# Ex 2: Solve a System of Equations Using the Elimination Method – WE WANT TO SOLVE THE SYSTEM
OF LINEAR EQUATIONS USING THE ELIMINATION METHOD
OR THE ADDITION METHOD. A SOLUTION TO A SYSTEM OF
EQUATIONS IS AN ORDERED PAIR OR AN X AND Y VALUE THAT WOULD SATISFY BOTH
OF THE GIVEN EQUATIONS. AND THE IDEA BEHIND
THE ELIMINATION METHOD IS YOU WANT TO ADD THESE
TWO EQUATIONS TOGETHER AND WHEN DOING SO, ELIMINATE
ONE OF THE VARIABLES. THE ONLY WAY WE CAN ELIMINATE
A VARIABLE WHEN WE ADD THESE EQUATIONS
TOGETHER IS IF THE X TERMS OR Y TERMS
ARE OPPOSITES. SO ONCE THE EQUATIONS
ARE IN STANDARD FORM, AS WE SEE HERE WITH THE X AND Y
TERMS ON THE LEFT AND THE CONSTANTS ON THE RIGHT, RIGHT AWAY WE SHOULD RECOGNIZED
THAT THE X TERMS AND Y TERMS ARE NOT OPPOSITES. EITHER ONE OR BOTH EQUATIONS
BY A CONSTANT SO THAT THEY ARE. IF YOU LOOK AT THE X TERMS, IF WE WANT THE X TERMS
TO BE OPPOSITES, SINCE WE HAVE 3X AND 7X WE’D
HAVE TO HAVE A -21X AND +21X. SO WE’D HAVE TO MULTIPLY THIS
FIRST EQUATION BY EITHER 7 OR -7 AND THE SECOND EQUATION BY +3
OR -3. SO WE’D HAVE TO MULTIPLY
BOTH EQUATIONS BY A CONSTANT, BUT IF YOU LOOK AT THE Y TERMS, HERE WE HAVE +1Y
AND HERE WE HAVE +5Y. IF THIS WAS -5Y, THE Y TERMS
WOULD BE OPPOSITES. SO WE CAN MAKE THE Y TERMS
OPPOSITES IF WE MULTIPLY THE FIRST
EQUATION BY -5. LET’S GO AHEAD AND DO THAT. WE’RE GOING TO MULTIPLY BOTH
SIDES OF THE EQUATION BY -5, WHICH WOULD GIVE US
-15X – 5Y=+50. BECAUSE THE Y TERM
IS ALREADY + 5Y WE’RE GOING TO LEAVE THE SECOND
EQUATION THE SAME. SO WE’LL HAVE 7X + 5Y=-18. AND NOW WHEN WE ADD THESE TWO
EQUATIONS TOGETHER NOTICE HOW WE HAVE -5Y + 5Y. THE Y TERMS WOULD HAVE A SUM
OF ZERO. THE SUM OF THE X TERMS WOULD BE
-8X AND OVER HERE WE HAVE +32. SO NOW WE CAN DIVIDE BOTH SIDES
BY -8, AND NOW WE KNOW X=-4. WE’RE NOT DONE, THOUGH, BECAUSE REMEMBER THE SOLUTION
IS AN ORDERED PAIR. WE KNOW X=-4, AND NOW WE HAVE
TO PERFORM SUBSTITUTION INTO EQUATION ONE OR EQUATION
TWO AND THEN SOLVE FOR Y. LET’S GO AHEAD AND USE EQUATION
ONE AND SUBSTITUTE -4 FOR X. WE WOULD HAVE 3 x -4 + Y=-10. SO THIS WOULD BE -12 + Y=-10. SO WE’D ADD 12 TO BOTH SIDES. AND HERE WE HAVE Y=+2. SO OUR SOLUTION IS X=-4
AND Y=2. AND SINCE WE HAVE ONE UNIQUE
SOLUTION, WE CAN SAY THE SYSTEM
IS CONSISTENT BECAUSE IT HAS AT LEAST
ONE SOLUTION AND THE EQUATIONS
ARE INDEPENDENT BECAUSE THEY ARE DIFFERENT. NOW LET’S GO AHEAD AND VERIFY
OUR SOLUTION. WE’LL SUBSTITUTE X=-4 AND
Y=2 INTO BOTH EQUATION ONE AND EQUATION TWO. SO FOR EQUATION ONE WE’D HAVE
3 x -4 + 2=-10. THIS WOULD BE -12 + 2=-10. SO THIS SATISFIES EQUATION ONE, AND NOW LET’S CHECK
EQUATION TWO. WE WOULD HAVE 7 x -4 + 5 x 2
=-18. SO WE HAVE -28 + 10=-18,
WHICH IS ALSO TRUE. AND WE’LL GO AHEAD AND STOP HERE
FOR THIS EXAMPLE. I HOPE YOU FOUND IT HELPFUL.

### 2 thoughts on “Ex 2: Solve a System of Equations Using the Elimination Method”

• November 28, 2015 at 6:36 pm
• 