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Chapter 1 Section 7-1 Linear Equations 7-1-1 © 2008 Pearson Addison-Wesley. All rights reserved Linear Equations • • • • Solving Linear Equations Special Kinds of Linear Equations Literal Equations and Formulas Models 7-1-2 © 2008 Pearson Addison-Wesley. All rights reserved Equation An equation is a statement that two algebraic expressions are equal. A linear equation in one variable involves only real numbers and one variable. 4x 7 8 and 5k 3 k 1 Linear equations 7-1-3 © 2008 Pearson Addison-Wesley. All rights reserved Linear Equation in One Variable An equation in the variable x is linear if it can be written in the form Ax + B = C where A, B, and C are real numbers, with A 0. 7-1-4 © 2008 Pearson Addison-Wesley. All rights reserved Terminology A linear equation in one variable is also called a firstdegree equation. If the variable in an equation is replaced by a real number that makes the statement of the equation true, then that number is a solution of the equation. An equation is solved by finding its solution set, the set of all answers. Equivalent equations are equations with the same solution set. 7-1-5 © 2008 Pearson Addison-Wesley. All rights reserved Solving a Linear Equation in One Variable Step 1 Clear fractions. Multiply both sides by a common denominator. Step 2 Simplify each side separately. Clear parentheses and combine like terms. Step 3 Isolate the variable terms on one side. Use the addition property of equality. 7-1-6 © 2008 Pearson Addison-Wesley. All rights reserved Solving a Linear Equation in One Variable Step 4 Transform so that the coefficient of the variable is 1. Use the multiplication property of equality. Step 5 Check. Substitute the solution into the original equation. 7-1-7 © 2008 Pearson Addison-Wesley. All rights reserved Example: Solve a Linear Equation using the Distributive Property Solve 3(k + 1) + 2k = 13. Solution 3k + 3 + 2k = 13 5k + 3 = 13 5k + 3 – 3 = 13 – 3 5k = 10 Distributive property. Combine like terms. Subtract 3. Combine like terms. 7-1-8 © 2008 Pearson Addison-Wesley. All rights reserved Example: Solve a Linear Equation using the Distributive Property Solution (continued) 5k 10 5 5 Divide by 5. k=2 Check that the solution set is {2} by substituting 2 for k in the original equation. 7-1-9 © 2008 Pearson Addison-Wesley. All rights reserved Example: Solve a Linear Equation with Fractions Solve x 7 2x 8 4. 6 2 Solution x 7 2x 8 6 6 4 2 6 x 7 2x 8 6 6 6 4 6 2 x 7 3(2 x 8) 24 Multiply by LCD, 6. Distributive property. Multiply. 7-1-10 © 2008 Pearson Addison-Wesley. All rights reserved Example: Solve a Linear Equation with Fractions Solution (continued) x + 7 + 6x – 24 = –24 7x – 17 = –24 7x – 17 + 17 = –24 + 17 7x = –7 7 x 7 7 7 x = –1 Distributive property. Combine like terms. Add 17. Combine like terms. Divide by 7. Now check. 7-1-11 © 2008 Pearson Addison-Wesley. All rights reserved Types of Linear Equations Type Conditional Identity Number of Solutions One Infinite; solution set {all real numbers} Contradiction None; solution set Final Line When Solving x = number True statement, such as 0 = 0. False statement, such as 0 = 1. 7-1-12 © 2008 Pearson Addison-Wesley. All rights reserved Example: Contradiction Solve 3k + 4 – 2k = k. Solution k+4=k k–k+4=k–k 4=0 Combine like terms. Subtract k. Combine like terms. Because the result is false, the equation has no solution. The solution set is , so the original equation is a contradiction. 7-1-13 © 2008 Pearson Addison-Wesley. All rights reserved Example: Identity Solve 3k + 4 – 2k = k + 4. Solution k+4=k+4 k–k+4=k–k 4=4 Combine like terms. Subtract k. Combine like terms. Because the result is true, the solution set is {all real numbers} and the original equation is an identity. 7-1-14 © 2008 Pearson Addison-Wesley. All rights reserved Literal Equations and Formulas An equation involving variables (or letters), such as y = mx + b is called a literal equation. Formulas are literal equations in which more than one letter is used to express a relationship. It may be necessary to solve for one of the variables in a formula. This process is called solving for a specified variable. 7-1-15 © 2008 Pearson Addison-Wesley. All rights reserved Solving for a Specified Variable Step 1 Transform the equation so that all terms containing the specified variable are on one side of the equation and all terms without the variable are on the other side. Step 2 If necessary, use the distributive property to combine the terms with the specified variable. 7-1-16 © 2008 Pearson Addison-Wesley. All rights reserved Solving for a Specified Variable Step 3 Divide both sides by the factor that is multiplied by the specified variable. 7-1-17 © 2008 Pearson Addison-Wesley. All rights reserved Example: Solving for a Specified Variable Solve the formula P = 2L + 2W for L. Solution P – 2W = 2L + 2W – 2W Subtract 2W. P – 2W = 2L Combine like terms. P 2W 2 L 2 2 Divide by 2. P 2W P L or W L. 2 2 7-1-18 © 2008 Pearson Addison-Wesley. All rights reserved Models A mathematical model is an equation (or inequality) that describes the relationship between two quantities. A linear model is a linear equation. 7-1-19 © 2008 Pearson Addison-Wesley. All rights reserved Example: Temperature Conversion The relationship between degrees Celsius (C) and degrees Fahrenheit (F) is modeled by the linear equation F = 1.8C + 32. What Celsius degree corresponds to a Fahrenheit reading of 50°? 7-1-20 © 2008 Pearson Addison-Wesley. All rights reserved Example: Temperature Conversion Solution Because F = 50, the equation becomes 50 = 1.8C + 32 500 = 18C + 320 Multiply by 10. 180 = 18C Subtract 320. C = 10. Divide by 18. Therefore, a reading of 50 degrees Fahrenheit corresponds to a reading of 10 degrees Celsius. 7-1-21 © 2008 Pearson Addison-Wesley. All rights reserved