- WELCOME TO AN EXAMPLE ON HOW
TO SOLVE A SYSTEM OF NONLINEAR EQUATIONS.
WE WANT TO FIND ALL THE SOLUTIONS TO THE SYSTEM
X SQUARED - 5Y = 6 AND X SQUARED + Y = 18.
WE'LL FIRST SOLVE THIS ALGEBRAICALLY
BY USING THE ELIMINATION OR SUBSTITUTION METHOD
AND THEN WE'LL CHECK OUR SOLUTIONS GRAPHICALLY.
AGAIN WE HAVE THE OPTION OF SOLVING THIS USING
SUBSTITUTION OR ELIMINATION.
I'M GOING TO GO AHEAD AND SOLVE THIS EXAMPLE USING SUBSTITUTION
AND IN THE NEXT EXAMPLE
WE USE ELIMINATION OR THE ADDITION METHOD.
SO LOOKING AT THIS FIRST EQUATION HERE,
NOTICE IF WE SOLVE IT FOR X SQUARED,
WE COULD PERFORM A SUBSTITUTION
FOR X SQUARED IN THE SECOND EQUATION.
SO LOOKING AT THIS FIRST EQUATION,
IF WE WERE TO ADD 5Y TO BOTH SIDES OF THIS EQUATION,
WE WOULD HAVE X SQUARED = 6 + 5Y.
AND SINCE X SQUARED = 6 + 5Y, WE CAN SUBSTITUTE 6 + 5Y
FOR X SQUARED IN THE SECOND EQUATION
WHICH WILL GIVE US AN EQUATION THAT JUST CONTAINS Y
AND THEN WE CAN SOLVE FOR Y AND PERFORM BACK SUBSTITUTION
TO FIND THE VALUES OF X.
SO PERFORMING THIS SUBSTITUTION WE WOULD HAVE 6 + 5Y + Y = 18.
COMBINING LIKE TERMS ON THE LEFT SIDE OF THE EQUATION,
5Y + Y OR 5Y + 1Y WOULD BE 6Y.
SO WE HAVE 6 + 6Y = 18.
NOW TO SOLVE FOR X WE'LL SUBTRACT 6
ON BOTH SIDES OF THE EQUATION.
SO THIS WOULD GIVE US 6Y = 18 - 6 IS 12,
DIVIDE BOTH SIDES BY 6 AND WE KNOW THAT Y MUST EQUAL +2.
BUT EACH SOLUTION CONSISTS OF AN X VALUE AND A Y VALUE
WHICH NOW PERFORM BACK SUBSTITUTION
TO FIND THEIR CORRESPONDING VALUES OF X.
LET'S GO AHEAD AND USE THIS FORM OF THE FIRST EQUATION
THAT'S ALREADY SOLVED FOR X SQUARED.
SO USING X SQUARED = 6 + 5Y,
WE'LL SUBSTITUTE 2 FOR Y AND THEN SOLVE FOR X.
SO IF Y = 2, THAT MEANS X SQUARED = 6 + 5 x 2.
WELL 5 x 2 IS 10, SO WE HAVE X SQUARED = 16.
WE SHOULD RECOGNIZE HERE WE'LL HAVE TWO SOLUTIONS FOR X
BOTH A POSITIVE AND NEGATIVE VALUE
AND SINCE WE HAVE X SQUARED TO UNDO THE SQUARING
WE'LL SQUARE ROOT BOTH SIDES OF THE EQUATION
AND WE'LL INCLUDE A + OR - SIGN ON THE RIGHT TO MAKE SURE
WE OBTAIN BOTH SOLUTIONS.
AND SINCE THE SQUARE ROOT OF 16 = 4, X = + OR -4 OR + OR -4.
SO WHEN Y = 2, X CAN BE EITHER + 4 OR A -4
SO WE HAVE TWO SOLUTIONS.
WE'LL GO AHEAD AND LIST THEM AS ORDERED PAIRS
WHEN X IS +4Y IS +2 OR WHEN X IS -4 Y IS ALSO +2.
REMEMBER WHEN GIVING THE SOLUTION AS AN ORDERED PAIR,
THE FIRST COORDINATE IS ALWAYS
THE X COORDINATE AND THE SECOND COORDINATE
IS ALWAYS THE Y COORDINATE.
LET'S GO AHEAD AND CHECK THIS GRAPHICALLY BY GRAPHING
BOTH OF THE ORIGINAL EQUATIONS ON THE COORDINATE PLANE
AND THESE TWO POINTS SHOULD BE WHERE THE TWO GRAPHS INTERSECT.
NOTICE HOW WE HAVE TWO POINTS OF INTERSECTION,
ONE POINT HERE AND ONE POINT HERE.
THE POINT IN THE FIRST QUADRANT HAS AN X COORDINATE OF +4
AND A Y COORDINATE OF +2.
AND THE POINT IN THE SECOND QUADRANT
HAS AN X COORDINATE OF -4 AND A Y COORDINATE OF +2.
SO THESE TWO POINTS OF INTERSECTION
VERIFY OUR SOLUTION IS CORRECT.
HOPE YOU FOUND THIS EXPLANATION HELPFUL.