Copyright © 2014, 2010, 2006 Pearson Education, Inc. 1 Chapter 2 Linear Functions and Equations

2 Copyright © 2014, 2010, 2006 Pearson Education, Inc. Linear Equations ♦ Learn about equations and recognize a linear equation ♦ Solve linear equations symbolically ♦ Sole linear equations graphically and numerically ♦ Solve problems involving percentages ♦ Apply problem-solving strategies 2.2

Copyright © 2014, 2010, 2006 Pearson Education, Inc. 3 Equations An equation is a statement that two mathematical expressions are equal. Some examples of equations are:

Copyright © 2014, 2010, 2006 Pearson Education, Inc. 4 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 © 2014, 2010, 2006 Pearson Education, Inc. 5 Types of Equations in One Variable Contradiction – An equation for which there is no solution. Example: 2x + 3 = 5 + 4x – 2x Simplifies to 2x + 3 = 2x + 5 Simplifies to 3 = 5 FALSE statement – there are no values of x for which 3 = 5. The equation has NO SOLUTION.

Copyright © 2014, 2010, 2006 Pearson Education, Inc. 6 Types of Equations in One Variable Identity – An equation for which every meaningful value of the variable is a solution. Example: 2x + 3 = 3 + 4x – 2x Simplifies to 2x + 3 = 2x + 3 Simplifies to 3 = 3 TRUE statement – no matter the value of x, the statement 3 = 3 is true. The solution is ALL REAL NUMBERS

Copyright © 2014, 2010, 2006 Pearson Education, Inc. 7 Types of Equations in One Variable Conditional Equation – An equation that is satisfied by some, but not all, values of the variable. Example 1: 2x + 3 = 5 + 4x Simplifies to 2x – 4x = 5 – 3 Simplifies to  2x = 2 Solution of the equation is: x =  1 Example 2: x 2 = 1 Solutions of the equation are: x =  1, x = 1

Copyright © 2014, 2010, 2006 Pearson Education, Inc. 8 Linear Equations in One Variable A linear equation in one variable is an equation that can be written in the form ax + b = 0, where a and b are real numbers with a ≠ 0. If an equation is not linear, then we say it is a nonlinear equation.

Copyright © 2014, 2010, 2006 Pearson Education, Inc. 9 Linear Equations in One Variable Rules of algebra can be used to write any linear equation in the form ax + b = 0. A linear equation has exactly one solution Examples of linear equations:

Copyright © 2014, 2010, 2006 Pearson Education, Inc. 10 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 © 2014, 2010, 2006 Pearson Education, Inc. 11 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 © 2014, 2010, 2006 Pearson Education, Inc. 12 Solve the equation 3(x – 4) = 2x – 1. Check your answer. Solution Apply the distributive property Example: Solving a linear equation symbolically

Copyright © 2014, 2010, 2006 Pearson Education, Inc. 13 The solution is 11. Check the answer. Example: Solving a linear equation symbolically

Copyright © 2014, 2010, 2006 Pearson Education, Inc. 14 Solve the linear equation Solution To eliminate fractions, multiply each side (or term in the equation) by the LCD, 12. Example: Eliminating fractions

Copyright © 2014, 2010, 2006 Pearson Education, Inc. 15 The solution Example: Eliminating fractions

Copyright © 2014, 2010, 2006 Pearson Education, Inc. 16 Solve the linear equation. Solution To eliminate fractions, multiply each side (or term in the equation) by the LCD, 12. Example: Eliminating decimals

Copyright © 2014, 2010, 2006 Pearson Education, Inc. 17 To eliminate decimals, multiply each side (or term in the equation) by 100. The solution Example: Eliminating decimals

Copyright © 2014, 2010, 2006 Pearson Education, Inc. 18 Intersection-of-Graphs Method The intersection-of-graphs method can be used to solve an equation graphically. STEP 1:Set y 1 equal to the left side of the equation, and set y 2 equal to the right side of the equation. STEP 2:Graph y 1 and y 2. STEP 3:Locate any points of intersection. The x-coordinates of these points correspond to solutions to the equation.

Copyright © 2014, 2010, 2006 Pearson Education, Inc. 19 Solve graphically and symbolically. Solution Graph Intersect at (2, 3), solution is 2. Example: Solving an equation graphically and symbolically

Copyright © 2014, 2010, 2006 Pearson Education, Inc. 20 Solution is 2, agrees with graphical. Example: Solving an equation graphically and symbolically

Copyright © 2014, 2010, 2006 Pearson Education, Inc. 21 Solve numerically to the nearest tenth. Enter Solution Make a table for y 1, incrementing by 1. This will show the solution is located in the interval 1 < x < 2. Example: Solving an equation numerically

Copyright © 2014, 2010, 2006 Pearson Education, Inc. 22 Solution lies in 1.43 < x < 1.44 Make a table for y 1, start at 1, increment by 0.1. Make a table for y 1, start at 1.4, increment by Solution lies in 1.4 < x < 1.5 Solution is 1.4 to the nearest tenth Example: Solving an equation numerically

Copyright © 2014, 2010, 2006 Pearson Education, Inc. 23 The area of a trapezoid with bases a and b and height h is given by Solve this equation for b. Multiply each side by 2 Solution Divide each side by h Subtract a from each side isolates b Example: Solving for a variable

Copyright © 2014, 2010, 2006 Pearson Education, Inc. 24 Solving Application Problems STEP 1:Read the problem and make sure you understand it. Assign a variable to what you are being asked. 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 © 2014, 2010, 2006 Pearson Education, Inc. 25 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? 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 11 miles per hour 1 – x: Time spent running at 9 miles per hour Solution Example: Solving an application involving motion

Copyright © 2014, 2010, 2006 Pearson Education, Inc. 26 STEP 2:d = rt; total distance is 10.1 STEP 3:Solve symbolically Example: Solving an application involving motion

Copyright © 2014, 2010, 2006 Pearson Education, Inc. 27 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. 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. STEP 4:check the solution Example: Solving an application involving motion

Copyright © 2014, 2010, 2006 Pearson Education, Inc. 28 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? 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 Solution Example: Mixing acid in chemistry

Copyright © 2014, 2010, 2006 Pearson Education, Inc. 29 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: Example: Mixing acid in chemistry

Copyright © 2014, 2010, 2006 Pearson Education, Inc. 30 STEP 3:Solve: divide each side by 0.13 Example: Mixing acid in chemistry

Copyright © 2014, 2010, 2006 Pearson Education, Inc. 31 STEP 3:We should add about 200 milliliters of water. Concentration or ≈ 13% 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. Example: Mixing acid in chemistry