Copyright © by Holt, Rinehart and Winston. 36 Holt GeometryAll rights reserved.

Name Date Class

LESSON

t � 6.5 � t � 3t � 1.3 � t (Subtr. Prop. of �)6.5 � 2t � 1.3 (Simplify.)6.5 � 1.3 � 2t � 1.3 + 1.3 (Add. Prop. of �)7.8 � 2t (Simplify.)

7.8 ___ 2 � 2t __

2 (Div. Prop. of �)

3.9 � t (Simplify.) t � 3.9 (Symmetric Prop. of �)

1 1–4 ft

7 � 2� (Simplify.)

7 __ 2 � 2� __

2 (Div. Prop. of �)

3 1 __ 2 � � (Simplify.)

� � 3 1 __ 2 (Symmetric Prop. of �)

Solve each equation. Show all your steps and write a justification for each step.

1. 1 __ 5

(a � 10) � �3 2. t � 6.5 � 3t � 1.3

5 [ 1 __ 5 (a � 10) ] � 5(�3) (Mult. Prop. of �)

a � 10 � �15 (Simplify.) a � 10 � 10 � �15 � 10 (Subtr. Prop. of �) a � –25 (Simplify.)

3. The formula for the perimeter P of a rectangle with length � and width w is P = 2(� � w). Find the length of the rectangle shown here if the perimeter is 9 1 __

2 feet.

Solve the equation for � and justify each step. Possible answer:

P � 2(� � w) (Given)

9 1 __ 2 � 2(� � 1 1 __

4 ) (Subst. Prop. of �)

9 1 __ 2 � 2� � 2 1 __

2 (Distrib. Prop.)

9 1 __ 2 � 2 1 __

2 � 2� � 2 1 __

2 � 2 1 __

2 (Subtr. Prop. of �)

Write a justification for each step. 4.

2 6 3 3

7 3

HJ � HI � IJ Seg. Add. Post. 7x � 3 � (2x � 6) � (3x � 3) Subst. Prop. of � 7x � 3 � 5x � 3 Simplify. 7x � 5x � 6 Add. Prop. of � 2x � 6 Subtr. Prop. of � x � 3 Div. Prop. of �

Identify the property that justifies each statement.

5. m � n, so n � m. 6. �ABC � �ABC

Symmetric Prop. of � Reflexive Prop. of � 7.

_ KL �

_ LK 8. p � q and q � �1, so p � �1.

Reflexive Prop. of � Transitive Prop. of � or Subst.

Practice BAlgebraic Proof2-5

Copyright © by Holt, Rinehart and Winston. 67 Holt GeometryAll rights reserved.

Copyright © by Holt, Rinehart and Winston. 35 Holt GeometryAll rights reserved.

Name Date Class

LESSON

For Exercises 1–12, write the letter of each property next to its definition. The letters a, b, and c represent real numbers.

1. If a � b, then b � a. F

2. If a � b, then ac � bc. C

3._AB �

_AB J

4. a � a E

5. If a � b, then a � c � b � c. A

6. a (b � c) � ab � ac I

7. If a � b and b � c, then a � c. G

8. If �P � �Q, then�Q � �P. K

9. If �A � �B and �B � �C,then �A � �C. L

10. If a � b and c � 0, then a__c � b__

c . D

11. If a � b, then b can be substituted for ain any expression. H

12. If a � b, then a � c � b � c. B

13. Cali measures her textbook and finds that it is 8 inches wide. She wants to know how many centimeters wide her textbook is. The formula to convert inches to centimeters is 2.54i � c, where i is the length in inches and c is the length in centimeters. Fill in the blanks to find the answer. The justifications will guide you.

2.54i � c Given equation

2.54( 8 ) � c Substitution Property of Equality

20.32 � c Simplify.

c � 20.32 Symmetric Property of Equality

14. Write a justification for each step.

71–3� � 1� ��

11

DE � EF � DF Seg. Add. Post.

