3.1 Solving Two-Step Equations Goal: Solve two-step equations. Notice in Example 1 that you isolate x by working backward. First you subtract from each side and then you divide. Example 1 Using Subtraction and Division to Solve Solve 4x 9 7. Check your solution. 4x 9 7 Write original equation. 4x 9 7 Subtract from each side. 4x 4x 16 Divide each side by. x Answer: The solution is. Check: 4x 9 7 Write original equation. 4( ) 9 7 Substitute for x. 7. Solve the equation. Check your solution. 1. 3x 8 26 2. 21 4x 7 Chapter 3 Pre-Algebra Notetaking Guide 43
3.1 Solving Two-Step Equations Goal: Solve two-step equations. Notice in Example 1 that you isolate x by working backward. First you subtract from each side and then you divide. Example 1 Using Subtraction and Division to Solve Solve 4x 9 7. Check your solution. 4x 9 7 Write original equation. 4x 9 9 7 9 Subtract 9 from each side. 4x 16 4x 16 Divide each side by 4. 4 4 x 4 Answer: The solution is 4. Check: 4x 9 7 Write original equation. 4( 4 ) 9 7 Substitute for x. 7 7 Solution checks. Solve the equation. Check your solution. 1. 3x 8 26 2. 21 4x 7 6 7 Chapter 3 Pre-Algebra Notetaking Guide 43
Example 2 Using Addition and Multiplication to Solve Solve 3 x 4 1. Check your solution. 3 x 4 1 Write original equation. 3 x 4 1 Add to each side. 3 x x 3 ( ) Multiply each side by. x Answer: The solution is. Check: 3 x 4 1 Write original equation. 3 4 1 Substitute for x. 1. Solve the equation. Check your solution. 3. 4 x 7 2 4. 8 5 b 3 44 Chapter 3 Pre-Algebra Notetaking Guide
Example 2 Using Addition and Multiplication to Solve Solve 3 x 4 1. Check your solution. 3 x 4 1 Write original equation. x 4 4 1 4 Add 4 to each side. 3 3 x 3 3 3 x 3 ( 3 ) Multiply each side by 3. x 9 Answer: The solution is 9. Check: x 4 1 3 Write original equation. 9 3 4 1 Substitute for x. 1 1 Solution checks. Solve the equation. Check your solution. 3. 4 x 7 2 4. 8 5 b 3 36 55 44 Chapter 3 Pre-Algebra Notetaking Guide
Example 3 Solving an Equation with Negative Coefficients Solve 2 3x 17. Check your solution. 2 3x 17 Write original equation. 2 3x 17 Subtract from each side. 3x 3x 15 Divide each side by. x Answer: The solution is. Check: 2 3x 17 Write original equation. 2 3( ) 17 Substitute for x. 17. Solve the equation. Check your solution. 5. 3 2y 19 6. 5 4 m Chapter 3 Pre-Algebra Notetaking Guide 45
Example 3 Solving an Equation with Negative Coefficients Solve 2 3x 17. Check your solution. 2 3x 17 Write original equation. 2 3x 2 17 2 Subtract 2 from each side. 3x 15 3x 15 Divide each side by 3. 3 3 x 5 Answer: The solution is 5. Check: 2 3x 17 Write original equation. 2 3( 5 ) 17 Substitute for x. 17 17 Solution checks. Solve the equation. Check your solution. 5. 3 2y 19 6. 5 4 m 8 9 Chapter 3 Pre-Algebra Notetaking Guide 45
3.2 Solving Equations Having Like Terms and Parentheses Goal: Solve equations using the distributive property. Example 1 Writing and Solving an Equation Baseball Game A group of five friends are going to a baseball game. Tickets for the game cost $12 each, or $60 for the group. The group also wants to eat at the game. Hot dogs cost $2.75 each and bottled water costs $1.25 each. The group has a total budget of $85. If the group buys the same number of hot dogs and bottles of water, how many can they afford to buy? Solution Let n represent the number of hot dogs and the number of bottles of water. Then 2.75n represents the cost of n hot dogs and 1.25n represents the cost of n bottles of water. Write a verbal model. Substitute. Combine like terms. Subtract from each side. n Divide each side by. Answer: The answer must be a whole number. Round down so the budget is not exceeded. The group can afford to buy hot dogs and bottles of water. 46 Chapter 3 Pre-Algebra Notetaking Guide
