- WE WANT TO SOLVE THE SYSTEM OF EQUATIONS
USING THE SUBSTITUTION METHOD,
WHICH IS AN ALGEBRAIC METHOD FOR SOLVING A SYSTEM.
THE SOLUTION TO A SYSTEM OF EQUATIONS
IS AN ORDERED PAIR, OR X AND Y VALUE,
THAT SATISFIES BOTH EQUATIONS.
SO TO KEEP THINGS ORGANIZED LET'S CALL THIS EQUATION ONE
AND THIS EQUATION TWO.
THE FIRST STEP IN THE SUBSTITUTION METHOD
IS TO SOLVE ONE OF THE EQUATIONS FOR ONE OF THE VARIABLES.
NOTICE IF WE LOOK AT THE FIRST EQUATION
WE CAN PRETTY EASILY SOLVE THIS EQUATION FOR X
JUST BY ADDING 4Y TO BOTH SIDES OF THE EQUATION,
SO LET'S START BY DOING THAT.
SO HERE WE'LL ADD 4Y TO BOTH SIDES OF THE EQUATION.
THIS WOULD BE ZERO, SO NOW WE HAVE X = 4 Y - 8.
THE SECOND STEP IS TO USE THIS EQUATION
AND PERFORM SUBSTITUTION INTO THE OTHER EQUATION
OR IN THIS CASE EQUATION TWO.
IF WE KNOW X MUST EQUAL 4Y - 8,
WE CAN SUBSTITUTE 4Y - 8 INTO THE SECOND EQUATION FOR X.
SO BY PERFORMING THIS SUBSTITUTION EQUATION TWO
WOULD BE 3 x,
INSTEAD OF X WE'D HAVE 4Y - 8 + 8Y = 24.
AND NOTICE HOW NOW WE HAVE AN EQUATION WITH ONE VARIABLE,
SO NOW WE'LL SOLVE THIS EQUATION FOR Y
AND THEN PERFORM THAT SUBSTITUTION TO DETERMINE X.
SO THE FIRST STEP HERE IS TO DISTRIBUTE.
SO WE'LL HAVE 12Y - 24 + 8Y = 24.
HERE WE HAVE TWO Y TERMS.
12Y + 8Y = 20Y.
SO WE HAVE 20Y - 24 = 24.
NOW WE HAVE A BASIC TWO STEP EQUATION,
SO WE'LL ADD 24 TO BOTH SIDES.
HERE WE HAVE 20Y = 48 AND NOW WE'LL DIVIDE BOTH SIDES BY 20.
SO WE HAVE Y = 48/20,
BUT THIS DOES SIMPLIFY.
THESE BOTH HAVE A COMMON FACTOR OF 4.
48 IS 4 x 12 AND 20 IS 4 x 5.
THESE 4 SIMPLIFY TO 1 AND WE'RE LEFT WITH Y = 12/5.
NOW REMEMBER THE SOLUTION CONSISTS OF AN ORDERED PAIR
OR AN X AND Y VALUE.
SO NOW WE KNOW THAT Y IS GOING TO 12/5,
BUT WE STILL HAVE TO GO BACK AND DETERMINE THE VALUE OF X.
TO DO THIS I THINK WE'LL USE THE FORM OF EQUATION ONE
THAT'S ALREADY SOLVED FOR X.
SO USING EQUATION ONE AND Y = 12/5,
WE WOULD HAVE X = 4 x Y OR 4 x 12/5 - 8.
SO WE CAN THINK OF THIS AS 4/1,
BUT NOTHINGS GOING TO SIMPLIFY HERE
SO WE'RE GOING TO HAVE X = 48/5 MINUS,
THEN HERE WE CAN THINK OF 8 AS 8/1.
WE KNOW WE HAVE TO HAVE A COMMON DENOMINATOR
SO I'LL MULTIPLY THIS BY 5/5. SO WE'D HAVE 48/5 - 40/5 = 8/5.
SO WE KNOW THE VALUE FOR X MUST BE 8/5.
SO HERE WE HAVE ONE UNIQUE SOLUTION,
X = 8/5 AND Y = 12/5,
SO WE CAN ALSO SAY THAT THIS SYSTEM IS CONSISTENT
AND THE EQUATIONS ARE INDEPENDENT.
NORMALLY IT'S A GOOD IDEA TO GO BACK
AND PERFORM SUBSTITUTION
TO CHECK THESE SOLUTIONS IN BOTH EQUATIONS,
BUT I'M NOT GOING TO SHOW THAT FOR THIS EXAMPLE.
YOU MAY WANT TO JUST TO VERIFY THEM.