� 1__3

x � 1 � � 7 � 11 Subst.

1__3

x � 8 � 11 Simplify.

1__3

x � 3 Subtr. Prop. of �

x � 9 Mult. Prop. of �

2-5Practice AAlgebraic Proof

A. Addition Property of Equality

B. Subtraction Property of Equality

C. Multiplication Property of Equality

D. Division Property of Equality

E. Reflexive Property of Equality

F. Symmetric Property of Equality

G. Transitive Property of Equality

H. Substitution Property of Equality

I. Distributive Property

J. Reflexive Property of Congruence

K. Symmetric Property of Congruence

L. Transitive Property of Congruence

Copyright © by Holt, Rinehart and Winston. 36 Holt GeometryAll rights reserved.

Name Date Class

LESSON

t � 6.5 � t � 3t � 1.3 � t (Subtr. Prop. of �)6.5 � 2t � 1.3 (Simplify.)6.5 � 1.3 � 2t � 1.3 + 1.3 (Add. Prop. of �)7.8 � 2t (Simplify.)7.8___2

� 2t__2

(Div. Prop. of �)

3.9 � t (Simplify.)t � 3.9 (Symmetric Prop. of �)

�

11–4 ft

7 � 2� (Simplify.)

7__2

� 2� __2

(Div. Prop. of �)

31__2

� � (Simplify.)

� � 3 1__2

(Symmetric Prop. of �)

Solve each equation. Show all your steps and write a justification for each step.

1. 1__5

(a � 10) � �3 2. t � 6.5 � 3t � 1.3

5[1__5

(a � 10) ] � 5(�3) (Mult. Prop. of �)

a � 10 � �15 (Simplify.)a � 10 � 10 � �15 � 10 (Subtr. Prop. of �)a � –25 (Simplify.)

3. The formula for the perimeter P of a rectangle with length � and width w is P = 2(� � w). Find the length of the rectangle shown here if the perimeter is 9 1__

2 feet.

Solve the equation for � and justify each step. Possible answer:

P � 2(� � w) (Given)

9 1__2

� 2(� � 1 1__4

) (Subst. Prop. of �)

9 1__2

� 2� � 2 1__2

(Distrib. Prop.)

9 1__2

� 2 1__2

� 2� � 2 1__2

� 2 1__2

(Subtr. Prop. of �)

Write a justification for each step. 4.

2� � 6 3� � 3� ��

7� � 3

HJ � HI � IJ Seg. Add. Post.

7x � 3 � (2x � 6) � (3x � 3) Subst. Prop. of �

7x � 3 � 5x � 3 Simplify.

7x � 5x � 6 Add. Prop. of �

2x � 6 Subtr. Prop. of �

x � 3 Div. Prop. of �

Identify the property that justifies each statement.

5. m � n, so n � m. 6. �ABC � �ABC

Symmetric Prop. of � Reflexive Prop. of � 7.

_KL �

_LK 8. p � q and q � �1, so p � �1.

Reflexive Prop. of � Transitive Prop. of � or Subst.

Practice BAlgebraic Proof2-5

Copyright © by Holt, Rinehart and Winston. 37 Holt GeometryAll rights reserved.

Name Date Class

LESSON

�XYZ � �ABC (Given)�ZYX � �XYZ (Reflexive Prop. of �)�ZYX � �ABC (Trans. Prop. of �)m�ZYX � m�ABC (Def. of �)�ABD � �CBD (Def. of � bisector)m�ABD � m�CBD (Def. of �)m�ABC � m�ABD� m�CBD (� Add. Post.)m�ABC � m�CBD� m�CBD (Subst. Prop. of �)m�ABC � 2m�CBD (Simplify.)m�ZYX � 2m�CBD (Subst. Prop. of �)

Practice CAlgebraic Proof

Solve Exercises 1 and 2. Write justifications for each step in your solutions.