3.2 Solving Equations Having Like Terms and Parentheses Goal: Solve equations using the distributive property. Example 1 Writing and Solving an Equation Baseball Game A group of five friends are going to a baseball game. Tickets for the game cost $12 each, or $60 for the group. The group also wants to eat at the game. Hot dogs cost $2.75 each and bottled water costs $1.25 each. The group has a total budget of $85. If the group buys the same number of hot dogs and bottles of water, how many can they afford to buy? Solution Let n represent the number of hot dogs and the number of bottles of water. Then 2.75n represents the cost of n hot dogs and 1.25n represents the cost of n bottles of water. Write a verbal model. Cost of hot dogs Cost of Cost of bottled water tickets Total budget 2.75n 1.25n 60 85 Substitute. 4n 4n 60 85 Combine like terms. 60 60 85 60 Subtract 60 from each side. 4n 25 4n 25 4 4 Divide each side by 4. n 6 1 4 Answer: The answer must be a whole number. Round down so the budget is not exceeded. The group can afford to buy 6 hot dogs and 6 bottles of water. 46 Chapter 3 Pre-Algebra Notetaking Guide
Example 2 Solving Equations Using the Distributive Property Solve the equation. a. 24 6(2 x) b. 2(7 4x) 10 Solution a. 24 6(2 x) Write original equation. 24 Distributive property 24 Subtract from each side. Divide each side by. x b. 2(7 4x) 10 Write original equation. 10 Distributive property 10 Add to each side. Divide each side by. x Chapter 3 Pre-Algebra Notetaking Guide 47
Example 2 Solving Equations Using the Distributive Property Solve the equation. a. 24 6(2 x) b. 2(7 4x) 10 Solution a. 24 6(2 x) Write original equation. 24 12 6x Distributive property 24 12 12 6x 12 Subtract 12 from each side. 36 6x 36 6 6x 6 Divide each side by 6. 6 x b. 2(7 4x) 10 Write original equation. 14 8x 10 Distributive property 14 8x 14 10 14 Add 14 to each side. 8x 24 8x 24 Divide each side by 8. 8 8 x 3 Chapter 3 Pre-Algebra Notetaking Guide 47
Example 3 Combining Like Terms After Distributing Solve 6x 4(x 1) 14. 6x 4(x 1) 14 Write original equation. 6x 14 Distributive property 14 Combine like terms. 14 Subtract from each side. Divide each side by. x Solve the equation. Check your solution. 1. 20 5(3 x) 2. 4y 14 3y 28 3. 3(6 2x) 12 4. 5x 2(x 3) 30 48 Chapter 3 Pre-Algebra Notetaking Guide
Example 3 Combining Like Terms After Distributing Solve 6x 4(x 1) 14. 6x 4(x 1) 14 Write original equation. 6x 4x 4 14 Distributive property 2x 4 14 Combine like terms. 2x 4 4 14 4 Subtract 4 from each side. 2x 10 2x 10 Divide each side by 2. 2 2 x 5 Solve the equation. Check your solution. 1. 20 5(3 x) 2. 4y 14 3y 28 7 6 3. 3(6 2x) 12 4. 5x 2(x 3) 30 5 8 48 Chapter 3 Pre-Algebra Notetaking Guide
3.3 Solving Equations with Variables on Both Sides Goal: Solve equations with variables on both sides. Example 1 Solving an Equation with the Variable on Both Sides Solve 5n 7 9n 21. 5n 7 9n 21 Write original equation. 5n 7 9n 21 Subtract from each side. 7 21 7 21 Subtract from each side. Divide each side by. n Answer: The solution is. Example 2 An Equation with No Solution Solve 3(2x 1) 6x. 3(2x 1) 6x Write original equation. 6x Distributive property Notice that this statement true because the number 6x. The equation has. As a check, you can continue solving the equation. 6x Subtract from each side. The statement. true, so the equation has Chapter 3 Pre-Algebra Notetaking Guide 49
