Download 2.4 Reasoning with Properties from Algebra

2.4 Reasoning with Properties from
Algebra
Algebraic Properties of Equality
• 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, and c≠0,
then a/c = b/c.
Writing reasons
5 x  18  3x  2
2 x  18  2
2 x  20
x  10
GIVEN
Subtraction Property of Equality
Addition Property of Equality
Division Property of Equality
Solve
• Solve 5x – 18 = 3x + 2 and explain each step
in writing.
5x – 18 = 3x + 2
2x – 18 = 2
2x = 20
x = 10
Subtraction p. of e.
Addition p. of e.
Division p. of e.
More properties of equality
• Reflexive property: For any real number a,
a=a.
• Symmetric property: If a=b, then b=a.
• Transitive property: If a=b and b=c, then a=c.
• Substitution property: If a=b, then a can be
substituted for b in any equation or
expression.
Writing Reasons
55 z  3(9 z  12)  64
Given
55z  27 z  36  64
Distr. Property
28z  36  64
Combine Like Terms
28z  28
Add POE
z  1
Div POE
Properties of Equality
Reflexive
Segment
Length
AB = AB
Symmetric If AB = CD,
then CD =
AB.
Transitive If AB = CD
and CD = EF,
then AB=EF.
Angle Measure
m<A = m<A
If m<A = m<B,
then
m<B=m<A.
If m<A = m<B
and
m<B=m<C,
then
m<A=m<C.




A
B
C
D
Given AB=CD, show that AC=BD
Statements
Reasons
AB=CD
Given
AB + BC = CD + BC
Addition Prop of Equality
AB + BC = AC
Segment Addition Postulate
BC + CD = BD
AC = BD
Segment Addition Postulate
Substitution Prop of Equality
2
4
3
1
Given: m1  m2  m3  180
m1  m2  93
m3  m4  180
Find: m4
Review
• Let p be “a shape is a triangle” and let q be
“it has an acute angle”.
– Write the contrapositive of p q.
– Write the inverse of p q.