To solve a multi-step 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 4 from both sides. 11x = 33 33 11 11x = Divide both sides by 11. 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 ? 

Check It Out: Example 1 Solve. 9x + 5 + 4x – 2 = 42 13x + 3 = 42 Combine like terms. – 3 – 3 Subtract 3 from both sides. 13x = 39 39 13 13x = Divide both sides by 13. x = 3

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

Check It Out: Example 2A Solve. + = – 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

Check It Out: Example 2A Continued – 5 –5 Subtract 5 from both sides. 3n = –6 3n 3 –6 = Divide both sides by 3. n = –2

Lesson Quiz Solve. 6x + 3x – x + 9 = 33 + = x = 3 x = 28 + = 3. 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 5 8 x 8 33 8 x = 28 $8.50

Answers 1 - 7 x= -2.2 2. w = 2.75 x = 11 b = -7 5. m = 1 6. x = 25

Additional Example 1A: Solving Equations with Variables on Both Sides Course 3 11-3 Solving Equations with Variables on Both Sides Additional Example 1A: Solving Equations with Variables on Both Sides Solve. 4x + 6 = x 4x + 6 = x – 4x – 4x Subtract 4x from both sides. 6 = –3x 6 –3 –3x = Divide both sides by –3. –2 = x

Solving Equations with Variables on Both Sides Course 3 11-3 Solving Equations with Variables on Both Sides Check your solution by substituting the value back into the original equation. For example, 4(-2) + 6 = -2 or -2 = -2. Helpful Hint

Additional Example 1B: Solving Equations with Variables on Both Sides Course 3 11-3 Solving Equations with Variables on Both Sides Additional Example 1B: Solving Equations with Variables on Both Sides Solve. 9b – 6 = 5b + 18 9b – 6 = 5b + 18 – 5b – 5b Subtract 5b from both sides. 4b – 6 = 18 + 6 + 6 Add 6 to both sides. 4b = 24 4b 4 24 = Divide both sides by 4. b = 6

Solving Equations with Variables on Both Sides Course 3 11-3 Solving Equations with Variables on Both Sides Check It Out: Example 1A Solve. 5x + 8 = x 5x + 8 = x – 5x – 5x Subtract 5x from both sides. 8 = –4x 8 –4 –4x = Divide both sides by –4. –2 = x

Solving Equations with Variables on Both Sides Course 3 11-3 Solving Equations with Variables on Both Sides Check It Out: Example 1B Solve. 3b – 2 = 2b + 12 3b – 2 = 2b + 12 – 2b – 2b Subtract 2b from both sides. b – 2 = 12 + 2 + 2 Add 2 to both sides. b = 14

Solving Equations with Variables on Both Sides Course 3 11-3 Solving Equations with Variables on Both Sides Additional Example 2A: Solving Multi-Step Equations with Variables on Both Sides Solve. 10z – 15 – 4z = 8 – 2z - 15 10z – 15 – 4z = 8 – 2z – 15 6z – 15 = –2z – 7 Combine like terms. + 2z + 2z Add 2z to both sides. 8z – 15 = – 7 + 15 +15 Add 15 to both sides. 8z = 8 8z 8 8 = Divide both sides by 8. z = 1

Solving Equations with Variables on Both Sides Course 3 11-3 Solving Equations with Variables on Both Sides Check It Out: Example 2A Solve. 12z – 12 – 4z = 6 – 2z + 32 12z – 12 – 4z = 6 – 2z + 32 8z – 12 = –2z + 38 Combine like terms. + 2z + 2z Add 2z to both sides. 10z – 12 = 38 + 12 +12 Add 12 to both sides. 10z = 50 10z 50 10 = Divide both sides by 10. z = 5

Check It Out: Example 2B Continued Course 3 11-3 Solving Equations with Variables on Both Sides Check It Out: Example 2B Continued 26y + 18 = 24y – 18 – 24y – 24y Subtract 24y from both sides. 2y + 18 = – 18 – 18 – 18 Subtract 18 from both sides. 2y = –36 –36 2 2y = Divide both sides by 2. y = –18

Additional Example 3: Business Application Course 3 11-3 Solving Equations with Variables on Both Sides Additional Example 3: Business Application Daisy’s Flowers sell a rose bouquet for $39.95 plus $2.95 for every rose. A competing florist sells a similar bouquet for $26.00 plus $4.50 for every rose. Find the number of roses that would make both florist’s bouquets cost the same price.