3.3 Solving Equations with Variables on Both Sides Goal: Solve equations with variables on both sides. Example 1 Solving an Equation with the Variable on Both Sides Solve 5n 7 9n 21. 5n 7 9n 21 Write original equation. 5n 7 5n 9n 21 5n Subtract 5n from each side. 7 4n 21 7 21 4n 21 21 Subtract 21 from each side. 28 4n 28 4n Divide each side by 4. 4 4 7 n Answer: The solution is 7. Example 2 An Equation with No Solution Solve 3(2x 1) 6x. 3(2x 1) 6x Write original equation. 6x 3 6x Distributive property Notice that this statement is not true because the number 6x cannot be equal to 3 more than itself. The equation has no solution. As a check, you can continue solving the equation. 6x 3 6x 6x 6x Subtract 6x from each side. 3 0 The statement 3 0 is not true, so the equation has no solution. Chapter 3 Pre-Algebra Notetaking Guide 49
Example 3 Solving an Equation with All Numbers as Solutions Solve 4(x 2) 4x 8. 4(x 2) 4x 8 Write original equation. 4x 8 Distributive property Notice that for all values of x, the statement. The equation has. 4x 8 is Solve the equation. Check your solution. 1. 3n 6 5n 20 2. 12 x 4(3x 1) 3. 3(2n 4) 2(3n 6) 4. 2x 7 2x 13 50 Chapter 3 Pre-Algebra Notetaking Guide
Example 3 Solving an Equation with All Numbers as Solutions Solve 4(x 2) 4x 8. 4(x 2) 4x 8 Write original equation. 4x 8 4x 8 Distributive property Notice that for all values of x, the statement true 4x 8. The equation has every number as a solution. 4x 8 is Solve the equation. Check your solution. 1. 3n 6 5n 20 2. 12 x 4(3x 1) 13 no solution 3. 3(2n 4) 2(3n 6) 4. 2x 7 2x 13 all numbers 5 50 Chapter 3 Pre-Algebra Notetaking Guide
Example 4 Solving an Equation to Find a Perimeter Geometry Find the perimeter of the square. x 6 Solution 1. A square has four sides of equal length. Write an equation and solve for x. Write equation. Subtract from each side. 3x Divide each side by. x 2. Find the length of one side by substituting for x in either expression. 3x 3( ) Substitute for x and multiply. 3. To find the perimeter, multiply the length of one side by. p Answer: The perimeter of the square is units. 5. Find the perimeter of the square. 3x 8 5x Chapter 3 Pre-Algebra Notetaking Guide 51
Example 4 Solving an Equation to Find a Perimeter Geometry Find the perimeter of the square. x 6 Solution 1. A square has four sides of equal length. Write an equation and solve for x. 3x x 6 3x Write equation. 3x x x 6 x Subtract x from each side. 2x 6 2x 6 Divide each side by 2. 2 2 x 3 2. Find the length of one side by substituting 3 for x in either expression. 3x 3( 3 ) 9 Substitute for x and multiply. 3. To find the perimeter, multiply the length of one side by 4. 4 p 9 36 Answer: The perimeter of the square is 36 units. 5. Find the perimeter of the square. 3x 8 5x 80 units Chapter 3 Pre-Algebra Notetaking Guide 51
Focus On Algebra Use after Lesson 3.3 Rewriting Equations and Formulas Goal: Rewrite literal equations and formulas. Vocabulary Literal equation Example 1 Solving a Literal Equation Solve ax bx c for x. Then use the solution to solve 5x 2x 12. Solution 1. Solve ax bx c for x. Write original equation. When solving you calculate a literal a equation, slope, be follow sure to the use same the x- steps and y- you would take coordinates when solving of the a specific two points equation in the of that same form. order. c Distributive property x Assume. Divide each side by. 2. Use the solution to solve 5x 2x 12. x Solution of literal equation Substitute for a, for b, and for c. Answer: The solution of 5x 2x 12 is. Solve the literal equation for x. Then use the solution to solve the specific equation. x 1. x b c; 1 3 a 2. ax c bx; 5x 9 4x 4 52 Chapter 3 Pre-Algebra Notetaking Guide
