1. 2.2 Linear Equations
Learn about equations and recognize a linear equation
Solve linear equations symbolically
Solve linear equations graphically
Solve linear equations numerically
Solve problems involving percentages
Solve for a variable
Apply problem-solving strategies

2. Equations
An equation is a statement that two mathematical expressions are equal. Some examples of equations are:

3. Solutions to Equations
To solve an equation means to find all the values of the variable that make the equation a true statement. Such values are called solutions. The set of all solutions is the solution set. Solutions to an equation satisfy the equation. Two equations are equivalent if they have the same solution set.

4. Types of Equations in One Variable (1 of 3)
Contradiction − An equation for which there is no solution.

5. Types of Equations in One Variable (2 of 3)
Identity − An equation for which every meaningful value of the variable is a solution.

6. Types of Equations in One Variable (3 of 3)
Conditional Equation − An equation that is satisfied by some, but not all, values of the variable.

7. Linear Equations in One Variable (1 of 2)
A linear equation in one variable is an equation that can be written in the form ax + b = 0, where a and b are constants with a ≠ 0. If an equation is not linear, then we say it is a nonlinear equation.

8. Linear Equations in One Variable (2 of 2)
Examples of linear equations: x − 12 = 0 2(1 − 4x) = 16x 2x − 4 = − x x − 5 + 3(x − 1) = 0

9. Symbolic Solutions
Linear equations can be solved symbolically, and the solution is always exact. To solve a linear equation symbolically, we usually apply the properties of equality to the given equation and transform it into an equivalent equation that is simpler.

10. Properties of Equality
Addition Property of Equality If a, b, and c are real numbers, then a = b is equivalent to a + c = b + c. Multiplication Property of Equality If a, b, and c are real numbers with c ≠ 0, then a = b is equivalent to ac = bc.

11. Example: Solving a linear equation symbolically (1 of 2)
Solve the equation 3(x − 4) = 2x − 1. Check your answer. Solution Apply the distributive property

12. Example: Solving a linear equation symbolically (2 of 2)
The solution is 11. Check the answer.

13. Example: Eliminating fractions (1 of 2)
Solve the linear equation
Solution To eliminate fractions, multiply each side (or term in the equation) by the LCD, 12.

14. Example: Eliminating fractions (2 of 2)

15. Example: Eliminating decimals (1 of 2)
Solve the linear equation. 0.03(z − 3) − 0.5(2z + 1) = 0.23

16. Example: Eliminating decimals (2 of 2)
Solution: To eliminate decimals, multiply each side (or term in the equation) by 100.

17. Example: Solving an equation graphically and symbolically (1 of 3)

18. Example: Solving an equation graphically and symbolically (2 of 3)

19. Example: Solving an equation graphically and symbolically (3 of 3)
Solution is 2, agrees with graphical.

20. x-intercept method
If necessary, begin by rewriting the given equation in the form h(x) = 0. That is, using techniques of algebra, rewrite the given equation so that one side is equal to 0. Graph y = h(x). The x-coordinate of any x-intercept of the resulting graph is a solution to the given equation.

21. Example: Using the x-intercept method (1 of 2)
Solve −4x + 9 = 2(x − 2) + 1 by applying the x-intercept method. Solution

22. Example: Using the x-intercept method (2 of 2)
Thus, h(x) = −6x Next, graph y = −6x + 12.
There is one x-intercept: (2, 0), so the only solution to the given equation is 2.

23. Example: Solving an equation numerically (1 of 3)
Make a table for y1, incrementing by 1. This will show the solution is located in the interval 1 < x < 2.

24. Example: Solving an equation numerically (2 of 3)
Make a table for y1, start at 1, increment by 0.1. Solution lies in 1.4 < x < 1.5 Make a table for y1, start at 1.4, increment by < x < 1.44 Solution is 1.4 to the nearest tenth.

25. Example: Solving an equation numerically (3 of 3)

26. Example: Solving an application involving percentages (1 of 2)
A survey found that 76% of bicycle riders do not wear helmets. (Source: Opinion Research Corporation for Glaxo Wellcome, Inc.) a. Find a formula ƒ(x) for a function that computes the number of people who do not wear helmets among x bicycle riders. b. There are approximately 38.7 million riders of all ages who do not wear helmets. Find the total number of bicycle riders.

27. Example: Solving an application involving percentages (2 of 2)
Solution a. A function ƒ that computes 76% of x is given by ƒ(x) = 0.76x. b. We must find the x-value for which ƒ(x) = 38.7 million, or solve the linear equation 0.76x = 38.7.

28. Example: Solving for a variable

29. Solving Application Problems
STEP 1: Read the problem and make sure you understand it. Assign a variable to what you are being asked to find. If necessary, write other quantities in terms of the variable. STEP 2: Write an equation that relates the quantities described in the problem. You may need to sketch a diagram and refer to known formulas. STEP 3: Solve the equation and determine the solution. STEP 4: Look back and check your solution. Does it seem reasonable?

30. Example: Solving an application involving motion (1 of 3)
In 1 hour an athlete traveled 10.1 miles by running first at 8 miles per hour and then at 11 miles per hour. How long did the athlete run at each speed? Solution STEP 1: Let x represent the time in hours running at 8 mph, then 1 − x represents the time spent running at 11 mph. x: Time spent running at 8 miles per hour 1 − x: Time spent running at 11 miles per hour

31. Example: Solving an application involving motion (2 of 3)
STEP 2: d = rt; total distance is 10.1
STEP 3: Solve symbolically

32. Example: Solving an application involving motion (3 of 3)
STEP 3: The athlete runs 0.3 hour (18 min) at 8 miles per hour and 0.7 hour (42 min) at 11 miles per hour. STEP 4: Check the solution. 8(0.3) + 11(0.7) = 10.1 This sounds reasonable. The runner’s average speed was 10.1 miles per hour so the runner must have run longer at 11 miles per hour than at 8 miles per hour.

33. Example: Mixing acid in chemistry (1 of 4)
Pure water is being added to 153 milliliters of a 30% solution of hydrochloric acid. How much water should be added to dilute the solution to a 13% mixture? Solution STEP 1: Let x be the amount of pure water added to the 153 ml of 30% acid to make a 13% solution x: Amount of pure water to be added x + 153: Final volume of 13% solution

34. Example: Mixing acid in chemistry (2 of 4)
STEP 2: Pure water contains no acid, so the amount of acid before the water is added equals the amount of acid after water is added. Pure acid before is 30% of 153, pure acid after is 13% of x The equation is: 0.13(x + 153) = 0.30(153)

35. Example: Mixing acid in chemistry (3 of 4)
STEP 3: Solve: divide each side by 0.13

36. Example: Mixing acid in chemistry (4 of 4)
STEP 3: We should add about 200 milliliters of water. STEP 4: Initially the solution contains 0.30(153) = 45.9 milliliters of pure acid. If we add 200 milliliters of water to the 153 milliliters, the final solution is 353 milliliters, which includes 45.9 milliliters of pure acid.