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Geometry Lesson 2.4 Reasoning with Properties from Algebra Warm-Up: Describing Solutions Each equation below can be solved in one step Solve each one and write in words what you did to solve it Example: 2x = 10 x = 5 (divided both sides by 2) (a) 3x = 27 x = 9 (divided both sides by 3) (b) x + 6 = –17 x = -23 (subtracted 6 from both sides) (c) x – 9 = 18 x = 27 (added 9 to both sides) (d) –x = 4 x = 4 (multiplied both sides by -1) 1. Algebraic Properties of Equality Here are the formal properties that allow you to solve algebraic equations… The Property 1. Addition If a = b, then a + c = b + c 2. Subtraction If a = b, then a – c = b – c 3. Multiplication If a = b, then ac = bc If a = b and c ≠ 0, 4. Division then a ÷ c = b ÷ c Property of Equality Note: a, b, and c are real numbers Algebraic Properties of Equality, cont. Property of Equality 5. Reflexive 6. Symmetric 7. Transitive The Property For any real number, a = a If a = b, then b = a If a = b and b = c, then a = c If a = b, then a can be 8. Substitution substituted for b 9. Distributive a(b + c) = ab + ac Note: a, b, and c are real numbers 2. Using Properties of Equality You know how to solve equations, but can you state why? Example 1a: Solve 5x – 18 = 3x + 2 by listing each step with a reason Step in Solution 5x – 18 = 3x + 2 Reason Given 2x – 18 = 2 if a = b, then a – c = b – c 2x = 20 if a = b, then a + c = b + c x = 10 a b If a = b, and c 0, then c c Example 1b Solve 55x – 3(9x + 12) = -64 and write the REASON for each step Step in Solution 55x – 3(9x + 12) = -64 55x – 27x – 36 = -64 28x – 36 = -64 28x = -28 x = -1 Reason Given a(b + c) = ab + ac Combine like terms if a = b, then a + c = b + c a b If a = b, and c 0, then c c Practice 1a List a reason for each step in the solution of 3x + 12 = 8x – 18 Step in Solution 3x + 12 = 8x – 18 Reason Given 12 = 5x – 18 if a = b, then a – c = b – c 30 = 5x if a = b, then a + c = b + c x=6 a b if a = b, and c 0, then c c Practice 1b Solve 4(2x + 5) = 2(x – 5) and list a reason for each step Step in Solution 4(2x + 5) = 2(x – 5) 8x + 20 = 2x – 10 Reason Given a(b + c) = ac + ac 6x + 20 = –10 if a = b, then a – c = b – c 6x = –30 if a = b, then a – c = b – c x = –5 a b if a = b, and c 0, then c c 3. Properties of Length and Angle Measures ANGLE MEASURES For any A, For any AB, Reflexive AB = AB mA = mA If AB = CD, If mA = mB, Symmetric then CD = AB then mB = mA If AB = CD If mA = mB Transitive and CD = EF, and mB = mC, then AB = EF then mA = mC Property of Equality SEGMENT LENGTHS Example 2a: Segment Lengths In the diagram, AB = CD (equal lengths) Show that AC = BD A B C D Step AB = CD AC = (AB + BC) BD = (CD + BC) (AB + BC) = (CD + BC) AC = BD Reason Given if B between A&C, then AC = AB + BC if C between B&D, then BD = CD + BC if AB = CD, then AB + BC = CD + BC if a = b, then subst a for b Example 2b: Angle Measures In the diagram, mBEA = mCED Show that mAEC = mBED Step Reason mBEA = mCED Given if B on interior, then mAEC = mBEA + mBEC mBEA + mBEC = mAEC mBED = mCED + mBEC Same for C on interior mBEA + mBEC if a = b, then = mCED + mBEC a + c = b + c if a = b, then mAEC = mBED subst a for b Practice 2a Step WY = XZ WY = WX + XY XZ = YZ + XY Reason Given if X between W&Y, then WY = WX + XY if Y between X&Z, then XZ = YZ + XY WX + XY = YZ + XY if a = b, then subst a for b WX = YZ if a = b, then a – c = b – c Practice 2b In the diagram, mAEC = mDEB Show that mAEB = mCED Step Reason mAEC = mDEB Given if B on interior, then mAEC = mAEB + mBEC mAEB + mBEC = mAEC mDEB = mCED + mBEC Same for C on interior mAEB + mBEC if a = b, then = mCED + mBEC subst a for b if a = b, then mAEB = mCED a–c=b–c Assignment Ch. 2.4 (Pg. 99-101) #10-26 EVEN, #32