This document provides examples of solving linear equations with one variable. It begins with examples of solving equations with the variable on both sides, including subtracting like terms to isolate the variable. It then gives examples of solving more complex multi-step equations, showing the process of combining like terms and using properties of equality. Finally, it provides a word problem example involving costs for flower bouquets with different variables representing fees and costs per item.

1. Warm Up California Standards Lesson Presentation Preview

2. Warm Up Solve. 1. 2 x + 9 x – 3 x + 8 = 16 2. –4 = 6 x + 22 – 4 x 3. + = 5 4. – = 3 x = 1 x = –13 x = 34 x = 50 2 7 x 7 7 1 9 x 16 2 x 4 1 8

3. Extension of AF4.1 Solve two-step linear equations and inequalities in one variable over the rational numbers, interpret the solution or solutions in the context from which they arose, and verify the reasonableness of the results. Also covered: AF1.1 California Standards

4. Solve. 4 x + 6 = x Additional Example 1A: Solving Equations with Variables on Both Sides 4 x + 6 = x – 4 x – 4 x 6 = –3 x To collect the variable terms on one side, subtract 4x from both sides. Since x is multiplied by -3, divide both sides by – 3. – 2 = x 6 – 3 – 3 x – 3 =

5. You can always check your solution by substituting the value back into the original equation. Helpful Hint

6. Solve. 9 b – 6 = 5 b + 18 Additional Example 1B: Solving Equations with Variables on Both Sides 9 b – 6 = 5 b + 18 – 5 b – 5 b 4 b – 6 = 18 To collect the variable terms on one side, subtract 5b from both sides. Since b is multiplied by 4, divide both sides by 4. b = 6 + 6 + 6 4 b = 24 Since 6 is subtracted from 4b, add 6 to both sides. 4 b 4 24 4 =

7. Solve. 9 w + 3 = 9 w + 7 Additional Example 1C: Solving Equations with Variables on Both Sides 3 ≠ 7 There is no solution. There is no number that can be substituted for the variable w to make the equation true. 9 w + 3 = 9 w + 7 – 9 w – 9 w To collect the variable terms on one side, subtract 9w from both sides.

8. if the variables in an equation are eliminated and the resulting statement is false, the equation has no solution. Helpful Hint

9. Solve. 5 x + 8 = x Check It Out! Example 1A 5 x + 8 = x – 5 x – 5 x 8 = –4 x Since x is multiplied by – 4, divide both sides by – 4. – 2 = x To collect the variable terms on one side, subtract 5x from both sides. 8 – 4 – 4 x – 4 =

10. Solve. 3 b – 2 = 2 b + 12 3 b – 2 = 2 b + 12 – 2 b – 2 b b – 2 = 12 + 2 + 2 b = 14 Since 2 is subtracted from b, add 2 to both sides. Check It Out! Example 1B To collect the variable terms on one side, subtract 2b from both sides.

11. Solve. 3 w + 1 = 3 w + 8 1 ≠ 8 No solution. There is no number that can be substituted for the variable w to make the equation true. Check It Out! Example 1C 3 w + 1 = 3 w + 8 – 3 w – 3 w To collect the variable terms on one side, subtract 3w from both sides.

12. To solve more complicated equations, you may need to first simplify by combining like terms or clearing fractions. Then add or subtract to collect variable terms on one side of the equation. Finally, use properties of equality to isolate the variable.

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

14. Additional Example 2B: Solving Multi-Step Equations with Variables on Both Sides Multiply by the LCD, 20. 4 y + 12 y – 15 = 20 y – 14 16 y – 15 = 20 y – 14 Combine like terms. + – = y – y 5 3 4 3 y 5 7 10 y 5 3 4 3 y 5 7 10 + – = y – 20 ( ) = 20 ( ) y 5 3 4 3 y 5 7 10 + – y – 20 ( ) + 20 ( ) – 20 ( ) = 20( y ) – 20 ( ) y 5 3 y 5 3 4 7 10

15. Additional Example 2B Continued Add 14 to both sides. – 15 = 4 y – 14 – 1 = 4 y + 14 + 14 Divide both sides by 4. 16 y – 15 = 20 y – 14 – 16 y – 16 y Subtract 16y from both sides. – 1 4 4 y 4 = – 1 4 = y

