1. Chapter 1: Solving Linear Equations Sections 1.1-1.4
Solving Linear Equations is An Important and Fundamental Skill in Algebra.
In algebra, we are often presented with a problem where the answer is known, but part of the problem is missing.
The missing part of the problem is what we seek to find.

2. Examples
Solve: 𝑏−4=6 𝑏= 10 Add 4 to both sides. Solve: 8=4+𝑐 𝑐 =4 Subtract 4 from both sides. Solve: 𝑡+5=−3 𝑡 =−8 Subtract 5 from both sides.
Solve: 7𝑥=63
𝑥=9
Divide by 7 on both sides.
Solve: 9=𝑦−7
𝑦 =16
Add 7 to both sides.
Solve: −8𝑧=48
𝑧 =−6
Divide by -8 on both sides.

3. Examples
Solve: −𝑥=11 𝑥= 11 −1 =−11. Here −𝑥 means −1𝑥. Solve: −𝑦=−12 𝑦= −12 −1 =12.
Solve: 𝑦 −7 =4. 𝑥=4∙ −7 =−28.
Solve: −𝑥 5 =−2.
−𝑥=−2∙5=−10 𝑥= −10 −1 =10

4. Solve the equation 6𝑥+7=31 algebraically.
Show each step. You must use only two steps.
Step 1: +/− giving = 6x =
Step 2: ∗/÷ ÷ giving 𝑥 =
Solve: 𝑥+16=−4
Step 1: giving: 4x = -20
Step 2: ÷ 4 giving x = -5

5. Application Problems 7.2= 𝑡 5.7 𝑡=5.7∗7.2 =41.04
In a thunderstorm, the formula: 𝑀= 𝑡 5.7 gives the approximate distance, 𝑀, in miles, from a lightning strike if it takes 𝑡 seconds to hear the thunder after seeing the lightning. If you are 7.2 miles away from the lightning flash, how long will it take the sound of the thunder to reach you. Solution:
7.2= 𝑡 5.7 𝑡=5.7∗7.2 =41.04
Answer: It will take seconds for the sound to reach you.

6. Application Problems
After writing a $700 check to pay a bill, a student had a balance of $550 in his account. How much did he have in the account before he wrote the check? a) Let 𝑥 = his balance before writing the check. Write the equation you would use to solve this problem. Solution: 𝑥−700=550 b) Now solve your equation: 𝑥= 𝑥=1250 His starting balance was $

7. Application Problems
Chandra earns $11.50 per hour at her job. How many hours will she have to work to earn $ a) Let ℎ = the number of hours she works. Write the equation you would use to solve this problem. Solution: 11.5ℎ=690 b) Now solve your equation ℎ= =60 She would need to work 60 hours.

8. Combine Like Terms Before Solving the Equation
Like Terms: The terms that have the same variables with the same exponents. Example: Solve the following. −9=4𝑥−5𝑥+5 Here 4𝑥 and −5𝑥 are like terms. 4𝑥−5𝑥=−𝑥. So −9=−𝑥+5 −9−5=−𝑥 −14=−𝑥 14=𝑥

9. Collect The Variable Terms on One Side First
Solve the equation. 8𝑥+6=7𝑥 Solution: 6=7𝑥−8𝑥 6=−𝑥 6 −1 =𝑥 −6=𝑥

10. Solve the equation for the given variable:
4𝑥+11=−12𝑥−2 Solution: 4𝑥+12𝑥=−2−11 16𝑥=−13 𝑥=− If your answer is a fraction, write it in fraction form and reduce it completely. Do NOT convert to decimals.

11. Multiply Out First
Solve the equation for the given variable. 3(3𝑦+2)=−37 Solution: 3∙3𝑦+3∙2=−37 9𝑦+6=−37 9𝑦=−37−6 9𝑦=−43 𝑦=− 43 9 If your answer is a fraction, write it in reduced, fractional form. Do NOT convert the answer to a decimal.

12. Practice
Solve the equation for the given variable. 34=18−7(−4−5w) If your answer is a fraction, write it in reduced, fractional form. Do NOT convert the answer to a decimal.

