Algebra 1: Solving Equations Review (September 1)

1. Bellringer Simplify. 1. 15 x – 4 x 2. 5( x – 7) Solve. 3. x + 7 = 18 4. 5a – 6 = 44 11 x 5 x - 35 x = 11 x = 10 Standards: SPI 3102.1.3, 3102.2.2, CLE 3102.3.2

3. Solve equations in one variable that contain variable terms on both sides. Standards: SPI 3102.1.3, 3102.2.2, CLE 3102.3.2 Objective Standards: SPI 3102.1.3, 3102.2.2, CLE 3102.3.2

4. Solve the equation. Check your answer. Review… 16 = m – 8 + 8 + 8 24 = m Standards: SPI 3102.1.3, 3102.2.2, CLE 3102.3.2

5. Solve 2 a + 3 – 8 a = 8. Check It Out! Example 3a Use the Commutative Property of Addition. 2 a + 3 – 8 a = 8 2 a – 8 a + 3 = 8 – 6 a + 3 = 8 Combine like terms. Since 3 is added to –6a, subtract 3 from both sides to undo the addition. – 6 a = 5 Since a is multiplied by –6, divide both sides by –6 to undo the multiplication. – 3 – 3

6. Solve the equation. Check your answer. Review…. Since 8 is subtracted from y, add 8 to both sides to undo the subtraction. y – 8 = 24 + 8 + 8 y = 32 32 – 8 24 24 24  To check your solution, substitute 32 for y in the original equation. Check y – 8 = 24

7. Solve 26 = 4 a + 10. Review… 26 = 4 a + 10 16 = 4 a 4 = a Standards: SPI 3102.1.3, 3102.2.2, CLE 3102.3.2 – 10 – 10 16 = 4 a 4 4

8. Solve 5 t – 2 = –32. Review… 5 t – 2 = –32 5 t = –30 t = –6 Standards: SPI 3102.1.3, 3102.2.2, CLE 3102.3.2 + 2 + 2 5 t = –30 5 5

9. Solve . Check it Out! Example 1c First n is divided by 7. Then 2 is added. Work backward: Subtract 2 from both sides. Since n is divided by 7, multiply both sides by 7 to undo the division. n = 0 – 2 –2

10. Solve 7 n – 2 = 5 n + 6. Example 1: Solving Equations with Variables on Both Sides To collect the variable terms on one side, subtract 5n from both sides. 7 n – 2 = 5 n + 6 2 n – 2 = 6 Since n is multiplied by 2, divide both sides by 2 to undo the multiplication. 2 n = 8 n = 4 Standards: SPI 3102.1.3, 3102.2.2, CLE 3102.3.2 – 5 n –5 n + 2 + 2

11. Solve 4 b + 2 = 3 b . Check It Out! Example 1a To collect the variable terms on one side, subtract 3b from both sides. 4 b + 2 = 3 b b + 2 = 0 b = –2 Standards: SPI 3102.1.3, 3102.2.2, CLE 3102.3.2 – 3 b –3 b – 2 – 2

12. Solve 4 – 6 a + 4 a = –1 – 5(7 – 2 a ). Example 2: Simplifying Each Side Before Solving Equations Combine like terms. Distribute –5 to the expression in parentheses. 4 – 6 a + 4 a = –1 –5 (7 – 2 a ) 4 – 6 a + 4 a = –1 –5 (7) –5 (–2 a ) 4 – 6 a + 4 a = –1 – 35 + 10 a 4 – 2 a = –36 + 10 a 40 – 2 a = 10 a 40 = 12 a Since –36 is added to 10a, add 36 to both sides. To collect the variable terms on one side, add 2a to both sides. Standards: SPI 3102.1.3, 3102.2.2, CLE 3102.3.2, CLE 3102.3.2 +36 +36 + 2 a +2 a

13. Solve 4 – 6 a + 4 a = –1 – 5(7 – 2 a ). Example 2 Continued 40 = 12 a Since a is multiplied by 12, divide both sides by 12. Standards: SPI 3102.1.3, 3102.2.2, CLE 3102.3.2, CLE 3102.3.2

14. Solve 3 x + 15 – 9 = 2( x + 2). Check It Out! Example 2B Combine like terms. Distribute 2 to the expression in parentheses. 3 x + 15 – 9 = 2 ( x + 2) 3 x + 15 – 9 = 2 ( x ) + 2 (2) 3 x + 15 – 9 = 2 x + 4 3 x + 6 = 2 x + 4 x + 6 = 4 x = –2 To collect the variable terms on one side, subtract 2x from both sides. Since 6 is added to x, subtract 6 from both sides to undo the addition. Standards: SPI 3102.1.3, 3102.2.2, CLE 3102.3.2, CLE 3102.3.2 – 2x –2x – 6 – 6