Holt Algebra 1 Homework and Practice Workbook 1st Edition Chapter 2
Page 11 Problem 1 Answer
Given: −4x+7=11.
To Find: Solve the equation.
Method Used: Simplification using addition, subtraction, division or multiplication of integers.
We know, −4x+7=11
Subtract both sides by 7,
−4x+7−7=11−7
−4x=4
−4x=4
Divide both sides by −4,−4x
−4=4/−4
x=−1
Therefore, x=−1.
Page 11 Problem 2 Answer
Given: 17=5y−3.
To Find: Solve the equation.
Method Used: Simplification using addition, subtraction, division or multiplication of integers.
We know,
17=5y−3
Adding both sides by 3,
17+3=5y−3+3
20=5y
20=5y
Divide both sides by 5,
20/5=5y/5
y=4
Therefore, y=4.
Page 11 Problem 3 Answer
Given: −4=2p+10.
To Find: Solve the equation.
Method Used: Simplification using addition, subtraction, division or multiplication of integers.
We know, −4=2p+10
Subtract both sides by 10,
−4−10=2p+10−10
−14=2p
−14=2p
Divide both sides by 2,−14/2
=2p/2
p=−7
Therefore, p=−7.
Page 11 Problem 4 Answer
Given: 3m+4=1.
To Find: Solve the equation.
Method Used: Simplification using addition, subtraction, division or multiplication of integers.
We know, 3m+4=1
Subtract both sides by 4,
3m+4−4=1−4
3m=−3
3m=−3
Divide both sides by 3,3m/3
=−3/3
m=−1
Therefore, m=−1.
Page 11 Problem 5 Answer
Given: 12.5=2g−3.5.
To Find: Solve the equation.
Method Used: Simplification using addition, subtraction, division or multiplication of integers.
We know, 12.5=2g−3.5
Add both sides by 3.5,
12.5+3.5=2g−3.5+3.5
16=2g
16=2g
Divide both sides by 2,16/2
=2g/2
g=8
Therefore, g=8.
Page 11 Problem 6 Answer
Given: 7/9=2n+1/9.
To Find: Solve the equation.
Method Used: Simplification using addition, subtraction, division or multiplication of integers.
We know, 7/9=2n+1/9
Subtract both sides by 1/9,7/9−1/9
=2n+1/9−1/9
7−1/9=2n/6
9=2n/6
9=2n
Divide both sides by 2,6/9×2
=2n/2
n=1/3
Therefore, n=1/3.
Page 11 Problem 7 Answer
Given: −4/5t+2/5=2/3.
To Find: Solve the equation.
Method Used: Simplification using addition, subtraction, division or multiplication of integers.
We know, −4/5t+2/5=2/3
Subtract both sides by 2/5,−4/5t+2/5−2
5=2/3−2/5−4/5
t=10−6/15−4/5
t=4/15−4/5
t=4/15
Multiple both sides by −5/4,−4/5t×−5/4
=4/15×−5/4
t=−1/3
Therefore, t=−1/3.
Page 11 Problem 8 Answer
Given: −2(b+5)=−6.
To Find: Solve the equation.
Method Used: Simplification using addition, subtraction, division or multiplication of integers.
We know, −2(b+5)=−6
−2b−10=−6
Add both sides by 10,
−2b−10+10=−6+10
−2b=4
−2b=4
Divide both sides by −2,
−2b/−2=4/−2
b=−2
Therefore, b=−2.
Page 11 Problem 9 Answer
Given: 8=4(q−2)+4.
To Find: Solve the equation.
Method Used: Simplification using addition, subtraction, division or multiplication of integers.
We know,
8=4(q−2)+4
8=4q−8+4
8=4q−4
Add both sides by 4,
8+4=4q−4+4
12=4q
12=4q
Divide both sides, by 4,12/4=4q/4
q=3
Therefore, q=3.
Page 11 Problem 10 Answer
Given: 3x−8=−2.
To Find: The value of x−6.
Method Used: Simplification using addition, subtraction, division or multiplication of integers.
We know,
3x−8=−2
Add both sides by 8,
3x−8+8=−2+8
3x=6
Divide both sides by 3,3x/3=6/3
x=2
We know, x=2.
Substitute x=2 in x−6,
x−6=2−6
x−6=−4.
Therefore, the value of x−6=−4.
Page 11 Problem 11 Answer
Given: −2(3y+5)=−4.
To Find: The value of 5y.
Method Used: Simplification using addition, subtraction, division or multiplication of integers.
We know,
−2(3y+5)=−4
−6y−10=−4
Add both sides by 10,
−6y−10+10=−4+10
−6y=6
Divide both sides by −6,−6y/−6=6/−6
y=−1
We know, y=−1.
Substitute y=−1 in 5y,
5y=5×−1=−5
Therefore, the value of 5y=−5.
Page 11 Problem 12 Answer
Given: The two angles shown form a right angle.
To Find: Write and solve an equation to find the value of x.
Method Used: Simplification using addition, subtraction, division or multiplication of integers.
We know, the right angle is equal to 90∘.
So, as per the given information in the question,
3x−5+2x=90
5x−5=90
Add both sides by 5,
5x−5+5=90+5
5x=95
Divide both sides by 5,5x/5=95/5
x=19
Therefore, the value of x=19∘.
Page 11 Problem 13 Answer
Given: For her cellular phone service, Vera pays $32 a month, plus $0.75 for each minute over the allowed minutes in her plan.
Vera received a bill for $47 last month.
To Find: For how many minutes did she use her phone beyond the allowed minutes?
Method Used: Using the information in the question form a linear equation and simplify using addition, subtraction, division or multiplication of integers.
We know, Vera pays $32 a month, plus $0.75 for each minute over the allowed minutes in her plan. Vera received a bill for $47 last month.
Let us assume that Vera used her phone for x minutes beyond allowed minutes.
So, as per the question equation,
0.75x+32=47
Subtract both sides by 32,
0.75x+32−32=47−32
0.75x=15
0.75x=15
Divide both sides by 0.75
0.75x/0.75=15/0.75
x=20
Therefore, Vera used her phone for 20 minutes beyond allowed minutes.