1. ALGEBRA READINESS LESSON 10-6 Warm Up Lesson 10-6 Warm-Up

2. ALGEBRA READINESS LESSON 10-6 Warm Up Lesson 10-6 Warm-Up

3. ALGEBRA READINESS “Solving Equations With Variables on Both Sides” (10-6) (2-2) How can you solve an equation with variables on both sides? Tip: To solve an equation with variables on both sides, use the Addition or Subtraction Properties of Equality to get the variables on one side of the equal sign. Example: 5m + 3 = 3m – 9 -3m -3m 2m + 3 = - 9 - 3 - 3 2m = -12 m = -6 + 0 2222 -12 2 + 0 1 1 1 1 1

4. ALGEBRA READINESS Solve 9 + 2p = – 3 – 4p. 9 + 2p = – 3 – 4p + 4p + 4p Addition Property of Equality 9 + 6p = – 3 Combine like terms. – 9 – 9 Subtraction Property of Equality 6p = – 12 Simplify. 6p 6 – 12 6 = Division Property of Equality p = – 2 Simplify. Check9 + 2p = – 3 – 4p 9 + 2(–2) – 3 – 4(–2) Substitute –2 for p. 5 = 5 The solution checks. Solving Equations with Variables on Both Sides LESSON 10-6 Additional Examples + 0 0 + 1 1 1 2 1

5. ALGEBRA READINESS Solve 5x – 3 = 2x + 12. 5x – 3=2x + 12 Subtract 2x from each side to get rid of the x term from the right side. – 2x – 2x 3x – 3= 12Combine like terms. + 3 + 3Add 3 to each side to get rid of -3. 3x = 15Simplify. x=5Simplify. = Divide each side by 3 to isolate the x. 3x33x3 15 3 Solving Equations With Variables on Both Sides LESSON 10-6 Additional Examples 0 + + 0 1 1 1 5 1

6. ALGEBRA READINESS Each week you set aside $18 for a stereo and put the remainder of your weekly pay in a savings account. After 7 weeks, the amount you place in the savings account is 4.2 times your total weekly pay. How much do you earn each week? Words: Equation: (weekly amount – 18) weekly amount 7 times= 4.2 Let = the amount earned weekly. x 7 (x – 18) x = Solving Equations with Variables on Both Sides LESSON 10-6 Additional Examples 4.2

7. ALGEBRA READINESS (continued) 7(x – 18) = 4.2x You earn $45 per week. 7x – 7(18) = 4.2x Distributive Property Combine like terms. –126 = –2.8x – 7x – 7.0x Subtraction Property of Equality 45 = x Simplify. Division Property of Equality – 126 – 2.8 – 2.8x – 2.8 = Solving Equations with Variables on Both Sides LESSON 10-6 Additional Examples 7x – 126 = 4.2x Simplify. 0 1 1 1

8. ALGEBRA READINESS “Solving Equations With Variables on Both Sides” (10-6) (2-2) What happens when the variables on both sides of the equation cancel each other out? If the variable on both sides of an equation cancel each other out when you use the Addition or Subtraction Properties of Equality (Example: 2x = 2x), it will have an infinite number of solutions called an “identity” if both sides of the equation are equal (Example: 10 = 10) or no solution if both sides of the equation are not equal (Example: 10 = 7) Example: 3x + 9 = 3 + 3(x + 2) 3x + 9 = 3 + 3x + 3(2) Distributive Property 3x + 9 = 3x + 9 Simplify -3x -3x 9 = 9Identity (infiniite number of solutions) Proof: The following function table shows that any solution for x works. x3x + 9=3 + 3(x + 2) True Statement? 1053(105) + 9 = 324 =3 + 3(105 + 2) = 3 + 3(107) = 3 + 321 =324 Yes 324 = 324 1,42 0 3(1,420) + 9 = 4,269 =3 + 3(1,420 + 2) = 3 + 3(1,422) = 3 + 4,266 = 4,269 Yes 4,269 = 4,269 23,2 46 3(23,246) + 9 = 69,747 =3 + 3(23,246 + 2) = 3 + 3(23,248) = 3 + 69,744 = 69,747 Yes 69,747=69,7 47

9. ALGEBRA READINESS “Solving Equations With Variables on Both Sides” (10-6) (2-2) Example: 3x + 10 = 3 + 3(x + 2) 3x + 10 = 3 + 3x + 6 Distributive Property 3x + 10 = 3x + 9 Simplify -3x -3x 10 ≠ 9no solutions Proof: The following function table shows that no solution work for x. x3x + 10=3 + 3(x + 2) True Statement? 123(12) + 10 = 46 =3 + 3(12 + 2) = 3 + 3(14) = 3 + 42 =45 No 45 ≠ 45 1053(105) + 10 = 325 =3 + 3(105 + 2) = 3 + 3(107) = 3 + 321 =324 Yes 325 ≠324 1,42 0 3(1,420) + 10 = 4,270 =3 + 3(1,420 + 2) = 3 + 3(1,422) = 3 + 4,266 = 4,269 Yes 4,270 ≠ 4,269

10. ALGEBRA READINESS Solve each equation. b.–6z + 8=z + 10 – 7z –6z + 8=1z + 10 – 7zIdentity Property (1x = x) –6z + 8=–6z + 10Combine like terms. + 6z + 6zAdd 6z to each side. 8= 10Not true for any value of z! This equation has no solution. a. 4 – 4y=–2(2y – 2) The equation is true for every value of y, so the equation is an identity. 4 – 4y=–2(2y) – (-2)(2) Use the Distributive Property. + 4y= +4y Add 4y to each side. 4 = 4Always true! Solving Equations With Variables on Both Sides LESSON 10-6 Additional Examples 4 – 4y= -4y + 4 Use the Distributive Property. + 0 0 +

11. ALGEBRA READINESS Solve each equation. 1. 4(3u – 1) = 20 2. 5t – 4 = t – 8 3. 3(k – 8) = –k 2 –1 6 Solving Equations with Variables on Both Sides LESSON 10-6 Lesson Quiz