Lesson 6: Solutions of a Linear Equation

Lesson 6
NYS COMMON COR E MATHEMATICS CURRICULUM
8β€’4
Lesson 6: Solutions of a Linear Equation
Exit Ticket Sample Solutions
Transform the equation if necessary, and then solve to find the value of 𝒙 that makes the equation true.
1.
𝟏
πŸ“π’™ βˆ’ ( 𝒙 + πŸ‘ ) = ( πŸ—π’™ + πŸπŸ– ) βˆ’ πŸ“
πŸ‘
𝟏
πŸ“π’™ βˆ’ ( 𝒙 + πŸ‘ ) = ( πŸ—π’™+ πŸπŸ– ) βˆ’ πŸ“
πŸ‘
πŸ“π’™ βˆ’ 𝒙 βˆ’ πŸ‘ = πŸ‘π’™+ πŸ” βˆ’ πŸ“
πŸ’π’™ βˆ’ πŸ‘ = πŸ‘π’™+ 𝟏
πŸ’π’™ βˆ’ πŸ‘π’™ βˆ’ πŸ‘ = πŸ‘π’™βˆ’ πŸ‘π’™ + 𝟏
𝒙 βˆ’πŸ‘ = 𝟏
𝒙 βˆ’πŸ‘ +πŸ‘ = 𝟏 +πŸ‘
𝒙=πŸ’
2.
πŸ“ (πŸ‘π’™+ πŸ— ) βˆ’ πŸπ’™ = πŸπŸ“π’™ βˆ’ 𝟐(𝒙 βˆ’ πŸ“)
πŸ“ ( πŸ‘π’™ + πŸ— ) βˆ’ πŸπ’™ = πŸπŸ“π’™βˆ’ 𝟐 (𝒙 βˆ’ πŸ“ )
πŸπŸ“π’™ + πŸ’πŸ“ βˆ’ πŸπ’™ = πŸπŸ“π’™βˆ’ πŸπ’™+ 𝟏𝟎
πŸπŸ‘π’™ + πŸ’πŸ“ = πŸπŸ‘π’™+ 𝟏𝟎
πŸ’πŸ“ β‰  𝟏𝟎
Since πŸ’πŸ“ β‰  𝟏𝟎, the equation has no solution.
Problem Set Sample Solutions
Students practice using the distributive property to transform equations and solve.
Transform the equation if necessary, and then solve it to find the value of 𝒙 that makes the equation true.
1.
𝟏
𝒙 βˆ’ ( πŸ—π’™ βˆ’ 𝟏𝟎 ) + 𝟏𝟏 = πŸπŸπ’™ + πŸ‘ ( βˆ’πŸπ’™ + )
πŸ‘
𝟏
𝒙 βˆ’ ( πŸ—π’™ βˆ’ 𝟏𝟎 ) + 𝟏𝟏 = πŸπŸπ’™ + πŸ‘ ( βˆ’πŸπ’™ + )
πŸ‘
𝒙 βˆ’ πŸ—π’™ + 𝟏𝟎 + 𝟏𝟏 = πŸπŸπ’™ βˆ’ πŸ”π’™ + 𝟏
βˆ’πŸ–π’™ + 𝟐𝟏 = πŸ”π’™+ 𝟏
βˆ’πŸ–π’™ + πŸ–π’™ + 𝟐𝟏 = πŸ”π’™+ πŸ–π’™ + 𝟏
𝟐𝟏 = πŸπŸ’π’™ + 𝟏
𝟐𝟏 βˆ’ 𝟏 = πŸπŸ’π’™ + 𝟏 βˆ’ 𝟏
𝟐𝟎 = πŸπŸ’π’™
𝟐𝟎 πŸπŸ’
=
𝒙
πŸπŸ’ πŸπŸ’
𝟏𝟎
=𝒙
πŸ•
Lesson 6:
Date:
Solutions of a Linear Equation
12/9/15
© 2014 Common Core, Inc. Some rights reserved. commoncore.org
64
This work is licensed under a
Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 6
NYS COMMON COR E MATHEMATICS CURRICULUM
2.
8β€’4
𝟏
πŸ•π’™ + πŸ– (𝒙 + ) = πŸ‘ ( πŸ”π’™ βˆ’ πŸ— ) βˆ’ πŸ–
πŸ’
𝟏
πŸ•π’™ + πŸ– (𝒙 + ) = πŸ‘ ( πŸ”π’™βˆ’ πŸ— ) βˆ’ πŸ–
πŸ’
πŸ•π’™ + πŸ–π’™ + 𝟐 = πŸπŸ–π’™βˆ’ πŸπŸ• βˆ’ πŸ–
πŸπŸ“π’™+ 𝟐 = πŸπŸ–π’™βˆ’ πŸ‘πŸ“
πŸπŸ“π’™ βˆ’ πŸπŸ“π’™+ 𝟐 = πŸπŸ–π’™βˆ’ πŸπŸ“π’™ βˆ’ πŸ‘πŸ“
𝟐 = πŸ‘π’™ βˆ’ πŸ‘πŸ“
𝟐 + πŸ‘πŸ“ = πŸ‘π’™ βˆ’ πŸ‘πŸ“ + πŸ‘πŸ“
πŸ‘πŸ• = πŸ‘π’™
πŸ‘πŸ• πŸ‘
= 𝒙
πŸ‘
πŸ‘
πŸ‘πŸ•
=𝒙
πŸ‘
3.