Additional Example 3 Continued Course 3 11-3 Solving Equations with Variables on Both Sides Additional Example 3 Continued Let r represent the price of one rose. 39.95 + 2.95r = 26.00 + 4.50r Subtract 2.95r from both sides. – 2.95r – 2.95r 39.95 = 26.00 + 1.55r Subtract 26.00 from both sides. – 26.00 – 26.00 13.95 = 1.55r 13.95 1.55 1.55r 1.55 = Divide both sides by 1.55. 9 = r The two services would cost the same when using 9 roses.

Insert Lesson Title Here Course 3 11-3 Solving Equations with Variables on Both Sides Insert Lesson Title Here Lesson Quiz Solve. 1. 4x + 16 = 2x 2. 8x – 3 = 15 + 5x 3. x = x – 9 x = –8 x = 6 1 4 1 2 x = 36

Answers 1 - 9 x = 9 2. k = 10.2 d = 2 4. a = -4 5. x = 3 6. d = ¾ 7. x = 7 8. y = -2/5 9. x = 4

Solving Two-Step Inequalities Course 3 11-5 Solving Two-Step Inequalities Warm Up Solve. 1. 6x + 36 = 2x 2. 4x – 13 = 15 + 5x 3. 5(x – 3) = 2x + 3 x = –9 x = –28 x = 6

Additional Example 1A: Solving Two-Step Inequalities Course 3 11-5 Solving Two-Step Inequalities Additional Example 1A: Solving Two-Step Inequalities Solve and graph. 4x + 1 > 13 4x + 1 > 13 – 1 – 1 Subtract 1 from both sides. 4x > 12 4x 4 > 12 Divide both sides by 4. x > 3 1 2 3 4 5 6 7

Solving Two-Step Inequalities Course 3 11-5 Solving Two-Step Inequalities If both sides of an inequality are multiplied or divided by a negative number, the inequality symbol must be reversed. Remember!

Additional Example 1B: Solving Two-Step Inequalities Course 3 11-5 Solving Two-Step Inequalities Additional Example 1B: Solving Two-Step Inequalities Solve and graph. –9x + 7  25 –9x + 7  25 – 7 – 7 Subtract 7 from both sides. –9x  18 –9x –9  18 Divide each side by –9; change  to . x  –2 -6 -5 -4 -3 -2 -1 0

Solving Two-Step Inequalities Course 3 11-5 Solving Two-Step Inequalities Check It Out: Example 1A Solve and graph. 5x + 2 > 12 5x + 2 > 12 – 2 – 2 Subtract 2 from both sides. 5x > 10 5x 5 > 10 Divide both sides by 5. x > 2 1 2 3 4 5 6 7

Solving Two-Step Inequalities Course 3 11-5 Solving Two-Step Inequalities Check It Out: Example 1B –4x + 2  18 –4x + 2  18 – 2 – 2 Subtract 2 from both sides. –4x  16 –4x –4  16 Divide each side by –4; change  to . x  –4 -6 -5 -4 -3 -2 -1 0

Check your Understanding Course 3 11-5 Solving Two-Step Inequalities Insert Lesson Title Here Check your Understanding Solve and graph. 1. 4x – 6 > 10 2. 7x + 9 < 3x – 15 3. w – 3w < 32 1 2 3 4 5 6 7 x > 4 -10 -9 -8 -7 -6 -5 -4 x < –6 -18 -17 -16 -15 -14 -13 -12 w > –16

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