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Algebra 2
Solving Systems Algebraically
Lesson 3-2 Part 2
Goals
Goal
• To solve a linear systems
using elimination.
Rubric
Level 1 – Know the goals.
Level 2 – Fully understand the
goals.
Level 3 – Use the goals to
solve simple problems.
Level 4 – Use the goals to
solve more advanced problems.
Level 5 – Adapts and applies
the goals to different and more
complex problems.
Vocabulary
• Equivalent Systems
Essential Question
Big Idea: Solving Equations and
Inequalities
• When can you use elimination to solve a system of
linear equation?
Definition
• You can also solve systems of linear equations with
the elimination method.
• With elimination, you get rid of one of the variables
by adding the two equations of the system.
• You may have to multiply one or both equations by a
number to create variable terms that can be
eliminated.
• The elimination method is sometimes called the
addition method or linear combination.
Elimination Method
• Like substitution, the goal of elimination is to get one
equation that has only one variable. To do this by
elimination, you add the two equations in the system
together.
• To eliminate a variable, the coefficients of that
variable in the two equations must be additive
inverses. To achieve this, we use properties of algebra
to change the system to an equivalent system, one
with the same solution set. The three transformations
that produce an equivalent system are listed next.
Elimination Method
Transformations of a Linear
System
1. Interchange any two equations of the system.
2. Multiply or divide any equation of the system
by a nonzero real number.
3. Replace any equation of the system by the sum
of that equation and a multiple of another
equation in the system.
Elimination Method
• When adding the two equations in the
system together to eliminate one of the
variables, remember that an equation
stays balanced if you add equal
amounts to both sides.
• So, if 5x + 2y = 1, you can add 5x + 2y
to one side of an equation and 1 to the
other side and the balance is
maintained.
Elimination Method
Since –2y and 2y have opposite coefficients (additive
inverses), the y-term is eliminated. The result is one
equation that has only one variable: 6x = –18.
When you use the elimination method to solve a
system of linear equations, align all like terms in the
equations. Then determine whether any like terms
can be eliminated because they have opposite
coefficients.
Elimination Method
• In some cases, you will first need to
multiply one or both of the equations by a
number so that one variable has opposite
coefficients, so you can add to eliminate the
variable.
• Elimination is easiest when the equations are
in standard form.
Elimination Procedure
Step 1: Put the equations in
Standard Form.
Step 2: Determine which
variable to eliminate.
Standard Form: Ax + By = C
Look for variables that have the
same coefficient.
Step 3: Multiply the equations
and add the equations.
Solve for the variable.
Step 4: Plug back in to find the
other variable.
Substitute the value of the variable
into the equation.
Step 5: Check your solution.
Substitute your ordered pair into
BOTH equations.
Example:
x+y=5
3x – y = 7
Step 1: Put the equations in
Standard Form.
Step 2: Determine which
variable to eliminate.
Step 3: Add the equations.
They already are!
The y’s have the same
coefficient.
Add to eliminate y.
x+ y=5
(+) 3x – y = 7
4x = 12
x=3
Example: Continued
x+y=5
3x – y = 7
Step 4: Plug back in to find the
other variable.
Step 5: Check your solution.
x+y=5
(3) + y = 5
y=2
(3, 2)
(3) + (2) = 5
3(3) - (2) = 7
The solution is (3, 2). What do you think the answer would
be if you solved using substitution?
Example:
4x + y = 7
4x – 2y = -2
Step 1: Put the equations in
Standard Form.
Step 2: Determine which
variable to eliminate.
Step 3: Multiply the equations
and add the equations.
They already are!
The x’s have the same
coefficient.
Multiple the 2nd equation by (-1)
and add to eliminate x.
4x + y = 7
- 4x + 2y = 2
3y = 9
y=3
Example: Continued
4x + y = 7
4x – 2y = -2
Step 4: Plug back in to find the
other variable.
Step 5: Check your solution.
