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2. 2 2.4 Introduction to Equations

3. 3 What You Will Learn  Check whether a given value is a solution of an equation  Use properties of equality to form equivalent equations.  Use a verbal model to write an algebraic equation

4. 4 Checking Solutions of Equations

5. 5 Equations An equation is a statement that two algebraic expressions are equal. For example, x = 3, 5x – 2 = 8, = 7, and x 2 – 9 = 0 are equations. To solve an equation involving the variable x means to find all values of x for which the equation is true. Such values are called solutions.

6. 6 Equations For instance, x = 2 is a solution of the equation 5x – 2 = 8 because 5(2) – 2 = 8 is a true statement. The solutions of an equation are said to satisfy the equation.

7. 7 Checking Solutions of Equations To check whether a given value is a solution of an equation, substitute the value into the original equation. If the substitution results in a true statement, then the value is a solution of the equation. If the substitution results in a false statement, then the value is not a solution of the equation.

8. 8 Example 1 – Checking Solutions of an Equation Determine whether (a)x = –2 and (b) x = 2 are solutions of x 2 – 5 = 4x + 7. Solution: a. x 2 – 5 = 4x + 7 (–2) 2 – 5 4(–2) + 7 4 – 5 –8 + 7 –1 = –1 Because the substitution results in a true statement, you can conclude that x = –2 is a solution of the original equation. Write original equation. Substitute –2 for x. Simplify. Solution checks.

9. 9 Solution: b. x 2 – 5 = 4x + 7 (2) 2 – 5 4(2) + 7 4 – 5 8 + 7 –1 ≠ 15 Because the substitution results in a false statement, you can conclude that x = 2 is not a solution of the original equation. Write original equation. Substitute 2 for x. Simplify. Solution checks. cont’d Example 1 – Checking Solutions of an Equation

10. 10 Example 2 – Comparing Equations and Expressions Make a table comparing algebraic expressions and algebraic equations. Solution:

11. 11 Forming Equivalent Equations

12. 12 Forming Equivalent Equations It is helpful to think of an equation as having two sides that are in balance. Consequently, when you try to solve an equation, you must be careful to maintain that balance by performing the same operation(s) on each side.

13. 13 Forming Equivalent Equations Two equations that have the same set of solutions are called equivalent. For instance, the equations x = 3 and x – 3 = 0 are equivalent because both have only one solution—the number 3. When any one of the operations in the following list is applied to an equation, the resulting equation is equivalent to the original equation.

14. 14 Forming Equivalent Equations

15. 15 Example 3 – Forming Equivalent Equations The second and third operations in this list can be used to eliminate terms or factors in an equation. For example, to solve the equation x – 5 = 1, you need to eliminate the term –5 on the left side. This is accomplished by adding its opposite, 5, to each side. x – 5 = 1 x – 5 + 5 = 1 + 5 x + 0 = 6 x = 6 These four equations are equivalent, and they are called the steps of the solution. Write original equation. Add 5 to each side. Combine like terms. Solution

16. 16 Example 4 – Identifying Properties of Equality Identify the property of equality used to solve each equation. a. x – 5 = 0 x – 5 + 5 = 0 + 5 x = 5 Solution: The Addition Property of Equality is used to add 5 to each side of the equation in the second step. Adding 5 eliminates the term –5 from the left side of the equation. Original equation Add 5 to each side. Solution

17. 17 cont’d b. Solution: The Multiplication Property of Equality is used to multiply each side of the equation by 5 in the second step. Multiplying by 5 eliminates the denominator from the left side of the equation. Multiply each side by 5. Original equation Solution Example 4 – Identifying Properties of Equality

18. 18 cont’d c. Solution: The Multiplication Property of Equality is used to divide each side of the equation by 4 (or multiply each side by ) in the second step. Dividing by 4 eliminates the coefficient from the left side of the equation. Original equation Solution Divide each side by 4. Example 4 – Identifying Properties of Equality

19. 19 Writing Equations

20. 20 Writing Equations It is helpful to use two phases in constructing equations that model real-life situations, as shown below. In the first phase, you translate the verbal description into a verbal model. In the second phase, you assign labels and translate the verbal model into a mathematical model or an algebraic equation.

21. 21 Example 5 – Using a Verbal Model to Write an Equation Write an algebraic equation for the following problem. The total income that an employee received in a year was $40,950. How much was the employee paid each week? Assume that each weekly paycheck contained the same amount, and that the year consisted of 52 weeks.

22. 22 Example 5 – Using a Verbal Model to Write an Equation cont’d Solution: Verbal Model: Labels: Income for year = 40,950 (dollars) Weekly pay = x (dollars per week) Number of weeks = 52 (weeks) Algebraic Model: 40,950 = 52x

23. 23 Example 6 – Using a Verbal Model to Write an Equation Write an algebraic equation for the following problem. Returning to college after spring break, you travel 3 hours and stop for lunch. You know that it takes 45 minutes to complete the last 36 miles of the 180-mile trip. What was the average speed during the first 3 hours of the trip?

24. 24 Example 6 – Using a Verbal Model to Write an Equation cont’d Solution: Verbal Model: Labels: Distance= 180 – 36 = 144 (miles) Rate = r (miles per hour) Time = 3 (hours) Algebraic Model: 144 = 3r

25. 25 Example 7 – Using a Verbal Model to Write an Equation Write an algebraic equation for the following problem. Tickets for a concert cost $175 for each floor seat and $95 for each stadium seat. There were 2500 seats ont he main floor, and these were sold out. The total revenue from ticket sales were $865,000. How many stadium seats were sold?

26. 26 Example 7 – Using a Verbal Model to Write an Equation cont’d Solution: Verbal Model: Labels: Total Revenue= 865,000 (dollars) Price per floor seat = 175 (dollars per seat) Number of floor seats = 2500 (seats) Price per stadium seat = 95 (dollars per seat) Number of stadium seats = x (seats) Algebraic Model: 865,000 = 175(2500) + 95x