Focus On Algebra Use after Lesson 3.3 Rewriting Equations and Formulas Goal: Rewrite literal equations and formulas. Vocabulary Literal equation An equation such as ax b c, in which coefficients and constants have been replaced by letters When solving you calculate a literal a equation, slope, be follow sure to the use same the x- steps and y- you would take coordinates when solving of the a specific two points equation in the of that same form. order. Example 1 Solving a Literal Equation Solve ax bx c for x. Then use the solution to solve 5x 2x 12. Solution 1. Solve ax bx c for x. ax bx c Write original equation. (a b)x c Distributive property c x Assume (a b) 0. Divide each side by (a b). a b 2. Use the solution to solve 5x 2x 12. c x Solution of literal equation a b 12 Substitute 5 for a, 2 for b, and 12 for c. 5 2 4 Answer: The solution of 5x 2x 12 is 4. Solve the literal equation for x. Then use the solution to solve the specific equation. x 1. x b c; 1 3 a 2. ax c bx; 5x 9 4x 4 c x a(b c); 16 x ; 1 a b 52 Chapter 3 Pre-Algebra Notetaking Guide
Example 2 Rewriting an Equation Solve 3y 6x 12 for y. Solution 3y 6x 12 Write original equation. Add to each side. Multiply each side by. Use and simplify. Solve the equation for y. 3. 2y 8 14x 4. 18 6x 9y Example 3 Rewriting and Using a Geometric Formula The area A of a rectangle is given by the formula A lw where l is the length and w is the width. a. Solve the formula for the width w. b. Use the rewritten formula to find the width of the rectangle shown. A 640 ft 2 w I 40 ft Solution a. A lw Original formula. w each side by. Notice in Example 1 that you isolate x by working A 640 backward. ft 2 w First you subtract from each I side 40 and ft then you divide. b. Substitute for and for in the rewritten formula. w Write rewritten formula. Substitute for and for. Answer: The width is Chapter 3 Pre-Algebra Notetaking Guide 53
Example 2 Rewriting an Equation Solve 3y 6x 12 for y. Solution 3y 6x 12 Write original equation. 3y 6x 12 Add 6x to each side. 1 1 1 3 (6x 12) 3 (3y) Multiply each side by 3. y 2x 4 Use distributive property and simplify. Solve the equation for y. 3. 2y 8 14x 4. 18 6x 9y y 7x 4 y 2 2 3 x Example 3 Rewriting and Using a Geometric Formula The area A of a rectangle is given by the formula A lw where l is the length and w is the width. a. Solve the formula for the width w. b. Use the rewritten formula to find the width of the rectangle shown. A 640 ft 2 w Notice in Example 1 that you isolate x by working A 640 backward. ft 2 w First you subtract from each I side 40 and ft then you divide. I 40 ft Solution a. A lw Original formula. A w Divide each side by I. I b. Substitute 640 for A and 40 for I in the rewritten formula. A w Write rewritten formula. I 6 40 Substitute 640 for A and 40 for I. 40 16 Answer: The width is 16 feet. Chapter 3 Pre-Algebra Notetaking Guide 53
3.4 Solving Inequalities Using Addition or Subtraction Goal: Solve inequalities using addition or subtraction. Vocabulary Inequality: Solution of an inequality: Equivalent inequalities: Example 1 Writing and Graphing an Inequality Air Travel An airline allows passengers to carry on-board one piece of luggage. Luggage that exceeds 40 pounds cannot be carried on-board. Write an inequality that gives the weight of luggage that cannot be carried on-board. Solution Let w represent the weight of the luggage. Because the weight cannot exceed 40 pounds, the weight of luggage that cannot be carried on-board must be. Answer: The inequality is. Draw the graph below. 0 10 20 30 40 50 60 70 80 54 Chapter 3 Pre-Algebra Notetaking Guide
