1. Linear Equations Part 1
Linear equations model real world relationships and patterns. They are represented by a straight line on a graph. They contain a constant and variables that are sometimes multiplied by a coefficient.

2. Linear Equations Part 1
Linear equations model real world relationships and patterns. They are represented by a straight line on a graph. They contain a constant and variables that are sometimes multiplied by a coefficient.
EXAMPLES : 𝑦=𝑥
2𝑥+5=10
5𝑥−4=2𝑥

3. Linear Equations Part 1
Linear equations model real world relationships and patterns. They are represented by a straight line on a graph. They contain a constant and variables that are sometimes multiplied by a coefficient.
EXAMPLES : 𝑦=𝑥
2𝑥+5=10
5𝑥−4=2𝑥
Being able to model real world problems from verbal information into equation form is needed. Look for key words and phrases such as added to, subtracted from, times, etc. to create your equation.

4. Linear Equations Part 1
Linear equations model real world relationships and patterns. They are represented by a straight line on a graph. They contain a constant and variables that are sometimes multiplied by a coefficient.
EXAMPLES : 𝑦=𝑥
2𝑥+5=10
5𝑥−4=2𝑥
Being able to model real world problems from verbal information into equation form is needed. Look for key words and phrases such as added to, subtracted from, times, etc. to create your equation.
EXAMPLE : A tree was four feet tall when it was planted. It grows 2 feet per year. How many years will it take to grow to a height of 16 feet ?

5. Linear Equations Part 1
Linear equations model real world relationships and patterns. They are represented by a straight line on a graph. They contain a constant and variables that are sometimes multiplied by a coefficient.
EXAMPLES : 𝑦=𝑥
2𝑥+5=10
5𝑥−4=2𝑥
Being able to model real world problems from verbal information into equation form is needed. Look for key words and phrases such as added to, subtracted from, times, etc. to create your equation.
EXAMPLE : A tree was four feet tall when it was planted. It grows 2 feet per year. How many years will it take to grow to a height of 16 feet ?
4+2𝑦=16 ** the tree started at 4 feet ( constant of 4 )
** it grows 2 feet per year ( variable with coefficient of 2 )
** it grows to 16 feet ( constant of 16 as the result )

6. Linear Equations Part 1
Our objective for these equations is to solve for the given variable or variables that are not known.
There are specific steps you must follow to solve these equations. Let’s look at a few examples to get you started.

7. Linear Equations Part 1
Our objective for these equations is to solve for the given variable or variables that are not known.
There are specific steps you must follow to solve these equations. Let’s look at a few examples to get you started.
EXAMPLE #1 : Solve 2𝑥+12=6
With all linear equations we will :
1. distribute any outside coefficients into parentheses
2. combine any like terms on each side
3. get the variable isolated on one side of the equation

8. Linear Equations Part 1
Our objective for these equations is to solve for the given variable or variables that are not known.
There are specific steps you must follow to solve these equations. Let’s look at a few examples to get you started.
EXAMPLE #1 : Solve 2𝑥+12=6
With all linear equations we will :
1. distribute any outside coefficients into parentheses
2. combine any like terms on each side
3. get the variable isolated on one side of the equation
We do not need to distribute here and terms are already combined on both sides. To isolate the variable, move your constant first AWAY from the variable by doing the opposite operation.

9. Linear Equations Part 1
Our objective for these equations is to solve for the given variable or variables that are not known.
There are specific steps you must follow to solve these equations. Let’s look at a few examples to get you started.
EXAMPLE #1 : Solve 2𝑥+12=6
With all linear equations we will :
1. distribute any outside coefficients into parentheses
2. combine any like terms on each side
3. get the variable isolated on one side of the equation
We do not need to distribute here and terms are already combined on both sides. To isolate the variable, move your constant first AWAY from the variable by doing the opposite operation.
2𝑥+12=6
−12=−12 ** opposite operation
2𝑥=−6

10. Linear Equations Part 1
Our objective for these equations is to solve for the given variable or variables that are not known.
There are specific steps you must follow to solve these equations. Let’s look at a few examples to get you started.
EXAMPLE #1 : Solve 2𝑥+12=6
With all linear equations we will :
1. distribute any outside coefficients into parentheses
2. combine any like terms on each side
3. get the variable isolated on one side of the equation
We do not need to distribute here and terms are already combined on both sides. To isolate the variable, move your constant first AWAY from the variable by doing the opposite operation.
2𝑥+12=6
−12=−12 ** opposite operation
2𝑥=−6
2𝑥 2 = −6 2 ** again opposite operation
𝑥=−3

11. Linear Equations Part 1
EXAMPLE #2 : Solve 4−3 2𝑥+1 =17

12. Linear Equations Part 1 EXAMPLE #2 : Solve 4−3 2𝑥+1 =25
4−6𝑥−3=25 ** distribute any outside coefficients

13. Linear Equations Part 1 EXAMPLE #2 : Solve 4−3 2𝑥+1 =25
4−6𝑥−3=25 ** distribute any outside coefficients
1−6𝑥=25 ** combine any like terms on both sides

14. Linear Equations Part 1 EXAMPLE #2 : Solve 4−3 2𝑥+1 =25
4−6𝑥−3=25 ** distribute any outside coefficients
1−6𝑥=25 ** combine any like terms on both sides
− =−1 ** move constants first
−6𝑥=24