4x + y = 7
4x + (3) = 7
4x = 4
x=1
(1, 3)
4(1) + (3) = 7
4(1) - 2(3) = -2
Example: Multiplying One
Equation
2x + 2y = 6
3x – y = 5
Step 1: Put the equations in
Standard Form.
Step 2: Determine which
variable to eliminate.
They already are!
None of the coefficients are the
same!
Find the least common multiple
of each variable.
LCM = 6x, LCM = 2y
Which is easier to obtain?
2y
(you only have to multiply
the bottom equation by 2)
Example: Continued
2x + 2y = 6
3x – y = 5
Step 3: Multiply the equations
and solve.
Multiply the bottom equation by 2
2x + 2y = 6
2x + 2y = 6
(2)(3x – y = 5) (+) 6x – 2y = 10
8x
= 16
x=2
Step 4: Plug back in to find the
other variable.
2(2) + 2y = 6
4 + 2y = 6
2y = 2
y=1
Example: Continued
2x + 2y = 6
3x – y = 5
Step 5: Check your solution.
(2, 1)
2(2) + 2(1) = 6
3(2) - (1) = 5
Solving with multiplication adds one more step
to the elimination process.
Example: Multiplying Both
Equations
3x + 4y = -1
4x – 3y = 7
Step 1: Put the equations in
Standard Form.
Step 2: Determine which
variable to eliminate.
They already are!
Find the least common multiple
of each variable.
LCM = 12x, LCM = 12y
Which is easier to obtain?
Either! I’ll pick y because the signs
are already opposite.
Example: Continued
3x + 4y = -1
4x – 3y = 7
Step 3: Multiply the equations
and solve.
Multiply both equations
(3)(3x + 4y = -1)
9x + 12y = -3
(4)(4x – 3y = 7) (+) 16x – 12y = 28
25x
= 25
x=1
Step 4: Plug back in to find the
other variable.
3(1) + 4y = -1
3 + 4y = -1
4y = -4
y = -1
Example: Continued
3x + 4y = -1
4x – 3y = 7
Step 5: Check your solution.
(1, -1)
3(1) + 4(-1) = -1
4(1) - 3(-1) = 7
Your Turn:
Solve
3x – 4y = 10
by elimination.
x + 4y = –2
3x – 4y = 10
x + 4y = –2
4x + 0 = 8
4x = 8
4x = 8
4
4
x=2
Write the system so that like
terms are aligned.
Add the equations to
eliminate the y-terms.
Simplify and solve for x.
Divide both sides by 4.
Continued
x + 4y = –2
2 + 4y = –2
–2
–2
4y = –4
4y
–4
4
4
y = –1
(2, –1)
Write one of the original
equations.
Substitute 2 for x.
Subtract 2 from both sides.
Divide both sides by 4.
Write the solution as an
ordered pair.
Your Turn:
Solve
3x + 3y = 15
by elimination.
–2x + 3y = –5
3x + 3y = 15
–(–2x + 3y = –5)
3x + 3y = 15
+ 2x – 3y = +5
5x + 0 = 20
5x = 20
x=4
Add the opposite of each
term in the second
equation.
Eliminate the y term.
Simplify and solve for x.
Continued
3x + 3y = 15
3(4) + 3y = 15
12 + 3y = 15
–12
–12
3y = 3
y=1
(4, 1)
Write one of the original
equations.
Substitute 4 for x.
Subtract 12 from both sides.
Simplify and solve for y.
Write the solution as an
ordered pair.
Your Turn:
Solve the system by elimination.
x + 2y = 11
–3x + y = –5
x + 2y = 11
–2(–3x + y = –5)
x + 2y = 11
+(6x –2y = +10)
7x + 0 = 21
7x = 21
x=3
Multiply each term in the
second equation by –2 to
get opposite y-coefficients.
Add the new equation to
the first equation.
Simplify and solve for x.