1. Solve for m�3 in terms of m�1. 2. Solve for m�ZYX in terms of m�CBD.Given: �1 and �2 are complementary. Given: �XYZ � �ABC.

�2 and �3 are supplementary. __›BD is the angle bisector of �ABC.

m�1 � m�2 � 90� (Given)m�2 � m�3 � 180� (Given)m�2 � m�3 � (m�1 � m�2)� 180� � 90� (Subtr. Prop. of �)m�3 � m�1 � 90� (Simplify.)

m�3 � m�1 � 90� (Add. Prop. of �)

3. Use the Distributive Property to find (x � y )(c � d ). Write out all the steps. (Hint: Let (x � y ) � a.)

(x � y ) � a ; a (c � d ) � ac � ad ; ac � ad � c (x � y ) � d (x � y );

c (x � y ) � d (x � y ) � cx � cy � dx � dy ; (x � y )(c � d ) �

cx � cy � dx � dy

4. Explain logically how the Transitive Property of Equality can be derived from the Substitution Property of Equality and the Symmetric Property of Equality.

Possible answer: The Substitution Property states that if a � b, then b can

be substituted for a in any expression. Applying the Symmetric Property to

the Substitution Property shows that if b � a, then a can be substituted for

b in any expression. So if a � b and b � c, then a � c by the Substitution

Property, and this is also the Transitive Property.

5. Explain why there is no Substitution Property of Congruence.

Possible answer: Consider the points A(0, 1), B (1, 0), C (0, �1), and

D (�1, 0). For reflection across the x-axis, the image of_AB is

_CB.

_AB �

_AD, but you cannot conclude that the image of

_AD is

_CB for

reflection across the x-axis.

2-5

Copyright © by Holt, Rinehart and Winston. 38 Holt GeometryAll rights reserved.

Name Date Class

LESSON

� x � x Subtr. Prop. of � 5 � x Simplify.

x � 5 Sym. Prop. of �

y � 4_____7

(7) � 3(7) Mult. Prop. of �y � 4 � 21 Simplify.

�4 �4 Subtr. Prop. of �y � 17 Simplify.

4t � 12 � �20 Distr. Prop.� 12 � 12 Add Prop. of �

4t � �8 Simplify.

4t__4

� �8___4

Div. Prop. of �

t � �2 Simplify.

ReteachAlgebraic Proof

A proof is a logical argument that shows a conclusion is true. An algebraic proof uses algebraic properties, including the Distributive Property and the properties of equality.

Properties of Equality

Symbols Examples

Addition If a � b, then a � c � b � c. If x � �4, then x � 4 � �4 � 4.

Subtraction If a � b, then a � c � b � c. If r � 1 � 7, then r � 1 � 1 � 7 � 1.

Multiplication If a � b, then ac � bc. If k__2

� 8, then k__2

(2) � 8(2).

Division If a � 2 and c � 0, then a__c � b__

c . If 6 � 3t, then 6__3

� 3t__3

.

Reflexive a � a 15 � 15

Symmetric If a � b, then b � a. If n � 2, then 2 � n.

Transitive If a � b and b � c, then a � c. If y � 32 and 32� 9, then y � 9.

Substitution If a � b, then b can be substituted for a in any expression.

If x � 7, then 2x � 2(7).

When solving an algebraic equation, justify each step by using a definition, property, or piece of given information.

2(a � 1) � �6 Given equation

2a � 2 � �6 Distributive Property

� 2 � 2 Subtraction Property of Equality

2a � �8 Simplify.2a___2

� �8___2

Division Property of Equality

a � �4 Simplify.

Solve each equation. Write a justification for each step.

1. n__6

� 3 � 10 Given equation 2. 5 � x � 2x Given equation� 3 � 3 Add. Prop. of �

n__6

� 13 Simplify.

n__6

(6) � 13(6) Mult. Prop. of �n � 78 Simplify.

3.y � 4_____

7� 3 Given equation 4. 4(t � 3) � �20 Given equation

2-5