15. Linear Equations Part 1 EXAMPLE #2 : Solve 4−3 2𝑥+1 =25
4−6𝑥−3=25 ** distribute any outside coefficients
1−6𝑥=25 ** combine any like terms on both sides
− =−1 ** move constants first
−6𝑥=24
−6𝑥 −6 = 24 −6 ** divide by any coefficient
𝑥=−4

16. Linear Equations Part 1
EXAMPLE #3 : Solve 4−2𝑥=−10+5𝑥

17. Linear Equations Part 1 EXAMPLE #3 : Solve 4−2𝑥=−10+5𝑥
There are no coefficients to distribute or like terms on either side to combine. But variables and constants appear on both sides. Let’s move them to opposite sides. I like my variables on the left side and my constants on the right side.

18. Linear Equations Part 1 EXAMPLE #3 : Solve 4−2𝑥=−10+5𝑥
−5𝑥= −5𝑥 ** opposite operation
4−7𝑥=−10

19. Linear Equations Part 1 EXAMPLE #3 : Solve 4−2𝑥=−10+5𝑥
−5𝑥= −5𝑥 ** opposite operation
4−7𝑥=−10
− =−4 ** opposite operation
−7𝑥=−14

20. Linear Equations Part 1 EXAMPLE #3 : Solve 4−2𝑥=−10+5𝑥
−5𝑥= −5𝑥 ** opposite operation
4−7𝑥=−10
− =−4 ** opposite operation
−7𝑥=−14
−7𝑥 −7 = −14 −7
𝑥=2

21. Linear Equations Part 1
EXAMPLE #4 : Lisa uses the equation below to estimate the amount of taxes, 𝑇, she owes the
government.
𝑇=0.15 𝑑−
Lisa estimates that she owes the government $3455 in taxes. What is 𝑑, the total
dollars Lisa earned during the past year ?
a) $15,755
b) $20,550
c) $25,550
d) $30,755

22. Linear Equations Part 1
EXAMPLE #4 : Lisa uses the equation below to estimate the amount of taxes, 𝑇, she owes the
government.
𝑇=0.15 𝑑−
Lisa estimates that she owes the government $3455 in taxes. What is 𝑑, the total
dollars Lisa earned during the past year ?
a) $15,755
b) $20,550
c) $25,550
d) $30,755
Solution : Solve for 𝑑 where 𝑇=3455

23. Linear Equations Part 1
EXAMPLE #4 : Lisa uses the equation below to estimate the amount of taxes, 𝑇, she owes the
government.
𝑇=0.15 𝑑−
Lisa estimates that she owes the government $3455 in taxes. What is 𝑑, the total
dollars Lisa earned during the past year ?
a) $15,755
b) $20,550
c) $25,550
d) $30,755
Solution : Solve for 𝑑 where 𝑇=3455
3455=0.15 𝑑−

24. Linear Equations Part 1
EXAMPLE #4 : Lisa uses the equation below to estimate the amount of taxes, 𝑇, she owes the
government.
𝑇=0.15 𝑑−
Lisa estimates that she owes the government $3455 in taxes. What is 𝑑, the total
dollars Lisa earned during the past year ?
a) $15,755
b) $20,550
c) $25,550
d) $30,755
Solution : Solve for 𝑑 where 𝑇=3455
3455=0.15 𝑑−
3455=0.15𝑑− * distribute

25. Linear Equations Part 1
EXAMPLE #4 : Lisa uses the equation below to estimate the amount of taxes, 𝑇, she owes the
government.
𝑇=0.15 𝑑−
Lisa estimates that she owes the government $3455 in taxes. What is 𝑑, the total
dollars Lisa earned during the past year ?
a) $15,755
b) $20,550
c) $25,550
d) $30,755
Solution : Solve for 𝑑 where 𝑇=3455
3455=0.15 𝑑−
3455=0.15𝑑− * distribute
3455=0.15𝑑− * combine like terms

26. Linear Equations Part 1
EXAMPLE #4 : Lisa uses the equation below to estimate the amount of taxes, 𝑇, she owes the
government.
𝑇=0.15 𝑑−
Lisa estimates that she owes the government $3455 in taxes. What is 𝑑, the total
dollars Lisa earned during the past year ?
a) $15,755
b) $20,550
c) $25,550
d) $30,755
Solution : Solve for 𝑑 where 𝑇=3455
3455=0.15 𝑑−
3455=0.15𝑑− * distribute
3455=0.15𝑑− * combine like terms
= * isolate 𝑑
=0.15𝑑

27. Linear Equations Part 1
EXAMPLE #4 : Lisa uses the equation below to estimate the amount of taxes, 𝑇, she owes the
government.
𝑇=0.15 𝑑−
Lisa estimates that she owes the government $3455 in taxes. What is 𝑑, the total
dollars Lisa earned during the past year ?
a) $15,755
b) $20,550
c) $25,550
d) $30,755
Solution : Solve for 𝑑 where 𝑇=3455
3455=0.15 𝑑−
3455=0.15𝑑− * distribute
3455=0.15𝑑− * combine like terms
= * isolate 𝑑
=0.15𝑑
= 0.15𝑑 0.15
𝑑= ANSWER is C