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Reason Using Properties from Algebra
2.5
Looking back at algebra, we used different properties of real numbers to solve an
equation. These properties include:
Algebraic Properties of Equality (Let a, b, and c be real numbers)
Addition Property
If a=b, then a+c=b+c
Subtraction Property
If a=b, then a-c=b-c
Multiplication Property
If a=b, then ac=bc
Division Property
If a=b, then
Substitution Property
If a=b, then a can be substituted for b in
any equation or expression
π‘Ž
𝑐
=
𝑏
𝑐
Example: Solve 2π‘₯ + 5 = 20 βˆ’ 3π‘₯ and write a reason for each step.
2π‘₯ + 5 = 20 βˆ’ 3π‘₯
5π‘₯ + 5 = 20
5π‘₯ = 15
π‘₯=3
Given
Addition Property of Equality
Subtraction Property of Equality
Division Property of Equality
We can also use the Distributive Property: π‘Ž(𝑏 + 𝑐) = π‘Žπ‘ + π‘Žπ‘ where a, b, and c are
real numbers.
Example: Solve βˆ’4(11π‘₯ + 2) = 80 and write a reason for each step.
βˆ’4(11π‘₯ + 2) = 80
(11π‘₯ + 2) = βˆ’20
11π‘₯ = βˆ’22
π‘₯ = βˆ’2
Properties of Equality (true for all real numbers)
Reflexive Property of Equality
ο‚· Real Numbers: For any real number π‘Ž, π‘Ž = π‘Ž
ο‚· Segment Length: For any segment Μ…Μ…Μ…Μ…
𝐴𝐡, 𝐴𝐡 = 𝐴𝐡
ο‚· Angle Measure: For any angle ∠A, π‘šβˆ A = m∠A
Symmetric Property of Equality
ο‚· Real Numbers: For any real numbers π‘Ž and 𝑏, if π‘Ž = 𝑏, then 𝑏 = π‘Ž
ο‚· Segment Length: For any segments Μ…Μ…Μ…Μ…
𝐴𝐡 and Μ…Μ…Μ…Μ…
𝐢𝐷, if 𝐴𝐡 = 𝐢𝐷, then 𝐢𝐷 = 𝐴𝐡
ο‚· Angle Measure: For any angles ∠A and ∠B, if π‘šβˆ A = m∠B, then π‘šβˆ B = m∠A
Transitive Property of Equality
ο‚· Real Numbers: For any real numbers π‘Ž, 𝑏, and 𝑐, if π‘Ž = 𝑏 and 𝑏 = 𝑐, then π‘Ž = 𝑐
ο‚· Segment Length: For any segments Μ…Μ…Μ…Μ…
𝐴𝐡, Μ…Μ…Μ…Μ…
𝐢𝐷, and Μ…Μ…Μ…Μ…
𝐸𝐹 , if 𝐴𝐡 = 𝐢𝐷 and 𝐢𝐷 = 𝐸𝐹,
then 𝐴𝐡 = 𝐸𝐹
ο‚· Angle Measure: For any angles ∠A, ∠B, and ∠C, if π‘šβˆ A = m∠B and π‘šβˆ B = m∠C,
then π‘šβˆ A = m∠C
Example: (Use the properties of equality) You are designing a logo to sell daffodils.
Use the information given. Determine whether π‘šβˆ EBA = m∠DBC
π‘šβˆ 1 = π‘šβˆ 3
π‘šβˆ πΈπ΅π΄ = π‘šβˆ 2 + π‘šβˆ 3
π‘šβˆ πΈπ΅π΄ = π‘šβˆ 2 + π‘šβˆ 1
π‘šβˆ 1 + π‘šβˆ 2 = π‘šβˆ π·π΅πΆ
π‘šβˆ πΈπ΅π΄ = π‘šβˆ π·π΅πΆ
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