEREFORE WE HAVE AN INFINITE NUMBER OF SOLUTIONS,
SO WE CAN SAY THE SYSTEM IS CONSISTENT AND DEPENDENT,
AGAIN, CONSISTENT BECAUSE IT HAS A SOLUTION AND DEPENDENT
BECAUSE IT HAS AN INFINITE NUMBER OF SOLUTIONS.
SO BEFORE WE SOLVE OUR SYSTEM,
LET'S REVIEW THE ELIMINATION METHOD.
THE FIRST STEP IS TO SELECT ANY TWO EQUATIONS
AND ELIMINATE ANY ONE OF THE VARIABLES
FROM THE TWO EQUATIONS.
AND THEN WE'LL SELECT TWO OTHER EQUATIONS
AND ELIMINATE THE SAME VARIABLE AGAIN.
THIS WILL GIVE US TWO EQUATIONS WITH TWO UNKNOWNS
WHICH WE'LL SOLVE USING THE ELIMINATION METHOD
AND THEN PERFORM BACK SUBSTITUTION
TO SOLVE FOR ALL THE VARIABLES.
SO FOR THE FIRST STEP, LET'S GO AHEAD
AND LABEL THESE EQUATIONS SO WE CAN STAY ORGANIZED.
LET'S CALL THIS EQUATION "A", EQUATION B, AND EQUATION C.
WE FIRST WANT TO SELECT TWO EQUATIONS
AND ELIMINATE ONE OF THE VARIABLES.
IT DOESN'T MATTER WHICH EQUATIONS WE SELECT
OR WHICH VARIABLE WE WANT TO ELIMINATE.
LET'S GO AHEAD AND SELECT EQUATION B AND C
AND ELIMINATE THE Z TERMS.
NOTICE HOW THEY'RE ALREADY OPPOSITE SIGNS,
SO IF THIS WAS + 4Z WE COULD ADD THE TWO EQUATIONS TOGETHER
AND ELIMINATE Z.
SO LET'S GO AHEAD AND MULTIPLY EQUATION C BY 2 AND ADD IT TO B.
SO WE'LL HAVE 2 x EQUATION C + EQUATION B.
2 x C WOULD BE 2X - 2Y + 4Z = -8.
WE'RE GOING TO LEAVE EQUATION B THE SAME,
AND NOW WE'LL ADD THE TWO EQUATIONS.
NOTICE THAT THE Z TERMS ARE OPPOSITES,
THEREFORE THIS SUM WOULD BE ZERO.
SO WE HAVE 4X + Y = -4.
WE'LL USE THIS EQUATION AGAIN IN JUST A MOMENT.
NOW WE'LL SELECT TWO OTHER EQUATIONS,
BUT WE DO WANT TO ELIMINATE Z AGAIN.
SO NOW LET'S USE EQUATION "A" AND EQUATION C.
AGAIN, NOTICE HOW THE Z TERMS ARE OPPOSITES,
SO AGAIN, IF THIS WAS + 4Z
AND WE ADDED THE TWO EQUATIONS TOGETHER,
THE Z TERM WOULD BE ELIMINATED.
SO WE'RE GOING TO MULTIPLY EQUATION C BY 2 AGAIN
BUT NOW ADD IT TO EQUATION "A".
SO NOW WE'LL HAVE 2 x EQUATION C + EQUATION "A".
WE JUST MULTIPLIED EQUATION C BY 2.
IT WAS 2X - 2Y + 4Z = -8.
WE'RE GOING TO ADD THIS TO EQUATION "A"
WHICH IS 3X + Y - 4Z = -6
SO WE'LL ADD THESE TWO EQUATIONS TOGETHER.
NOTICE HOW THE Z TERMS AGAIN ARE OPPOSITES,
BECAUSE THEY SUM TO ZERO, SO WE HAVE 5X - Y = -14.
NOTICE WHAT WE'VE DONE NOW IS WE'VE CREATED TWO EQUATIONS
WITH TWO UNKNOWNS,
SO WE'LL SOLVE THIS AS A SYSTEM USING ELIMINATION.
SO NOW WE WANT TO SOLVE THE SYSTEM
4X + Y = -4 AND 5X - Y = -14.
