Ch. 2.4
Reasoning Algebra Properties
HMWK:
p. 100, #s 17 – 23 odd, 24 – 28 all
Game Plan:
Today I will be able to use Algebraic properties to
reason.
Warm-up:
Solve each equation. SHOW ALL STEPS.
1. x = 9
1. 3x = 27
2. x = -23
2. x+6 = -17
3. x = 27
3. x – 9 = 18
4. x = 9
4. (2/3)x = 6
Ch. 2.4
Reasoning Algebra Properties
Algebraic Properties
Let a, b, and c be real numbers.
Addition Property:
If a = b, then a + c = b + c
ex: a = 3, b = 3, c = 5
a = b, then a + c = b + c
3 = 3, 3 + 5 = 3 + 5
Ch. 2.4
Reasoning Algebra Properties
Algebraic Properties
Subtraction Property:
If a = b, then a – c = b – c
Ex:
Ch. 2.4
Reasoning Algebra Properties
Algebraic Properties
Multiplication Property:
If a = b, then ac = bc
Ex: a = 11, b = 11, c = 8
a = b, then ac = bc
11 = 11, 11(8) = 11(8)
Ch. 2.4
Reasoning Algebra Properties
Algebraic Properties
Division Property:
If a = b and c  0, then a/c = b/c.
Ex:
Ch. 2.4
Reasoning Algebra Properties
Examples
1. Solve 3x + 12 = 8x – 18 and write a
reason for each step.
2. A child’s dose (c) for a medicine with
an adult dose of 500mg can be
found by using the child’s age (a) in
years in the given formula. Solve
the formula for c and write a reason
for each step.
 24 
a 
c  1
 500 
R,S,T Properties of
Algebra
Ch. 2.4
Reasoning Algebra Properties
Reflexive: For any real number a,
a = a.
Symmetric: If a = b, then b = a.
Transitive: If a = b and b = c, then
a = c.
Substitution: If a = b, then a may
be substituted for b in any
equation or expression.
Ch. 2.4
Reasoning Algebra Properties
R,S,T Properties of Equality
Property
Segments
Angles
Reflexive
For any segment AB, For any angle A, mA
AB = AB
= mA.
Symmetric
If AB=CD, then
CD=AB
If mA = mB, then
mB = mA
Transitive
IF AB=CD and
CD=EF, then AB=EF.
If the mA = mB
and mB = mC,
then mA = mC.
Ch. 2.4
Reasoning Algebra Properties
Examples
Worksheet
Ch. 2.4
Reasoning Algebra Properties
Wrap-up
1. What properties do R, S, and T
stand for?
2. If AB = CD, and CD = EF, write a
valid statement about AB and
EF and give a reason.