Transcript Solving Systems Using Elimination

Solving Systems Using
Elimination
Section 6-3 Part 2
Goals
Goal
• To solve systems by adding
or subtracting to eliminate a
variable.
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
• Elimination Method
Further Elimination
• In Part 1 of this lesson, you was that to
eliminate a variable, its coefficients must
have a sum or difference of zero.
• 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 or subtract to
eliminate the variable.
Further 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 solve.
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: 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 One
Equation
x + 4y = 7
4x – 3y = 9
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 = 4x, LCM = 12y
Which is easier to obtain?
4x
(you only have to multiply
the top equation by -4 to make
them inverses)
Example: Continued
x + 4y = 7
4x – 3y = 9
Step 3: Multiply the equations
and solve.
Multiply the top equation by -4
(-4)(x + 4y = 7)
-4x – 16y = -28
4x – 3y = 9) (+) 4x – 3y = 9
-19y = -19
y=1
Step 4: Plug back in to find the
other variable.
x + 4(1) = 7
x+4=7
x=3
Example: Continued
x + 4y = 7
4x – 3y = 9
Step 5: Check your solution.
(3, 1)
(3) + 4(1) = 7
4(3) - 3(1) = 9
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.
3x + 2y = 6
–x + y = –2
3x + 2y = 6
3(–x + y = –2)
3x + 2y = 6
+(–3x + 3y = –6)
0
+ 5y = 0
5y = 0
y=0
Multiply each term in the second
equation by 3 to get opposite
x-coefficients.
Add the new equation to
the first equation.
Simplify and solve for y.
Continued
–x + y = –2
–x + 3(0) = –2
–x + 0 = –2
–x = –2
x=2
(2, 0)
Write one of the original
equations.
Substitute 0 for y.
Simplify and solve for x.
Write the solution as an
ordered pair.
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 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.
IDENTIFYING THE NUMBER OF SOLUTIONS
CONCEPT
NUMBER OF SOLUTIONS OF A LINEAR SYSTEM
SUMMARY
y
y
y
x
x
x
Lines intersect
Lines are parallel
Lines coincide
one solution
no solution
infinitely many solutions
Identifying The Number of
Solutions
• If both variable terms are eliminated as you
solve a system of equations, the answer is
either no solution or infinite solutions.
– No solution: get a false statement when solving
the system.
– Infinite solutions: get a true statement when
solving the system.
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
12 x 8 y  20
2)  3x  2 y 5

True Statement
Infinite Solutions
Summary
7.3 The Elimination Method
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.
7.3 The Elimination Method
Summary of Methods for Solving Systems
Example
Suggested
Method
Why
2x – 5y = –20
4x + 5y = 14
Elimination
eliminate ‘y’  5
Add the two equations
7.3 The Elimination Method
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
Joke Time
• How does an octopus go to war?
• Well-Armed!
• Why does a Moo-rock taste better than an Earth-rock?
• Because it’s a little meteor!
• What did the elder chimney say to the younger chimney?
• You’re too young to smoke!
Assignment
• 6-3 Part 2 Exercises Pg. 402 - 404: #6 – 34
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