2 2 Linear Equations Learn about equations and
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 Copyright © 2018, 2014, 2010 Pearson Education, Inc. All Rights Reserved
Equations An equation is a statement that two mathematical expressions are equal. Some examples of equations are: Copyright © 2018, 2014, 2010 Pearson Education, Inc. All Rights Reserved
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. Copyright © 2018, 2014, 2010 Pearson Education, Inc. All Rights Reserved
Types of Equations in One Variable (1 of 3) Contradiction − An equation for which there is no solution. Copyright © 2018, 2014, 2010 Pearson Education, Inc. All Rights Reserved
Types of Equations in One Variable (2 of 3) Identity − An equation for which every meaningful value of the variable is a solution. Copyright © 2018, 2014, 2010 Pearson Education, Inc. All Rights Reserved
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. Copyright © 2018, 2014, 2010 Pearson Education, Inc. All Rights Reserved
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. Copyright © 2018, 2014, 2010 Pearson Education, Inc. All Rights Reserved
Linear Equations in One Variable (2 of 2) Examples of linear equations: x − 12 = 0 2(1 − 4 x) = 16 x 2 x − 4 = − x x − 5 + 3(x − 1) = 0 Copyright © 2018, 2014, 2010 Pearson Education, Inc. All Rights Reserved
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. Copyright © 2018, 2014, 2010 Pearson Education, Inc. All Rights Reserved
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. Copyright © 2018, 2014, 2010 Pearson Education, Inc. All Rights Reserved
Example: Solving a linear equation symbolically (1 of 2) Solve the equation 3(x − 4) = 2 x − 1. Check your answer. Solution Apply the distributive property Copyright © 2018, 2014, 2010 Pearson Education, Inc. All Rights Reserved
Example: Solving a linear equation symbolically (2 of 2) The solution is 11. Check the answer. Copyright © 2018, 2014, 2010 Pearson Education, Inc. All Rights Reserved
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. Copyright © 2018, 2014, 2010 Pearson Education, Inc. All Rights Reserved
Example: Eliminating fractions (2 of 2) Copyright © 2018, 2014, 2010 Pearson Education, Inc. All Rights Reserved
Example: Eliminating decimals (1 of 2) Solve the linear equation. 0. 03(z − 3) − 0. 5(2 z + 1) = 0. 23 Copyright © 2018, 2014, 2010 Pearson Education, Inc. All Rights Reserved
Example: Eliminating decimals (2 of 2) Solution: To eliminate decimals, multiply each side (or term in the equation) by 100. Copyright © 2018, 2014, 2010 Pearson Education, Inc. All Rights Reserved
Example: Solving an equation graphically and symbolically (1 of 3) Copyright © 2018, 2014, 2010 Pearson Education, Inc. All Rights Reserved
Example: Solving an equation graphically and symbolically (2 of 3) Copyright © 2018, 2014, 2010 Pearson Education, Inc. All Rights Reserved
Example: Solving an equation graphically and symbolically (3 of 3) Solution is 2, agrees with graphical. Copyright © 2018, 2014, 2010 Pearson Education, Inc. All Rights Reserved
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. Copyright © 2018, 2014, 2010 Pearson Education, Inc. All Rights Reserved
Example: Using the x-intercept method (1 of 2) Solve − 4 x + 9 = 2(x − 2) + 1 by applying the x-intercept method. Solution Copyright © 2018, 2014, 2010 Pearson Education, Inc. All Rights Reserved
Example: Using the x-intercept method (2 of 2) Thus, h(x) = − 6 x + 12. Next, graph y = − 6 x + 12. There is one x-intercept: (2, 0), so the only solution to the given equation is 2. Copyright © 2018, 2014, 2010 Pearson Education, Inc. All Rights Reserved
Example: Solving an equation numerically (1 of 3) Make a table for y 1, incrementing by 1. This will show the solution is located in the interval 1 < x < 2. Copyright © 2018, 2014, 2010 Pearson Education, Inc. All Rights Reserved
Example: Solving an equation numerically (2 of 3) Make a table for y 1, start at 1, increment by 0. 1. Solution lies in 1. 4 < x < 1. 5 Make a table for y 1, start at 1. 4, increment by 0. 01. Solution lies in 1. 43 < x < 1. 44 Solution is 1. 4 to the nearest tenth. Copyright © 2018, 2014, 2010 Pearson Education, Inc. All Rights Reserved
Example: Solving an equation numerically (3 of 3) Copyright © 2018, 2014, 2010 Pearson Education, Inc. All Rights Reserved
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 approximately 38. 7 million riders of all ages who do not wear helmets. Find the total number of bicycle riders. Copyright © 2018, 2014, 2010 Pearson Education, Inc. All Rights Reserved
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. 76 x. b. We must find the x-value for which ƒ(x) = 38. 7 million, or solve the linear equation 0. 76 x = 38. 7. Copyright © 2018, 2014, 2010 Pearson Education, Inc. All Rights Reserved
Example: Solving for a variable Copyright © 2018, 2014, 2010 Pearson Education, Inc. All Rights Reserved
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? Copyright © 2018, 2014, 2010 Pearson Education, Inc. All Rights Reserved
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 Copyright © 2018, 2014, 2010 Pearson Education, Inc. All Rights Reserved
Example: Solving an application involving motion (2 of 3) STEP 2: d = rt; total distance is 10. 1 STEP 3: Solve symbolically Copyright © 2018, 2014, 2010 Pearson Education, Inc. All Rights Reserved
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. Copyright © 2018, 2014, 2010 Pearson Education, Inc. All Rights Reserved
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 Copyright © 2018, 2014, 2010 Pearson Education, Inc. All Rights Reserved
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 + 153. The equation is: 0. 13(x + 153) = 0. 30(153) Copyright © 2018, 2014, 2010 Pearson Education, Inc. All Rights Reserved
Example: Mixing acid in chemistry (3 of 4) STEP 3: Solve: divide each side by 0. 13 Copyright © 2018, 2014, 2010 Pearson Education, Inc. All Rights Reserved
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. Copyright © 2018, 2014, 2010 Pearson Education, Inc. All Rights Reserved
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