16. Solve. 12 z – 12 – 4 z = 6 – 2 z + 32 Check It Out! Example 2A 12 z – 12 – 4 z = 6 – 2 z + 32 + 12 +12 8 z – 12 = –2 z + 38 Combine like terms. + 2 z + 2 z Add 2z to both sides. 10 z – 12 = 38 10 z = 50 z = 5 Add 12 to both sides. Divide both sides by 10. 10 z 50 10 10 =

17. Multiply by the LCD, 24. 6 y + 20 y + 18 = 24 y – 18 26 y + 18 = 24 y – 18 Combine like terms. + + = y – Check It Out! Example 2B y 4 3 4 5 y 6 6 8 y 4 3 4 5 y 6 6 8 + + = y – 24 ( ) = 24 ( ) y 4 3 4 5 y 6 6 8 + + y – 24 ( ) + 24 ( ) + 24 ( ) = 24( y ) – 24 ( ) y 4 5 y 6 3 4 6 8

18. Subtract 18 from both sides. 2 y + 18 = – 18 2 y = –36 – 18 – 18 Divide both sides by 2. y = –18 26 y + 18 = 24 y – 18 – 24 y – 24 y Subtract 24y from both sides. Check It Out! Example 2B Continued – 36 2 2 y 2 =

19. Additional Example 3: Business Application Daisy’s Flowers sells 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 florists' bouquets cost the same price. What is the price? Daisy’s: c = 39.95 + 2.95 r Write an equation for each service. Let c represent the total cost and r represent the number of roses. total cost is flat fee plus cost for each rose Other: c = 26.00 + 4.50 r

20. Additional Example 3 Continued 39.95 + 2.95 r = 26.00 + 4.50 r Now write an equation showing that the costs are equal. – 2.95 r – 2.95 r 39.95 = 26.00 + 1.55 r Subtract 2.95r from both sides. – 26.00 – 26.00 Subtract 26.00 from both sides. 13.95 = 1.55 r Divide both sides by 1.55. 9 = r The two bouquets from either florist would cost the same when purchasing 9 roses. 13.95 1.55 1.55 r 1.55 =

21. Additional Example 3 Continued To find the cost, substitute 9 for r into either equation. Daisy’s: The cost for a bouquet with 9 roses at either florist is $66.50. c = 39.95 + 2.95 r c = 39.95 + 2.95( 9 ) c = 39.95 + 26.55 c = 66.5 Other florist: c = 26.00 + 4.50 r c = 26.00 + 4.50( 9 ) c = 26.00 + 40.50 c = 66.5

22. Check It Out! Example 3 Marla’s Gift Baskets sells a muffin basket for $22.00 plus $2.25 for every balloon. A competing service sells a similar muffin basket for $16.00 plus $3.00 for every balloon. Find the number of balloons that would make both baskets cost the same price. Marla’s: c = 22.00 + 2.25 b total cost is flat fee plus cost for each balloon Other: c = 16.00 + 3.00 b Write an equation for each service. Let c represent the total cost and b represent the number of balloons.

23. Check It Out! Example 3 Continued 22.00 + 2.25 b = 16.00 + 3.00 b Now write an equation showing that the costs are equal. – 2.25 b – 2.25 b 22.00 = 16.00 + 0.75 b Subtract 2.25b from both sides. – 16.00 – 16.00 Subtract 16.00 from both sides. 6.00 = 0.75 b Divide both sides by 0.75. 8 = b The two services would cost the same when purchasing a muffin basket with 8 balloons. 6.00 0.75 0.75 b 0.75 =

24. Lesson Quiz Solve. 1. 4 x + 16 = 2 x 2. 8 x – 3 = 15 + 5 x 3. 2(3 x + 11) = 6 x + 4 4. x = x – 9 5. An apple has about 30 calories more than an orange. Five oranges have about as many calories as 3 apples. How many calories are in each? x = 6 x = –8 no solution x = 36 An orange has 45 calories. An apple has 75 calories. 1 4 1 2