βˆ’πŸ’π’™βˆ’ 𝟐 ( πŸ–π’™ + 𝟏 ) = βˆ’( βˆ’πŸπ’™ βˆ’ 𝟏𝟎 )
βˆ’πŸ’π’™ βˆ’ 𝟐 ( πŸ–π’™ + 𝟏 ) = βˆ’( βˆ’πŸπ’™βˆ’ 𝟏𝟎)
βˆ’πŸ’π’™ βˆ’ πŸπŸ”π’™βˆ’ 𝟐 = πŸπ’™ + 𝟏𝟎
βˆ’πŸπŸŽπ’™βˆ’ 𝟐 = πŸπ’™ + 𝟏𝟎
βˆ’πŸπŸŽπ’™ + πŸπŸŽπ’™βˆ’ 𝟐 = πŸπ’™ + πŸπŸŽπ’™ + 𝟏𝟎
βˆ’πŸ = πŸπŸπ’™+ 𝟏𝟎
βˆ’πŸ βˆ’ 𝟏𝟎 = πŸπŸπ’™+ 𝟏𝟎 βˆ’ 𝟏𝟎
βˆ’πŸπŸ = πŸπŸπ’™
𝟏𝟐 𝟐𝟐
βˆ’
=
𝒙
𝟐𝟐 𝟐𝟐
πŸ”
βˆ’
=𝒙
𝟏𝟏
4.
𝟏𝟏( 𝒙+ 𝟏𝟎) = πŸπŸ‘πŸ
𝟏𝟏 (𝒙 + 𝟏𝟎 ) = πŸπŸ‘πŸ
𝟏
𝟏
( ) 𝟏𝟏 (𝒙 + 𝟏𝟎 ) = ( ) πŸπŸ‘πŸ
𝟏𝟏
𝟏𝟏
𝒙 + 𝟏𝟎 = 𝟏𝟐
𝒙 + 𝟏𝟎 βˆ’ 𝟏𝟎 = 𝟏𝟐 βˆ’ 𝟏𝟎
𝒙=𝟐
5.
𝟏
𝟏
πŸ‘πŸ•π’™ + βˆ’ (𝒙 + ) = πŸ— ( πŸ’π’™ βˆ’ πŸ• ) + πŸ“
𝟐
πŸ’
𝟏
𝟏
πŸ‘πŸ•π’™+ βˆ’ (𝒙 + ) = πŸ— ( πŸ’π’™ βˆ’ πŸ• ) + πŸ“
𝟐
πŸ’
𝟏
𝟏
πŸ‘πŸ•π’™ + βˆ’ 𝒙 βˆ’ = πŸ‘πŸ”π’™βˆ’ πŸ”πŸ‘ + πŸ“
𝟐
πŸ’
𝟏
πŸ‘πŸ”π’™ + = πŸ‘πŸ”π’™βˆ’ πŸ“πŸ–
πŸ’
𝟏
πŸ‘πŸ”π’™ βˆ’ πŸ‘πŸ”π’™ + = πŸ‘πŸ”π’™βˆ’ πŸ‘πŸ”π’™βˆ’ πŸ“πŸ–
πŸ’
𝟏
β‰  βˆ’πŸ“πŸ–
πŸ’
This equation has no solution.
Lesson 6:
Date:
Solutions of a Linear Equation
12/9/15
© 2014 Common Core, Inc. Some rights reserved. commoncore.org
65
This work is licensed under a
Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 6
NYS COMMON COR E MATHEMATICS CURRICULUM
6.
8β€’4
πŸ‘ (πŸπ’™βˆ’ πŸπŸ’) + 𝒙 = πŸπŸ“ βˆ’ (βˆ’πŸ—π’™βˆ’ πŸ“)
πŸ‘ ( πŸπ’™ βˆ’ πŸπŸ’ ) + 𝒙 = πŸπŸ“ βˆ’ (βˆ’πŸ—π’™ βˆ’ πŸ“)
πŸ”π’™ βˆ’ πŸ’πŸ + 𝒙 = πŸπŸ“ + πŸ—π’™+ πŸ“
πŸ•π’™ βˆ’ πŸ’πŸ = 𝟐𝟎 + πŸ—π’™
πŸ•π’™ βˆ’ πŸ•π’™ βˆ’ πŸ’πŸ = 𝟐𝟎 + πŸ—π’™βˆ’ πŸ•π’™
βˆ’πŸ’πŸ = 𝟐𝟎 + πŸπ’™
βˆ’πŸ’πŸ βˆ’ 𝟐𝟎 = 𝟐𝟎 βˆ’ 𝟐𝟎 + πŸπ’™
βˆ’πŸ”πŸ = πŸπ’™
βˆ’πŸ‘πŸ = 𝒙
7.
