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2 3.1 Solving Linear Equations

3 What You Will Learn  Solve a linear equation in standard form  Solve a linear equation in nonstandard form

4 Solving Linear Equations in Standard Form

5 This is an important step in your study of algebra. You were introduced to the rules of algebra, and you learned to use these rules to rewrite and simplify algebraic expressions. You have gained experience in translating verbal phrases and problems into algebraic forms. You are now ready to use these skills and experiences to solve equations.

6 Solving Linear Equations in Standard Form A linear equation in one variable x is an equation that can be written in the standard form ax + b = 0 a and b are real numbers with a ≠ 0

7 Solving Linear Equations in Standard Form A linear equation in one variable is also called a first-degree equation because its variable has an (implied) exponent of 1. Some examples of linear equations in standard form are 2x = 0, x – 7 = 0, 4x + 6 = 0, and Remember that to solve an equation involving x means to find all values of x that satisfy the equation.

8 Solving Linear Equations in Standard Form For the linear equation ax + b = 0, the goal is to isolate x by rewriting the equation in the form You begin with the original equation and write a sequence of equivalent equations, each having the same solution as the original equation. For instance, to solve the linear equation x – 2 = 0, you can add 2 to each side of the equation to obtain x = 2. Each equivalent equation is called a step of the solution. Isolate the variable x.

9 Example 1 – Solving Linear Equations in Standard Form Solve 3x – 15 = 0. Then check the solution. Solution: 3x – 15 = 0 3x – = x = 15 x = 5 It appears that the solution is x = 5. Write original equation. Add 15 to each side. Combine like terms. Divide each side by 3. Simplify.

10 Example 1 – Solving Linear Equations in Standard Form cont’d You can check this as follows. Check: 3x – 15 = 0 3(5) – – = 0 So, the solution is x = 5. Write original equation. Substitute 5 for x. Multiply. Solution checks.

11 In Example 1, be sure you see that solving an equation has two basic stages. The first stage is to find the solution (or solutions). The second stage is to check that each solution you find actually satisfies the original equation. You can improve your accuracy in algebra by developing the habit of checking each solution. A common question in algebra is “How do I know which step to do first to isolate x?” The answer is that you need practice. Solving Linear Equations in Standard Form

12 By solving many linear equations, you will find that your skill will improve. The key thing to remember is that you can “get rid of” terms and factors by using inverse operations. Here are some guidelines and examples. Guideline Equation Inverse Operation 1. Subtract to remove a sum. x + 3 = 0 Subtract 3 from each side. 2. Add to remove a difference. x – 5 = 0 Add 5 to each side. 3. Divide to remove a product. 4x = 20 Divide each side by Multiply to remove a quotient. Multiply each side by 8. Solving Linear Equations in Standard Form

13 A linear equation in one variable always has exactly one solution. You can show this with the following steps. ax + b = 0 ax + b – b = 0 – b ax = – b Original equation, with a ≠ 0 Subtract b from each side. Divide each side by a. Combine like terms. Simplify. Solving Linear Equations in Standard Form

14 It is clear that the last equation has only one solution, x = –b  a. Because the last equation is equivalent to the original equation, you can conclude that every linear equation in one variable written in standard form has exactly one solution. Solving Linear Equations in Standard Form

15 Example 2 – Solving Linear Equations in Standard Form a. b. Original equation Subtract 18 from each side Combine like terms Divide each side by 2 Simplify. Check in original equation Original equation Subtract 12 from each side Combine like terms Divide each side by 5 Simplify. Check in original equation

16 Example 2 – Solving Linear Equations in Standard Form c. Original equation Subtract 3 from each side Combine like terms Multiply each side by 3 Simplify. Check in original equation cont’d

17 Solving Linear Equations in Nonstandard Form

18 Solving Linear Equations in Nonstandard Form The definition of linear equation contains the phrase “that can be written in the standard form ax + b = 0.” This suggests that some linear equations may come in nonstandard or disguised form. A common type of linear equation is one in which the variable terms are not combined into one term. In such cases, you can begin the solution by combining like terms.

19 Example 3 – Solving a Linear Equation in Nonstandard Form Solve 3y + 8 – 5y = 4. Solution: 3y + 8 – 5y = 4 3y – 5y + 8 = 4 –2y + 8 = 4 –2y + 8 – 8 = 4 – 8 –2y = –4 y = 2 The solution is y = 2. Check this in the original equation. Write original equation. Group like terms. Combine like terms. Subtract 8 from each side. Combine like terms. Simplify. Divide each side by –2.

20 Example 4 – Solving a Linear Equation: Special Cases a. 2x + 3 = 2(x + 4) 2x + 3 = 2x + 8 2x – 2x + 3 = 2x – 2x ≠ 8 Because 3 does not equal 8, you can conclude that the original equation has no solution. Original equation Distributive Property Subtract 2x from each side. Simplify.

21 Example 4 – Solving a Linear Equation: Special Cases cont’d b. 4(x + 3) = 4x x + 12 = 4x x – 4x + 12 = 4x – 4x = 12 Because the last equation is true for any value of x, you can conclude that the original equation has infinitely many solutions. This type of equation is called an identity. Original equation Distributive Property Subtract 4x from each side. Simplify.

22 Example 5 – Geometry: Finding Dimensions You have 96 feet of fencing to enclose a rectangular pen for your dog. To provide sufficient running space for the dog to exercise, the pen is to be three times as long as it is wide. Find the dimensions of the pen. Solution: Begin by drawing and labeling a diagram Labels & Diagrams:

23 Example 5 – Geometry: Finding Dimensions cont’d The perimeter of a rectangle is the sum of twice its length and twice its width. Verbal Model: Equation: 96 = 2(3x) + 2x You can solve this equation as follows. 96 = 6x + 2x 96 = 8x 12 = x So, the width of the pen is 12 feet, and its length is 3(12) = 36 feet. Multiply. Combine like terms. Simplify. Divide each side by 8.

24 Example 6 – Finding the Number of Stadium Seats Sold Tickets for a concert cost $175 for each floor seat and $95 for each stadium seat. There were 2500 seats on the main floor, and these were sold out. The total revenue from ticket sales was $865,000. How many stadium seats were sold? Solution: 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) Equation:865,000 = 175(2500) + 95x

25 Example 6 – Finding the Number of Stadium Seats Sold You can solve the equations as follows. 865,000 = 175(2500) + 95x 865,000 = 437, x 865,000 – 437,500 = 437,500 – 437, x 427,500 = 95x 4500 = x There were 4500 stadium seats sold. To check this, go back to the original statement of the problem. cont’d Write equation Simplify Subtract Combine like terms Divide each side by 95 Simplify

26 Example 7 – Finding Your Gross Pay per Paycheck Write an algebraic equation that represents the following problem. Then solve the equation and answer the question. You have accepted a job offer at an annual salary of $40,830. This salary includes a year-end bonus of $750. You are paid twice a month. What will your gross pay be for each paycheck?

27 Example 7 – Finding Your Gross Pay per Paycheck cont’d Solution: Verbal Model: Labels: Income for year= 40,830 (dollars) Amount of each paycheck= x (dollars) Bonus = 750 (dollars) Equation: 40,830 = 24x Write equation 40,080 = 24x Subtract 750 from each side Divide each side by = x Simplify Each paycheck will be $1670. Check this in the original statement of the problem.

28 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