Page 500 #15-32 ANSWERS

Student Progress Chart Lesson Reflection

Today’s Learning Goal Assignment Learn to solve multistep equations.

Pre-Algebra HW Page 504 #12-24

Solving Multistep Equations 10-2 Solving Multistep Equations Warm Up Problem of the Day Lesson Presentation Pre-Algebra

Solving Multistep Equations Pre-Algebra 10-2 Solving Multistep Equations Warm Up Solve. 1. 3x = 102 2. = 15 3. z – 100 = –1 4. 1.1 + 5w = 98.6 x = 34 y 15 y = 225 z = 99 w = 19.5

Problem of the Day Ana has twice as much money as Ben, and Ben has three times as much as Clio. Together they have $160. How much does each person have? Ana, $96; Ben, $48; Clio, $16

Today’s Learning Goal Assignment Learn to solve multistep equations.

To solve a complicated equation, you may have to simplify the equation first by combining like terms.

Additional Example 1: Solving Equations That Contain Like Terms Solve. 8x + 6 + 3x – 2 = 37 11x + 4 = 37 Combine like terms. – 4 – 4 Subtract to undo addition. 11x = 33 33 11 11x = Divide to undo multiplication. x = 3

Additional Example 1 Continued Check 8x + 6 + 3x – 2 = 37 8(3) + 6 + 3(3) – 2 = 37 ? Substitute 3 for x. 24 + 6 + 9 – 2 = 37 ? 37 = 37 ? 

Try This: Example 1 Solve. 9x + 5 + 4x – 2 = 42 13x + 3 = 42 Combine like terms. – 3 – 3 Subtract to undo addition. 13x = 39 39 13 13x = Divide to undo multiplication. x = 3

Try This: Example 1 Continued Check 9x + 5 + 4x – 2 = 42 9(3) + 5 + 4(3) – 2 = 42 ? Substitute 3 for x. 27 + 5 + 12 – 2 = 42 ? 42 = 42 ? 

If an equation contains fractions, it may help to multiply both sides of the equation by the least common denominator (LCD) to clear the fractions before you isolate the variable.

Additional Example 2: Solving Equations That Contain Fractions Solve. A. + = – 5n 4 7 4 3 4 Multiply both sides by 4 to clear fractions, and then solve. ( ) ( ) 7 4 –3 5n 4 + = 4 ( ) ( ) ( ) 5n 4 7 –3 4 + 4 = 4 Distributive Property. 5n + 7 = –3

Additional Example 2 Continued – 7 –7 Subtract to undo addition. 5n = –10 5n 5 –10 = Divide to undo multiplication. n = –2

The least common denominator (LCD) is the smallest number that each of the denominators will divide into. Remember!

Additional Example 2B: Solving Equations That Contain Fractions Solve. B. + – = x 2 7x 9 17 2 3 The LCD is 18. ( ) x 2 3 7x 9 17 18 + – = 18 Multiply both sides by the LCD. 18( ) + 18( ) – 18( ) = 18( ) 7x 9 x 2 17 3 Distributive Property. 14x + 9x – 34 = 12 23x – 34 = 12 Combine like terms.

Additional Example 2B Continued 23x – 34 = 12 Combine like terms. + 34 + 34 Add to undo subtraction. 23x = 46 = 23x 23 46 Divide to undo multiplication. x = 2

Additional Example 2B Continued Check x 2 7x 9 17 2 3 + – = 2 3 Substitute 2 for x. 7(2) 9 + – = (2) 17 ? 2 3 14 9 + – = 17 ? 2 3 14 9 + – = 17 ? 1 The LCD is 9. 6 9 14 + – = 17 ? 6 9 = ? 

Try This: Example 2A Solve. A. + = – 3n 4 5 4 1 4 Multiply both sides by 4 to clear fractions, and then solve. ( ) ( ) 5 4 –1 3n 4 + = 4 ( ) ( ) ( ) 3n 4 5 –1 4 + 4 = 4 Distributive Property. 3n + 5 = –1

Try This: Example 2A Continued – 5 –5 Subtract to undo addition. 3n = –6 3n 3 –6 = Divide to undo multiplication. n = –2

Try This: Example 2B Solve. B. + – = x 3 5x 9 13 1 3 The LCD is 9. ( ) x 3 1 5x 9 13 9 + – = 9( ) Multiply both sides by the LCD. 9( ) + 9( )– 9( ) = 9( ) 5x 9 x 3 13 1 Distributive Property. 5x + 3x – 13 = 3 8x – 13 = 3 Combine like terms.

Try This: Example 2B Continued 8x – 13 = 3 Combine like terms. + 13 + 13 Add to undo subtraction. 8x = 16 = 8x 8 16 Divide to undo multiplication. x = 2

Try This: Example 2B Continued Check x 3 5x 9 13 1 3 + – = 1 3 Substitute 2 for x. 5(2) 9 + – = (2) 13 ? 1 3 10 9 + – = 2 13 ? The LCD is 9. 3 9 10 + – = 6 13 ? 3 9 = ? 

Additional Example 3: Money Application When Mr. and Mrs. Harris left for the mall, Mrs. Harris had twice as much money as Mr. Harris had. While shopping, Mrs. Harris spent $54 and Mr. Harris spent $26. When they arrived home, they had a total of $46. How much did Mr. Harris have when he left home? Let h represent the amount of money that Mr. Harris had when he left home. So Mrs. Harris had 2h when she left home. h + 2h – 26 – 54 = 46 Mr. Harris $+ Mrs. Harris $ – Mr. Harris spent – Mrs. Harris spent = amount left

Additional Example 3 Continued 3h – 80 = 46 Combine like terms. + 80 +80 Add 80 to both sides. 3h = 126 3h 3 126 = Divide both sides by 3. h = 42 Mr. Harris had $42 when he left home.

Try This: Example 3 When Mr. and Mrs. Wesner left for the store, Mrs. Wesner had three times as much money as Mr. Wesner had. While shopping, Mr. Wesner spent $50 and Mrs. Wesner spent $25. When they arrived home, they had a total of $25. How much did Mr. Wesner have when he left home? Let h represent the amount of money that Mr. Wesner had when he left home. So Mrs. Wesner had 3h when she left home. h + 3h – 50 – 25 = 25 Mr. Wesner $ + Mrs. Wesner $ – Mr. Wesner spent – Mrs. Wesner spent = amount left

Try This: Example 3 Continued Combine like terms. + 75 +75 Add 75 to both sides. 4h = 100 4h 4 100 = Divide both sides by 4. h = 25 Mr. Wesner had $25 when he left home.

Lesson Quiz Solve. 1. 6x + 3x – x + 9 = 33 2. –9 = 5x + 21 + 3x x = 3 3. + = 5. Linda is paid double her normal hourly rate for each hour she works over 40 hours in a week. Last week she worked 52 hours and earned $544. What is her hourly rate? x = 3 x = –3.75 5 8 x 8 33 8 x = 28 x = 1 9 16 4. – = 6x 7 2x 21 25 21 $8.50