3.4 Solving Inequalities Using Addition or Subtraction Goal: Solve inequalities using addition or subtraction. Vocabulary Inequality: An inequality is a statement formed by placing an inequality symbol, such as < or >, between two expressions. Solution of an inequality: Equivalent inequalities: The solution of an inequality with a variable is the set of all numbers that produce true statements when substituted for the variable. Equivalent inequalities are inequalities that have the same solution. Example 1 Writing and Graphing an Inequality Air Travel An airline allows passengers to carry on-board one piece of luggage. Luggage that exceeds 40 pounds cannot be carried on-board. Write an inequality that gives the weight of luggage that cannot be carried on-board. Solution Let w represent the weight of the luggage. Because the weight cannot exceed 40 pounds, the weight of luggage that cannot be carried on-board must be greater than 40 pounds. Answer: The inequality is w > 40. Draw the graph below. 0 10 20 30 40 50 60 70 80 54 Chapter 3 Pre-Algebra Notetaking Guide
The addition and subtraction properties of inequality are also true for inequalities involving and. Addition and Subtraction Properties of Inequality Words Adding or subtracting the same number on each side of an inequality produces an equivalent inequality. Algebra If a < b, then a c < b c and a c < b c. If a > b, then a c > b c and a c > b c. Example 2 Solving an Inequality Using Subtraction Solve m 9 12. Graph and check your solution. m 9 12 Write original inequality. m 9 12 Subtract from each side. m Answer: The solution is m. Draw the graph below. 1 0 1 2 3 4 5 6 7 Check: Choose any number less than or equal to the number into the original inequality.. Substitute m 9 12 Write original inequality. 9? 12 Substitute 0 for m. 12. You can read an inequality from left to right as well as from right to left. For instance, "2 is greater than x" can also be read "x is less than 2." Algebraically, this means that 2 x can also be written as x 2. Example 3 Solving an Inequality Using Addition Solve 7 < x 11. Graph your solution. 7 < x 11 Write original inequality. 7 < x 11 Add to each side. < x Answer: The solution is < x, or. Draw the graph below. 1 0 1 2 3 4 5 6 7 Chapter 3 Pre-Algebra Notetaking Guide 55
The addition and subtraction properties of inequality are also true for inequalities involving and. Addition and Subtraction Properties of Inequality Words Adding or subtracting the same number on each side of an inequality produces an equivalent inequality. Algebra If a < b, then a c < b c and a c < b c. If a > b, then a c > b c and a c > b c. Example 2 Solving an Inequality Using Subtraction Solve m 9 12. Graph and check your solution. m 9 12 Write original inequality. m 9 9 12 9 Subtract 9 from each side. m 3 Answer: The solution is m 3. Draw the graph below. 1 0 1 2 3 4 5 6 7 Check: Choose any number less than or equal to the number into the original inequality. 3. Substitute m 9 12 Write original inequality. 0 9? 12 Substitute 0 for m. 9 12. Solution checks You can read an inequality from left to right as well as from right to left. For instance, "2 is greater than x" can also be read "x is less than 2." Algebraically, this means that 2 x can also be written as x 2. Example 3 Solving an Inequality Using Addition Solve 7 < x 11. Graph your solution. 7 < x 11 Write original inequality. 7 11 < x 11 11 Add 11 to each side. 4 < x Answer: The solution is 4 < x, or x > 4. Draw the graph below. 1 0 1 2 3 4 5 6 7 Chapter 3 Pre-Algebra Notetaking Guide 55
