SECTION 7.3
“Solving Linear Systems by
Elimination”
WHAT’S IMPORTANT:
-- Be able to solve a system of linear
equations by linear combinations.
USING ALGEBRAIC METHODS TO SOLVE SYSTEMS
If neither variable has a coefficient of 1 or –1, you can still use
substitution. In such cases, however, the linear combination
method may be better. The goal of this method is to add the
equations to obtain an equation in one variable.
USING ALGEBRAIC METHODS TO SOLVE SYSTEMS
THE LINEAR COMBINATION METHOD
1
Multiply one or both equations by a constant to
obtain coefficients that differ only in sign for one
of the variables.
2
Add revised equations from Step 1. Combine like
terms to eliminate one of the variables. Solve for
remaining variable.
3
Substitute value obtained in Step 2 into either
original equation and solve for other variable.
The Linear Combination Method: Multiplying One Equation
4x + 3y  16
2x – 3y  8
Equation 1
Equation 2
Add the equations
4x + 3y  16
2x – 3y  8
6x
 24
x4
4(4) + 3y  16
16 + 3y  16
- 16
- 16
3y  0 y  0
( 𝟒 ,𝟎 )
𝟏. x + 3y  6
x – 3y  12
Add the equations
Equation 1
Equation 2
𝟐. −𝒙 + 𝟒𝒚 = −𝟏
𝒙 − 𝟑𝒚 = 𝟏𝟏
Equation 1
Equation 2
Add the equations
x + 3y  6
−𝒙 + 𝟒𝒚 = −𝟏 𝒙 −
x – 3y  12
𝟑𝒚 = 𝟏𝟏
𝒚 = 𝟏𝟎
2x  18
(𝟒𝟏 , 𝟏𝟎)
x9
(𝟗 , −𝟏) −𝒙 + 𝟒 𝟏𝟎 = −𝟏
(9) + 3y  6
−𝒙 + 𝟒𝟎 = −𝟏
-9
-9
−𝟒𝟎 − 𝟒𝟎
−𝒙 = −𝟒𝟏
3y  - 3
𝒙 = 𝟒𝟏
𝒚 = −𝟏
2x – 4y  13
4x – 5y  8
Equation 1
Equation 2
Multiply the first equation by – 2 so that x-coefficients differ only in sign.
2x –4y  13
4x –5y  8
•
(–2)
–4x + 8y  – 26
4x – 5y  8
3y  –18
y  –6
Add the revised equations
and solve for y.
2x – 4y  13
2x – 4(–6)  13
2x + 24  13
- 24 -24
2x  -11
𝟏𝟏
𝒙=−
𝟐
Write Equation 1.
Substitute −𝟔 for y.
𝟏𝟏
−
𝟐
, −𝟔
𝟏. x + 3y  3
x + 6y  3
Equation 1
Equation 2
x + 3y  3
x + 6y  3
•
(–2)
– 2x − 𝟔𝒚  – 6
x + 6y  3
−𝒙 = −𝟑
x3
( 𝟑 ,𝟎 )
y0
𝟐. 𝒗 − 𝒘 = −𝟓
𝒗 + 𝟐𝒘 = 𝟒
𝒗 − 𝒘 = −𝟓
𝒗 + 𝟐𝒘 = 𝟒
( −𝟐 , 𝟑 )
3 + 3y  3
−3
−3
3y  0
Equation 1
Equation 2
• (2)
𝟐𝒗 − 𝟐𝒘 = −𝟏𝟎
𝒗 + 𝟐𝒘 = 𝟒
𝟑𝒗  – 6
v  −𝟐
−𝟐 −𝒘  −𝟓
+𝟐
+𝟐
𝒘= 𝟑
−𝒘  − 𝟑
Write an equation of the following:
1. A wedding planner placed an order for eight
centerpieces and five glasses, and the bill was
$106. Let 𝒙 represent the cost of one centerpiece,
and let 𝒚 represent the cost of one glass.
𝟖𝒙 + 𝟓𝒚 = 𝟏𝟎𝟔
2. The planner later realized the wedding was short
one centerpiece and six glasses. She placed
another order for the missing items. The order
came to $24.
𝒙 + 𝟔𝒚 = 𝟐𝟒
A wedding planner placed an order for eight
centerpieces and five glasses, and the bill was $106.
Let 𝒙 represent the cost of one centerpiece, and let
𝒚 represent the cost of one glass. The planner later
realized the wedding was short one centerpiece and
six glasses. She placed another order for the missing
items. The order came to $24. What was the cost of
each?
𝟖𝒙 + 𝟓𝒚 = 𝟏𝟎𝟔
𝒙 + 𝟔𝒚 = 𝟐𝟒
∙ (−8))
Glass = $𝟐
Centerpiece = $𝟏𝟐
𝟖𝒙 + 𝟓𝒚 = 𝟏𝟎𝟔
−𝟖𝒙 − 𝟒𝟖𝒚 = −𝟏𝟗𝟐
−𝟒𝟑𝒚 = −𝟖𝟔
𝒚=𝟐
𝒙 + 𝟔𝒚 = 𝟐𝟒
𝒙 + 𝟔(𝟐) = 𝟐𝟒
𝒙 + 𝟏𝟐 = 𝟐𝟒
𝒙 = 𝟏𝟐
𝟏. 𝒙 + 𝟐𝒚 = 𝟓
𝟓𝒙 − 𝒚 = 𝟑
Equation 1
Equation 2
𝒙 + 𝟐𝒚 = 𝟓
𝟓𝒙 − 𝒚 = 𝟑
• (2)
( 𝟏 ,𝟐 )
𝟐. 𝟑𝒃 + 𝟐𝒄 = 𝟒𝟔
Equation 1
𝒃 + 𝟓𝒄 = 𝟏𝟏
Equation 2
𝟑𝒃 + 𝟐𝒄 = 𝟒𝟔
𝒃 + 𝟓𝒄 = 𝟏𝟏
• (-3)
𝒙 + 𝟐𝒚 = 𝟓
𝟏𝟎𝒙 − 𝟐𝒚 = 𝟔
𝟏𝟏𝒙  11
x𝟏
𝟏 + 𝟐𝒚 = 𝟓
−𝟏
−𝟏
𝟐𝒚 = 𝟒
𝒚=𝟐
𝟑𝒃 + 𝟐𝒄 = 𝟒𝟔
−𝟑𝒃 − 𝟏𝟓𝒄 = −𝟑𝟑
−𝟏𝟑𝒄 = 𝟏𝟑
𝒄 = −𝟏
(𝟏𝟔 , −𝟏 )
𝟑𝒃 + 𝟐(−𝟏) = 𝟒𝟔
𝟑𝒃 − 𝟐 = 𝟒𝟔
𝒃 = 𝟏𝟔
𝟑𝒃 = 𝟒𝟖