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```Record and Practice Journal Answer Key
5. −
7
26
6. 1
5
22
Value When
8. − 45.45
7. 9.452
Expression
9
ft
10
x = 0
x = 1
10. −3
9. 7.9
a.
4
4
4
b.
−x + 1
1
0
c.
4x − 4
−4
0
d.
2x + 6
6
8
e.
5 x − 13
−13
−8
f.
− 2 x + 10
10
8
g.
x + 2
2
3
h.
2x − 1
−1
1
i.
− 2x + 2
2
0
j.
−x + 2
2
1
Chapter 3
Fair Game Review
1. 6
2. 12
3. 20
19
3
4.
or 4
4
4
5. 12
6. −1
7. 24 ft2
8. 8 + y
9. p − 6
11. 11c − 8
10. 7m
r
12. r −
2
13. 9( z + 4)
3.1 Activity
1. a.
Value When
Expression
6
A.
3x + 2 − x + 4
B.
5( x − 3) + 2
x = 0
x = 1
6
8
−13
−8
C.
x + 3 − ( 2 x + 1)
2
1
D.
− 4 x + 2 − x + 3x
2
0
E.
− (1 − x ) + 3
2
3
F.
2 x + x − 3x + 4
4
4
G.
4 − 3 + 2( x − 1)
−1
1
H.
2(1 − x + 4)
10
8
I.
5 − (4 − x + 2 x)
1
0
J.
5 x − (2 x + 4 − x)
−4
0
Big Ideas Math Red
A matches d; B matches e; C matches j;
D matches i; E matches g; F matches a;
G matches h; H matches f; I matches b;
J matches c.
b. Sample answer: The expressions in the right
column have no parentheses and do not have
more than one x-term or more than one constant
term.
To obtain the equivalent expressions, the like
terms are combined.
2. Answer should include, but is not limited to:
Students should describe the steps they can use to
simplify an expression. Then students should use
3 expressions from Activity 1 as examples. Finally,
students should write 3 new expressions as
exercises.
Record and Practice Journal Answer Key
3. Sample answer: You can simplify an algebraic
4. a.
expression by: (1) using the Distributive Property
to eliminate parentheses, (2) use the Commutative
and Associative Properties of Addition to get the
variable terms together and the terms without
variables together, (3) combine the variable terms
and combine the terms without variables.
3( x − 2) + 5 − x = 3 x − 6 + 5 − x (1)
= 3x − x − 6 + 5
(2)
= 2x − 1
(3)
(2 x
b.
c.
2. Terms: − 9, 2.5 y, − 0.7 y, 1, 6.4 y 2
Like terms: 2.5 y and − 0.7 y; − 9 and 1
6. 7 d − 5
7. 11g + 8k
5. − 0.7 w + 1.1
8. 4 p − 8
9. 20 w + 9m
)
(
)
− 6 ) + (3 x + 2 ) = 5 x − 4
)
(
)
+ 4) − ( x + 2) = x + 2
) (
)
+ 3) − ( 2 x − 1) = 2 x + 4
6. ( − x + 3) − ( − 2 x − 2) = x + 5
3.2 Practice
3.2 Activity
1. a. x + 3
+ 1) + ( x − 1) = 3 x
using algebraic tiles.
To add: Combine “like” tiles and remove zero
pairs. Write the resulting algebraic expression
modeled by the remaining tiles.
To subtract: Remove any “like” tiles from the
first expression and the second expression. If tiles
remain in the second expression, add zero pairs to
the first expression as necessary. Then remove any
“like” tiles. Write the resulting algebraic expression
modeled by the remaining tiles.
Like terms: 3x and − 7 x; 4 and − 6
2
3
)
5. Sample answer: Model the algebraic expressions
1. Terms: 3 x, 4, − 7 x, − 6
4. − 3m +
(
(
(4 x
3.1 Practice
3. 3a + 9
(
(2 x
d.
)
(
(2 x
4. Sample answer: It may be easier to evaluate an
algebraic expression for a given value when the
expression is simplified. For real-life problems,
it may be easier to recognize a value, like a total
cost, based on the parts of a simplified algebraic
expression.
(
b. 2 x − 2
c. 2 x + 3
2. a.
(x
+ 2) + ( x + 4) = 2 x + 6
b.
(x
− 5) + ( x − 2) = 2 x − 7
c.
(x
+ 5) + ( 2 x − 3) = 3x + 2
d. ( 2 x − 8) + (3 x + 5) = 5 x − 3
3. a.
(x
+ 3) − ( x + 1) = 2
b.
(x
− 4) − ( x − 3) = −1
c. ( 2 x + 5) − ( x − 1) = x + 6
d. 3 x − 4
1. 2 x + 4
2. − 2n − 1
3. − 4 y + 5
4. −13k − 7
5. 3 g − 4
6. 3r − 2
7. 2 x + 17
3.2 Extension Practice
1. 7(1 + 4)
2. 25(1 + 2)
3. 7(b − 1)
4. 8( a − 2)
5. 4( 2 x + 3)
6. 12( y + 2t )
7. 10( w + 5 z )
8. 2(5v + 6u )
9. 3(3a + 5b)
10.
1
(a − 1)
2
13.
3⎛
4⎞
⎜y − ⎟
10 ⎝
3⎠
d. (3 x − 7) − ( 2 x + 3) = x − 10
11.
1
(d − 3)
4
12.
