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LESSON 10-6 Warm Up Lesson 10-6 Warm-Up ALGEBRA READINESS LESSON 10-6 Warm Up Lesson 10-6 Warm-Up ALGEBRA READINESS “Solving Equations With Variables on Both Sides” (10-6) Tip: To solve an equation with variables on both sides, use the Addition How can you solve an (2-2) or Subtraction Properties of Equality to get the variables on one side of equation with variables on both sides? the equal sign. Example: 5m + 3 = 3m – 9 -3m -3m 2m + 3 = + 0 - 9 -3 -3 -12 1 2 +0= 1 2m 1 2 2 1 1m = -6 ALGEBRA READINESS Solving Equations with Variables on Both Sides LESSON 10-6 Additional Examples Solve 9 + 2p = – 3 – 4p. 9 + 2p = – 3 – 4p + 4p + 4p 9 + 6p = – 3 + 0 –9 –9 0 + 6p = – 12 1 6p 1 6 = – 12 2 6 1 1p = – 2 Addition Property of Equality Combine like terms. Subtraction Property of Equality Simplify. Division Property of Equality Simplify. Check 9 + 2p = – 3 – 4p 9 + 2(–2) – 3 – 4(–2) 5=5 Substitute –2 for p. The solution checks. ALGEBRA READINESS Solving Equations With Variables on Both Sides LESSON 10-6 Additional Examples Solve 5x – 3 = 2x + 12. 5x – 3 = 2x + 12 – 2x – 2x Subtract 2x from each side to get rid of the x term from the right side. 3x – 3 = 0 + 12 +3 +3 Combine like terms. Add 3 to each side to get rid of -3. 3x + 0 = Simplify. 5 1 3x 3 15 = 1 15 3 Divide each side by 3 to isolate the x. 1 1x = 5 Simplify. ALGEBRA READINESS Solving Equations with Variables on Both Sides LESSON 10-6 Additional Examples 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: 7 times (weekly amount – 18) = 4.2 • weekly amount Let x = the amount earned weekly. Equation: 7 • (x – 18) = 4.2 • x ALGEBRA READINESS Solving Equations with Variables on Both Sides LESSON 10-6 Additional Examples (continued) 7(x – 18) = 4.2x 7x – 7(18) = 4.2x Distributive Property 7x – 126 = 4.2x Simplify. – 7x – 7.0x 0 –126 = –2.8x 1 –126 –2.8x = –2.8 –2.8 45 = Subtraction Property of Equality Combine like terms. Division Property of Equality 1 1x Simplify. You earn $45 per week. ALGEBRA READINESS “Solving Equations With Variables on Both Sides” (10-6) If the variable on both sides of an equation cancel each other out when you use the What happens when (2-2) Addition or Subtraction Properties of Equality (Example: 2x = 2x), it will have an the variables on both sides of the equation cancel each other out? 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=9 Identity (infiniite number of solutions) Proof: The following function table shows that any solution for x works. x 3x + 9 = 3 + 3(x + 2) True Statement? 105 3(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 3 + 3(23,246 + 2) = 3 + 3(23,248) = 3 + 69,744 = 69,747 Yes 69,747=69,7 47 23,2 46 3(23,246) = +9= 69,747 ALGEBRA READINESS “Solving Equations With Variables on Both Sides” (10-6) Example: 3x + 10 = 3 + 3(x + 2) (2-2) 3x + 10 = 3 + 3x + 6 Distributive Property 3x + 10 = 3x + 9 -3x -3x 10 ≠ 9 Simplify no solutions Proof: The following function table shows that no solution work for x. x 3x + 10 = 3 + 3(x + 2) = 3 + 3(12 + 2) = 3 + 3(14) = 3 + 42 =45 True Statement? 12 3(12) + 10 = 46 No 45 ≠ 45 105 3(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 ALGEBRA READINESS Solving Equations With Variables on Both Sides LESSON 10-6 Additional Examples Solve each equation. a. 4 – 4y = –2(2y – 2) 4 – 4y = –2(2y) – (-2)(2) 4 – 4y = -4y + 4 + 4y = +4y Use the Distributive Property. Use the Distributive Property. Add 4y to each side. 4+0 = 0 Always true! + 4 The equation is true for every value of y, so the equation is an identity. b. –6z + 8 = z + 10 – 7z –6z + 8 = 1z + 10 – 7z Identity Property (1x = x) –6z + 8 = –6z + 10 + 6z + 6z 0 + 8 = 0 + 10 Combine like terms. Add 6z to each side. Not true for any value of z! This equation has no solution. ALGEBRA READINESS Solving Equations with Variables on Both Sides LESSON 10-6 Lesson Quiz Solve each equation. 1. 4(3u – 1) = 20 2 2. 5t – 4 = t – 8 –1 3. 3(k – 8) = –k 6 ALGEBRA READINESS