πŸ– (𝟐𝐱 + πŸ— ) = πŸ“πŸ”
πŸ– ( πŸπ’™ + πŸ— ) = πŸ“πŸ”
𝟏
𝟏
( ) πŸ– ( πŸπ’™ + πŸ— ) = ( ) πŸ“πŸ”
πŸ–
πŸ–
πŸπ’™ + πŸ— = πŸ•
πŸπ’™ + πŸ— βˆ’ πŸ— = πŸ• βˆ’ πŸ—
πŸπ’™ = βˆ’πŸ
𝟏
𝟏
( ) πŸπ’™ = ( ) βˆ’ 𝟐
𝟐
𝟐
𝒙 = βˆ’πŸ
Lesson 6:
Date:
Solutions of a Linear Equation
12/9/15
© 2014 Common Core, Inc. Some rights reserved. commoncore.org
66
This work is licensed under a
Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 6
NYS COMMON COR E MATHEMATICS CURRICULUM
8β€’4
Lesson 7: Classification of Solutions
Exit Ticket Sample Solutions
Give a brief explanation as to what kind of solution(s) you expect the following linear equations to have. Transform the
equation into a simpler form if necessary.
8.
πŸ‘ (πŸ”π’™+ πŸ– ) = πŸπŸ’ + πŸπŸ–π’™
If I use the distributive property on the left side, I notice that the coefficients of 𝒙 are the same, and the constants
are the same. Therefore, this equation has infinitely many solutions.
9.
𝟏𝟐( 𝒙+ πŸ– ) = πŸπŸπ’™ βˆ’ πŸ“
If I use the distributive property on the left side, I notice that the coefficients of 𝒙 are different, and the constants are
different. Therefore, this equation has a unique solution.
10. πŸ“π’™ βˆ’ πŸ– = 𝟏𝟏 βˆ’ πŸ•π’™+ πŸπŸπ’™
If I collect the like terms on the right side, I notice that the coefficients of 𝒙 are the same, but the constants are
different. Therefore, this equation has no solution.
Problem Set Sample Solutions
Students apply their knowledge of solutions to linear equations by writing equations with unique solutions, no solutions ,
and infinitely many solutions.
𝟏
𝟐
11. Give a brief explanation as to what kind of solution(s) you expect for the linear equation πŸπŸ–π’™ + = πŸ”(πŸ‘π’™ + πŸπŸ“).
Transform the equation into a simpler form if necessary.
If I use the distributive property on the right side of the equation, I notice that the coefficients of 𝒙 are the same, but
the constants are different. Therefore, this equation has no solutions.
12. Give a brief explanation as to what kind of solution(s) you expect for the linear equation πŸ– βˆ’ πŸ—π’™ = πŸπŸ“π’™+ πŸ• + πŸ‘π’™.
Transform the equation into a simpler form if necessary.
If I collect the like terms on the right side of the equation, then I notice that the coefficients of 𝒙 are different and so
are the constants. Therefore, this equation will have a unique solution.
13. Give a brief explanation as to what kind of solution(s) you expect for the linear equation πŸ“ (𝒙 + πŸ— ) = πŸ“π’™ + πŸ’πŸ“.
Transform the equation into a simpler form if necessary.
This is an identity under the distributive property. Therefore, this equation will have infinitely many solutions.
14. Give three examples of equations where the solution will be unique, that is, only one solution is possible.
Accept equations where the coefficients of 𝒙 on each side of the equal sign are unique.
Lesson 6:
Date:
Solutions of a Linear Equation
12/9/15
© 2014 Common Core, Inc. Some rights reserved. commoncore.org
67
This work is licensed under a
Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 6
NYS COMMON COR E MATHEMATICS CURRICULUM
8β€’4
15. Solve one of the equations you wrote in Problem 4, and explain why it is the only solution.
Verify that students solved one of the equations. They should have an explanation that includes the statement that
there is only one possible number that could make the equation true. They may have referenced one of the simpler
forms of their transformed equation to make their case.
16. Give three examples of equations where there will be no solution.
Accept equations where the coefficients of 𝒙 on each side of the equal sign are the same, and the constants on each
side are unique.
17. Attempt to solve one of the equations you wrote in Problem 6, and explain why it has no solution.
Verify that students have solved one of the equations. They should have an explanation that includes the statement
about getting a false equation, e.g., πŸ” β‰  𝟏𝟎.
18. Give three examples of equations where there will be infinitely many solutions.
Accept equations where the coefficients of 𝒙 and constants on each side of the equal sign are the same.
19. Attempt to solve one of the equations you wrote in Problem 8, and explain why it has infinitely many solutions.
Verify that students have solved one of the equations. They should have an explanation that includes the statement
about the linear expressions being exactly the same, an identity; therefore, any rational number 𝒙 would make the
equation true.
Lesson 6:
Date:
Solutions of a Linear Equation
12/9/15
© 2014 Common Core, Inc. Some rights reserved. commoncore.org
68
This work is licensed under a
Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.