5⎛
4⎞
⎜s + ⎟
6⎝
5⎠
14. 1.1( x + 9)
Big Ideas Math Red
7
Record and Practice Journal Answer Key
15. 3.4(c + 3)
16. − 2(3 x − 5)
1⎛
9⎞
17. − ⎜ − y + ⎟
3⎝
2⎠
18. ( 2 x + 3) ft
3.3 Activity
1. a.
3.3 Practice
1. −38
2. − 27
4. −13.1
5. 5
3. −37
21
40
6. −8.72
7. x − 42 = − 50; − 8
8. 32 = z + 9; 23
=
10. −198
9. −\$1.8 million
=
3
ft
4
3.4 Activity
1. a.
=
=
=
=
=
red; 3 yellow; 3; So, x = −1.
b. z = 8
2. a. n = − 7
c. b = − 8
c. p = 4
=
d. t = −10
=
b. y = −15
3; red; 3; x = − 4
d. z = − 4
3. a. − 4 = x + 1; x = −5
c. x − 5 = − 4; x = 1
b. x − 3 = 3; x = 6
d. 5 = x − 2; x = 7
4. a. x + 57 = 19; − 38°F
b. Sample answer: The melting point of bromine
is 19 degrees Fahrenheit. This is 57 degrees
Fahrenheit more than the melting point of
mercury. So, subtract 57 from 19 to get
19 − 57 = − 38. So, the melting point of
mercury is − 38°F.
The method in part (a) is the algebraic solution
because it assigns a variable to the unknown
quantity. The method in part (b) is the arithmetic
solution because it is a direct numerical solution
that does not use any variables.
5. Sample answer: Use algebra tiles to model the
equation. Then get the variable tile by itself by
adding tiles and removing zero pairs.
subtract to get the variable by itself. To solve a
subtraction equation, add to get the variable by
itself.
b. k = − 4
c. t = −5
d. m = − 4
e. h = − 4
2. a. 8 = 4 x, x = 2
b. 6 x = −12, x = − 2
c. −10 = 2 x, x = −5
d. 3 x = −18, x = − 6
3. a. 32 x = 400; 12.5 sec; Divide the distance by the
rate to get
b.
c.
400
= 12.5.
32
x
= − 96; − 345.6°F; Multiply the melting point
3.6
of radon by 3.6 to get − 96 • 3.6 = − 345.6.
3
x = − 6; − \$8 million; Multiply this year’s
4
3
profit by the reciprocal of to get
4
4
−6 •
= − 8.
3
Sample answer: The method that uses the equation
is the algebraic solution because it assigns a
variable to the unknown quantity. The other method
is the arithmetic solution because it is a direct
numerical solution that does not use any variables.
7. The value of x changes or varies, so x can equal
both 2 and 3 in two different problems.
8. Sample answer: The weather varies from day to day.
The amount of food a person eats in a day varies.
8
Big Ideas Math Red
Record and Practice Journal Answer Key
4. Sample answer: Multiplication can be used to solve
an equation that uses division and division can be
used to solve an equation that uses multiplication.
The equation x ÷ ( −3) = − 6 uses division, so
multiply each side by −3 to obtain x = 18. The
equation 2 x = −10 uses multiplication, so divide
each side by 2 to obtain x = − 5.
3.4 Practice
1. −30
3. 30
4. − 0.25
5. −50
6. 1.1
x
= 7; −56
−8
b. \$38.92
5. Add enough 1 or −1 tiles to each side to remove the
zero pairs and leave only variable tiles on one side.
Form as many equal groups of 1 or −1 tiles as there
are variable tiles. One of these groups shows the
value of the variable.
6. Subtract 1 from each side.
7. x = 25; The steps are the same.
3
2. −
4
7.
4. a. 25 points
8. −12 x = 60; −5
3.5 Practice
1. −3
2. 6
3. − 7
4. 3
5. − 6.9
6. 1
7. a. A = 2 w − 3
9. 65 cups
2
9
b. 7 meters
Chapter 4
3.5 Activity
Fair Game Review
1.
=
1.
2.
=
−6
−5
−4
−3
−2
−1
0
−7
−6
−5
−4
−3
−2
−1
−1
0
1
2
3
4
5
4
5
6
7
8
9
10
3.
=
4.
=
5.
=
−2.5 −2.4 −2.3 −2.2 −2.1 −2 −1.9
6.
=
0
3; 3; 3; 2; 2; x = −1
1
5
2
5
3
5
4
5
1
6
5
7.
413 414 415 416 417 418 419
2.
2x − 3 = −5
Write the equation.
2x − 3 + 3 = −5 + 3
2x = − 2
Simplify.
−2
2x
=
2
2
Divide each side by 2.
8. >
9. >
10. >
11. <
12. <
13. <
14. your friend; 5.6 ft is about 5 ft and 7 in.
4.1 Activity
1. a. 5, 6; The numbers are greater than or equal to 5;
a. adding 3 to each side
yes; The phrase at least implies that the number
can be equal to 5; Sample answer: 7, 8, 9, 10
b. − 5, − 6, − 7; The numbers are less than − 4; no;
The phrase less than implies that the number
cannot be equal to − 4; Sample answer: − 8, − 9,
−10, −11
b. They are the same.
c. 4, 5, 6; The numbers are greater than 3; no; The
x = −1
So, x = −1.
3. a. 2 x + 2 = − 6; x = − 4
b. −13 = 3 x − 4; x = −3