13. Solve for 𝑥 algebraically.
1 𝑥+8 −1=4 𝑥−7 Solution: 𝑥+8−1=4𝑥−28 𝑥+7=4𝑥−28 𝑥−4𝑥=−28−7 −3𝑥=−35 𝑥= −35 −3 𝑥= 35 3

14. Solve for 𝑦.
−6𝑦+1 −9𝑦−9 =−6−8 3−𝑦 Solution: −6𝑦−9𝑦−9=−6−24+8𝑦 −15𝑦−9=−30+8𝑦 −15𝑦−8𝑦=−30+9 −23𝑦=−21 𝑦= −21 −23 𝑦= 21 23

15. Practice
Solve for 𝑥 3(𝑥+7)−7=−4(𝑥+6)

16. Practice
Solve the equation for 𝑥 8(𝑥+3)+2=−8(𝑥−10)−8

17. Practice
Solve the equation for the given variable. 10𝑥−1=9(𝑥+10)−9

18. 3 Cases on the Solutions After we simplify the equation:
If the result is like 𝑥=9, then the equation has one solution or finite solutions.
if the result has no more variables and is a true statement, such as 9=9, the equation has infinite solutions;
if the result has no more variables and is a false statement, such as 4=10, the equation has no solution.

19. Examples
Solve the equation for the given variable. (type "DNE" for no solution or ∞ or infinite solutions) 3(5𝑥+6)+4(4𝑥+6)+5=31𝑥+47 Solution: 15𝑥+18+16𝑥+24+5=31𝑥+47 31𝑥+47=31𝑥+47 31𝑥−31𝑥+47=47 47=47 This is a true statement, that means all real numbers are solution. There are infinite solutions.

20. Example: Solve the equation for the given variable
Example: Solve the equation for the given variable. (type "no solution" if the equation has no solution and type "all real numbers" if the equation is valid for all possible x-values)
7(6𝑥+6)+9(9𝑥+6)+5=123𝑥+102 Solution: 42𝑥+42+81𝑥+54+5=123𝑥 𝑥+101=123𝑥 𝑥−123𝑥+101= =102 This is a false statement since 101 is not equal to 102. Then the equation has no solution.

21. Practice: Solve the equation for the given variable
Practice: Solve the equation for the given variable. (type "no solution" if the equation has no solution and type "all real numbers" if the equation is valid for all possible x-values)
9(9𝑥+9)+4(5𝑥+2)+3=101𝑥+92

22. Multiply by LCD on Both Sides to Get Rid of the Denominators
Example: Solve the equation for the given variable: −6𝑏+8 7 =−5 Solution: There is only one denominator: 7. Multiply by 7 on both sides to get −6𝑏+8 7 ∙7=−5∙7 −6𝑏+8=−35 −6𝑏=−35−8 −6𝑏=−43 𝑏= −43 −6 𝑏= 43 6

23. Example: Solve the following.
1 6 𝑥− 1 2 =− 3 10
Solution: LCD (6, 2, 10) = 𝑥∙30− 1 2 ∙30=− 3 10 ∙30 5𝑥−15=−9 5𝑥=−9+15 5𝑥=6 𝑥= 6 5

24. Example: Solve the following.
𝑦 5 + 𝑦 2 = 7 10
Solution: LCD (5, 2, 10)=10 𝑦 5 ∙10+ 𝑦 2 ∙10= 7 10 ∙10 2𝑦+5𝑦=7 7𝑦=7 𝑦= 7 7 𝑦=1

25. Example: Solve the following.
If you are using a calculator, make sure it is in decimal mode for this problem.
Solution:
1.2 𝑥 𝑥=1.2 𝑥− 𝑥+1.2∗ 𝑥=1.2𝑥−1.2∗ 𝑥 𝑥=1.2𝑥− 𝑥+1.56=1.2𝑥 𝑥−1.2𝑥=5.4− 𝑥=3.84 𝑥= =4.8

26. Application Problem
In certain deep parts of oceans, the pressure of sea water, 𝑃, in pounds per square foot, at a depth of 𝑑 feet below the surface, is given by the following equation: 𝑃=12+ 8𝑑 13 . If a scientific team uses special equipment to measures the pressure under water and finds it to be 612 pounds per square foot, at what depth is the team making their measurements? Solution: 612=12+ 8𝑑 −12= 8𝑑 = 8𝑑 ∗13=8𝑑 =𝑑 97.5=𝑑 Answer: The team is measuring at 97.5 feet below the surface.