NOW, WE KIND OF LUCKED OUT HERE
BECAUSE NOTICE HOW THE Y TERMS ARE ALREADY OPPOSITES,
SO WE CAN ADD THESE TWO EQUATIONS TOGETHER
AND ELIMINATE Y.
SOMETIMES THIS WON'T BE THE CASE,
AND WE'D HAVE TO MULTIPLY ONE OR BOTH OF THE EQUATIONS
BY A CONSTANT SO EITHER THE X OR Y TERMS WOULD BE OPPOSITES,
SIMILAR TO WHAT WE DID OVER HERE FOR Z.
HERE WE CAN GO AHEAD AND ADD THESE EQUATIONS.
THIS WOULD BE ZERO, SO WE HAVE 9X = -18.
DIVIDE BOTH SIDES BY 9,
SO NOW WE KNOW THAT THE X-COORDINATE OF OUR SOLUTION
MUST BE -2.
BUT REMEMBER, WE'RE NOT DONE.
WE'RE TRYING TO FIND THE X, Y, AND Z VALUE
THAT SATISFY THIS SYSTEM,
SO NOW I'LL PERFORM BACK SUBSTITUTION TO FIND Y AND Z.
TO DO THIS, WE'LL START BY FINDING AN EQUATION
THAT HAS JUST X AND ONE OTHER VARIABLE.
SO AGAIN, TO KEEP ORGANIZED LET'S CALL THIS EQUATION D
AND CALL THIS EQUATION E.
NOTICE IF WE USED EQUATION D,
WE COULD SOLVE IT FOR Y BY SUBTRACTING 4X ON BOTH SIDES.
WE CAN SAY Y = -4X - 4.
WE CAN SUBSTITUTE -2 FOR X AND SOLVE FOR Y.
SO Y = -4 x -2 - 4.
THIS WOULD 8 - 4, SO NOW WE KNOW Y = 4.
NOW, WE STILL HAVE TO FIND Z.
TO DO THIS, WE'LL GO BACK TO ONE OF THE ORIGINAL EQUATIONS
AND SUBSTITUTE VALUES FOR X AND Y SO THAT WE CAN FIND Z.
LET'S GO AHEAD AND USE EQUATION C.
SO USING EQUATION C WE KNOW X IS -2,
SO WE'D HAVE -2 - Y, BUT Y IS 4, SO - 4 + 2Z = -4.
WELL, THIS WOULD BE -6 + 2Z = -4.
ADD 6 TO BOTH SIDES.
WE HAVE 2Z = 2, SO Z = 1.
SO NOTICE HOW THIS SYSTEM HAS ONE SOLUTION,
MEANING THAT THESE THREE PLANES WOULD INTERSECT IN THE POINT
WHERE THE X-COORDINATE IS -2, THE Y-COORDINATE IS 4,
AND THE Z-COORDINATE IS 1.
AND IN FACT, MANY TIMES YOU'LL SEE THE SOLUTION
GIVEN AS AN ORDERED TRIPLE,
SO WE COULD SAY THE SOLUTION IS THE POINT (-2,4,1).
AND WE'LL GO AHEAD AND TAKE A LOOK AT THIS IN JUST A MOMENT.
AGAIN, IF WE GRAPH THIS IN SPACE WE WOULD HAVE THREE PLANES,
AND WE'RE SAYING THAT THE POINT OF INTERSECTION
WOULD BE THIS POINT HERE.
WELL, HERE IS THE GRAPH OF THE THREE PLANES IN SPACE.
THIS POINT RIGHT HERE IS OUR SOLUTION
BECAUSE IT'S THE ONLY POINT THAT'S ON ALL THREE PLANES.
LET'S GO AHEAD AND ROTATE THIS TO THE OTHER SIDE,
SO WE CAN SEE IT AGAIN.
AGAIN, IT WOULD BE THIS POINT THAT WE SEE RIGHT HERE
WHICH WE FOUND USING THE ELIMINATION METHOD.
I HOPE YOU FOUND THIS EXAMPLE HELPFUL.
WE'LL TAKE A LOOK AT SEVERAL MORE EXAMPLES
IN THE NEXT FEW VIDEOS.