Continued
x + 2y = 11
3 + 2y = 11
–3
–3
2y = 8
y=4
(3, 4)
Write one of the original
equations.
Substitute 3 for x.
Subtract 3 from each side.
Simplify and solve for y.
Write the solution as an
ordered pair.
Your Turn:
Solve the system by elimination.
–5x + 2y = 32
2x + 3y = 10
2(–5x + 2y = 32)
5(2x + 3y = 10)
–10x + 4y = 64
+(10x + 15y = 50)
19y = 114
y=6
Multiply the first equation
by 2 and the second
equation by 5 to get
opposite x-coefficients
Add the new equations.
Simplify and solve for y.
Continued
2x + 3y = 10
2x + 3(6) = 10
2x + 18 = 10
–18 –18
2x = –8
x = –4
(–4, 6)
Write one of the original
equations.
Substitute 6 for y.
Subtract 18 from both sides.
Simplify and solve for x.
Write the solution as an
ordered pair.
Your Turn:
Solve the system by elimination.
2x + 5y = 26
–3x – 4y = –25
3(2x + 5y = 26)
+(2)(–3x – 4y = –25)
Multiply the first equation
by 3 and the second
equation by 2 to get
opposite x-coefficients
6x + 15y = 78
+(–6x – 8y = –50) Add the new equations.
0 + 7y = 28
Simplify and solve for y.
y =4
Your Turn:
2x + 5y = 26
2x + 5(4) = 26
2x + 20 = 26
–20 –20
2x
= 6
x=3
(3, 4)
Write one of the original
equations.
Substitute 4 for y.
Subtract 20 from both
sides.
Simplify and solve for x.
Write the solution as an
ordered pair.
A Linear System with Infinite Solutions
Show
that this linear system
M
ETHOD: Elimination
has infinitely many solutions.
–2x  y  3
– 4 x  2y  6
Equation 1
Equation 2
You can multiply Equation 1 by –2.
4x – 2y  – 6
– 4 x  2y  6
0  0
Multiply Equation 1 by –2.
Write Equation 2.
Add Equations. True statement!
The variables are eliminated and you are left with a statement that is true
regardless of the values of x and y. This result tells you that the linear system
has infinitely many solutions.
A Linear System with No Solution
Show that this linear system
METHOD: Elimination
has no solution.
2x  y  5
2x  y  1
Equation 1
Equation 2
You can multiply Equation 1 by –1.
-2x - y  -5
2x  y  1
0 -4
Multiply Equation 1 by –1.
Write Equation 2.
Add Equations. False statement!
The variables are eliminated and you are left with a statement that is false
regardless of the values of x and y. This result tells you that the linear system
has no solution.
Your Turn:
Solve the systems using elimination.
2 x 5 y 7

1)  2 12
 y  5 x  5
False Statement
No Solution
12x 8 y 20
2) 
 3x 2 y 5
True Statement
Infinite Solutions
Summary
Substitution
Summary of Methods for Solving Systems
Example
6x + y = 10
y=5
Suggested
Method
Substitution
Why
The value of one variable is
known and can easily be
substituted into the other
equation.
Elimination
Summary of Methods for Solving Systems
Example
Suggested
Method
Why
2x – 5y = –20
4x + 5y = 14
Elimination
eliminate ‘y’  5
Add the two equations
Elimination
Summary of Methods for Solving Systems
Example
Suggested
Method
Why
9a – 2b = –11
8a + 4b = 25
Elimination
eliminate ‘b’  4
Multiply first equation by 2
Add the equations
Essential Question
Big Idea: Solving Equations and
Inequalities
• When can you use elimination to solve a system of
linear equation?
• Use elimination when the system contains a pair
of additive inverses. Then add to eliminate a
variable. You can also multiply one or both
equations by nonzero numbers to make an
equivalent system with additive inverses.
Assignment
• Section 3-2 Part 2, Pg. 159 – 161; #1 – 6 all,
8 – 48 even.