Solve the inequality. Graph and check your solution. 1. m 7 < 13 1 0 1 2 3 4 5 6 7 2. a 4 5 1 0 1 2 3 4 5 6 7 3. x 2 3 1 0 1 2 3 4 5 6 7 4. 6 < z 7 1 0 1 2 3 4 5 6 7 56 Chapter 3 Pre-Algebra Notetaking Guide
Solve the inequality. Graph and check your solution. 1. m 7 < 13 m < 6 1 0 1 2 3 4 5 6 7 2. a 4 5 a 1 1 0 1 2 3 4 5 6 7 3. x 2 3 x 5 1 0 1 2 3 4 5 6 7 4. 6 < z 7 1 < z or z > 1 1 0 1 2 3 4 5 6 7 56 Chapter 3 Pre-Algebra Notetaking Guide
3.5 Solving Inequalities Using Multiplication or Division Goal: Solve inequalities using multiplication or division. The multiplication properties of inequality are also true for inequalities involving >,, and. Multiplication Property of Inequality Words Multiplying each side of an inequality by a positive number produces an equivalent inequality. Multiplying each side of an inequality by a negative number and reversing the direction of the inequality symbol produces an equivalent inequality. Algebra If a < b and c > 0, then ac If a < b and c < 0, then ac bc. bc. Example 1 Solving an Inequality Using Multiplication m Solve > 2. 4 m > 2 4 m 4 p 2 m Write original inequality. Multiply each side by. Reverse inequality symbol. Solve the inequality. Graph your solution. t b 1. < 3 2. 1 5 8 9 10 11 12 13 14 15 16 9 8 7 6 5 4 3 2 Chapter 3 Pre-Algebra Notetaking Guide 57
3.5 Solving Inequalities Using Multiplication or Division Goal: Solve inequalities using multiplication or division. The multiplication properties of inequality are also true for inequalities involving >,, and. Multiplication Property of Inequality Words Multiplying each side of an inequality by a positive number produces an equivalent inequality. Multiplying each side of an inequality by a negative number and reversing the direction of the inequality symbol produces an equivalent inequality. Algebra If a < b and c > 0, then ac If a < b and c < 0, then ac < > bc. bc. Example 1 Solving an Inequality Using Multiplication m Solve > 2. 4 4 m > 2 4 m 4 < 4 p 2 m < 8 Write original inequality. Multiply each side by 4. Reverse inequality symbol. Solve the inequality. Graph your solution. t b 1. < 3 2. 1 5 8 t < 15 b 8 9 10 11 12 13 14 15 16 9 8 7 6 5 4 3 2 Chapter 3 Pre-Algebra Notetaking Guide 57
The division properties of inequality are also true for inequalities involving >,, and. Division Property of Inequality Words Dividing each side of an inequality by a positive number produces an equivalent inequality. Dividing each side of an inequality by a negative number and reversing the direction of the inequality symbol produces an equivalent inequality. Algebra If a < b and c > 0, then a c If a < b and c < 0, then a c b c. b c. Example 2 Solve 11t 33. 11t 33 11t t Solving an Inequality Using Division 33 Write original inequality. Divide each side by. Reverse inequality symbol. Solve the inequality. Graph your solution. 3. 4y 36 4. 3x > 12 3 4 5 6 7 8 9 10 9 8 7 6 5 4 3 2 58 Chapter 3 Pre-Algebra Notetaking Guide
