Page 189 - Math Course 2 (Book 1)

Systems of Equations: Elimination

Mo. 6

Lesson 3
Let’s Begin

KEY CONCEPTS:
1. Solve systems of equations algebraically Elimination Using Addition
by using elimination with addition.
2. Solve systems of equations algebraically
by using elimination. Example

Use elimination to solve the system of equations.
–3x + 4y = 12
MO. 6 - L3a 3x – 6y = 18

Solving Systems of Equations Since the coef cients of the x terms, –3 and 3, are
by Elimination additive inverses, you can eliminate the x terms by
adding the equations.

Vocabulary A-Z –3x + 4y = 12 Write the equations in
column form and add.
Let us learn some vocabulary (+) 3x –6y = 18
Notice that the x variable
–2y = 30
is eliminated.

Elimination –2y = 30 Divide each side by –2.
–2 = –2
For two algebra equations with 2 different
variables, elimination is used to eliminate one
variable to solve for another variable. y = –15 Simplify.

y = x + 1 y = –x Now substitute –15 for y in either equation to f nd
the value of x.

y = x + 1 –3x + 4y = 12 First equation
+ y = –x
Eliminate X –3x + 4(–15) = 12 Replace y with –15.
2y = 1
1
y = –3x – 60 = 12 Simplify.
2
–3x – 60 + 60 = 12 + 60 Add 60 to each side.

y = –x –3x = 72 Simplify.
Substitute –3x 72
1 = Divide each side by –3.
– = x –3 –3
2
x = –24 Simplify.
1 1
–
,
Solution( )
2 2
Answer The solution is (–24, –15).

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