The division properties of inequality are also true for inequalities involving >,, and. Division Property of Inequality Words Dividing each side of an inequality by a positive number produces an equivalent inequality. Dividing each side of an inequality by a negative number and reversing the direction of the inequality symbol produces an equivalent inequality. Algebra If a < b and c > 0, then a c If a < b and c < 0, then a c < > b c. b c. Example 2 Solve 11t 33. 11t 33 11t 11 t Solving an Inequality Using Division 33 11 3 Write original inequality. Divide each side by 11. Reverse inequality symbol. Solve the inequality. Graph your solution. 3. 4y 36 4. 3x > 12 y 9 x < 4 3 4 5 6 7 8 9 10 9 8 7 6 5 4 3 2 58 Chapter 3 Pre-Algebra Notetaking Guide
3.6 Solving Multi-Step Inequalities Goal: Solve multi-step inequalities. Example 1 Writing and Solving a Multi-Step Inequality Charity Walk You are participating in a charity walk. You want to raise at least $500 for the charity. You already have $175 by asking people to pledge $25 each. How many more $25 pledges do you need? Solution Let p represent the number of additional pledges. Write a verbal model. p Substitute. Subtract each side. from Divide each side by. p Answer: You need at least more $25 pledges. 1. Look back at Example 1. Suppose you wanted to raise at least $620 and you already have raised $380 by asking people to pledge $20 each. How many more $20 pledges do you need? Chapter 3 Pre-Algebra Notetaking Guide 59
3.6 Solving Multi-Step Inequalities Goal: Solve multi-step inequalities. Example 1 Writing and Solving a Multi-Step Inequality Charity Walk You are participating in a charity walk. You want to raise at least $500 for the charity. You already have $175 by asking people to pledge $25 each. How many more $25 pledges do you need? Solution Let p represent the number of additional pledges. Write a verbal model. Money already raised Amount Additional per pledge p pledges Minimum desired amount 175 175 25p 500 Substitute. 25p 175 500 175 Subtract 175 each side. 25p 325 from 25p 325 25 25 Divide each side by. 25 p 13 Answer: You need at least 13 more $25 pledges. 1. Look back at Example 1. Suppose you wanted to raise at least $620 and you already have raised $380 by asking people to pledge $20 each. How many more $20 pledges do you need? at least 12 more $20 pledges Chapter 3 Pre-Algebra Notetaking Guide 59
Example 2 Solving a Multi-Step Inequality x 9 < 7 3 Original inequality x 9 < 7 Add to each side. 3 x < 3 x 3 x p Multiply each side by. Reverse inequality symbol. Solve the inequality. Then graph the solution. x 2. 2x 9 < 25 3. 3 2 4 3 4 5 6 7 8 9 10 3 4 5 6 7 8 9 10 4. 2 > 4 x 5. 2 x 4 9 9 8 7 6 5 4 3 2 3 4 5 6 7 8 9 10 60 Chapter 3 Pre-Algebra Notetaking Guide
Example 2 x 9 < 7 3 Original inequality x 9 9 < 7 9 Add 9 to each side. 3 3 Solving a Multi-Step Inequality x < 3 x 3 > x > 2 3 6 p 2 Multiply each side by 3. Reverse inequality symbol. Solve the inequality. Then graph the solution. x 2. 2x 9 < 25 3. 3 2 4 x < 8 x 4 3 4 5 6 7 8 9 10 3 4 5 6 7 8 9 10 4. 2 > 4 x 5. 2 x 4 9 x > 6 x 10 9 8 7 6 5 4 3 2 3 4 5 6 7 8 9 10 60 Chapter 3 Pre-Algebra Notetaking Guide
3 Words to Review Give an example of the vocabulary word. Inequality Solution of an inequality Equivalent inequalities Literal equation Review your notes and Chapter 3 by using the Chapter Review on pages 156 159 of your textbook. Chapter 3 Pre-Algebra Notetaking Guide 61
3 Words to Review Give an example of the vocabulary word. Inequality x > 0 Solution of an inequality The solution of m 9 12 is m 3. Equivalent inequalities 2x < 10 and x > 5 Literal equation ax by c Review your notes and Chapter 3 by using the Chapter Review on pages 156 159 of your textbook. Chapter 3 Pre-Algebra Notetaking Guide 61