Holt Algebra 1

Holt Algebra 1 Homework and Practice Workbook 1st Edition Chapter 2

Holt Algebra 1 Homework And Practice Workbook Chapter 2 Exercise 2.6 Solutions Page 14 Problem 1 Answer

Given: The ratio of freshmen to sophomores in a drama club is 5: 6. There are 18 sophomores in the drama club.

To Find: How many freshmen are there?

Method Used: Simplification using addition, subtraction, division, or multiplication of integers.

The ratio of freshmen to sophomores in a drama club is Freshman/Sophomores=5/6.

There are 18 sophomores in the drama club.

Freshman/Sophomores=5/6 Freshman

18=5/6

⇒Freshman=5/6×18

⇒Freshman=15

​Number of freshman=15

Read and Learn More Holt Algebra 1 Homework and Practice Workbook Solutions

Holt Algebra 1 Chapter 2 Page 14 Problem 2 Answer

Given:  Four pounds of apples cost1.96

To Find: unit rate

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

As we have four pounds of apples cost 196

Cost of 4 apples=1.96

Cost of 1 apples=1.96/4

Cost of 1 apple=0.49

​Therefore, cost of 1 apple=0.49 dollars

Holt Algebra 1 Chapter Page 14 Problem 3 Answer

Chapter 2 Exercise 2.6 Answers Holt Algebra 1 Practice Workbook 1st Edition Page 14 Problem 4 Answer

Given: A giraffe can run 32 miles per hour.

To Find: What is this speed in feet per second?

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

According to question, a giraffe can run 32 miles per hour.

Using 1 mile =5280 feet and 1 hour =60×60 second

we get Then a giraffe can run

32×miles hour

=32×5280

60×60×feet second

=47 feet per second

​32×miles per hour =47 feet per second

Holt Algebra 1 Chapter Page 14 Problem 5 Answer

Given: proportion is y/4=10/8

To Find: value ofthe  variable

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

Cross-multiplying the given proportion

Y/4=10/8

⇒8y=10×4

⇒y=10×4/8

⇒y=5

​Required value is y=5

Step-By-Step Solutions For Exercise 2.6 Chapter 2 Holt Algebra 1 Homework Workbook Page 14 Problem 6 Answer

Given: proportion is 2/x=30−6

To Find: value of the variable

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

Cross-multiplying the given proportion

2/x=30−6

⇒−12=30×x

⇒x=−12/30

⇒x=−2/5

⇒x=−1/2.5

⇒x=−0.4

​The required value is x=−0.4

Holt Algebra 1 Chapter Page 14 Problem 7 Answer

Given: proportion is 3/12=−24/m

To Find: value of the variable

Method Used: Simplification using addition, subtraction, division, or multiplication of integers.

Cross-multiplying the given proportion

3/12=−24m

⇒1/4 =−24m

⇒m=−24×4

⇒m=−96

​Required value ism=−96

Exercise 2.6 Solutions For Chapter 2 Holt Algebra 1 Homework And Practice Workbook Page 14 Problem 8 Answer

Given: proportion is 3t

10=1/2

To Find: the value of the variable

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

Cross-multiplying the given proportion

3t/10=1/2

⇒3t×2=10

⇒t=10/6

⇒t=1.67

​The required value is t=1.67

Holt Algebra 1 Chapter Page 14 Problem 9 Answer

Given: proportion is 7x=1/0.5

To Find: value of variable

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

Cross multiplying the given proportion

7/x=1/0.5

⇒7×0.5=x

⇒x=3.5

​Required value is x=3.5

Examples Of Problems From Exercise 2.6 Chapter 2 In Holt Algebra 1 Workbook Page 14 Problem 10 Answer

Given: The scale of his model to the actual car is 1:10

and his model is 181/2 inches long

To Find: How long is the actual car?

Method Used: use of formula of scale

Here, 1 inch model/10 inches actual=18.5 inch model/x inches actual

x=18.5×10/1

=185 inches

​Actual length is=185 inches

Holt Algebra 1 Chapter Page 14 Problem 11 Answer

Given: The scale on a map of Virginia shows that 1 centimeter represents 30 miles.

The actual distance from Richmond, VA to Washington, DC is 110 miles.

To Find: how many centimeters are between the two cities?

Method Used: use of formula of scale.

Scale is given as dimension on map actual dimension

Here, 1 centimeters represents 30 miles then

30 miles =1 cm

1 mile =1/30cm

110 miles =1/30×110

110 miles =3.67 cm

​Required distance in map is110 miles =3.67 cm

Holt Algebra 1 Homework and Practice Workbook 1st Edition Chapter 2

Holt Algebra 1 Homework and Practice Workbook Chapter 2 Exercise 2.5 Solutions Page 13 Problem 1 Answer

Given: C=2πr

To Find: Solve the equation for r.

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

Given equation can be written as

C=2πr

⇒C/2π

=r

⇒r=C/2π

Required equation is r=C/2π

Read and Learn More Holt Algebra 1 Homework and Practice Workbook Solutions

Chapter 2 Exercise 2.5 answers Holt Algebra 1 Practice Workbook 1st Edition Page 13 Problem 2 Answer

Given: y=mx+b

To Find: Solve the equation for m.

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

Given equation can be written as

y=mx+b

⇒y−b=mx

⇒m=y−b/x

Required equation ism=y−b/x

Step-By-Step Solutions For Exercise 2.5 Chapter 2 Holt Algebra 1 Homework Workbook Page 13 Problem 3 Answer

Given: 4c=d

To Find: Solve the equation for c.

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

Given equation can be written as

4c=d

⇒c=d/4

Required equation is c=d/4

Holt Algebra 1 Chapter 2 Page 13 Problem 4 Answer

Exercise 2.5 Solutions For Chapter 2 Holt Algebra 1 Homework And Practice Workbook Page 13 Problem 5 Answer

Given: 2p+5r=q

To Find: Solve the equation for p.

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

Given equation can be written as

2p+5r=q

⇒2p=q−5r

⇒p=q−5r/2

Required equation is p=q−5r/2

Holt Algebra 1 Chapter 2 Page 13 Problem 6 Answer

Given: −10=xy+z

To Find: Solve the equation for x.

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

Given equation can be written as

−10=xy+z

⇒−10−z=xy

⇒x=−(10+z)/y

Required equation is x=−(10+z)/y

Common Core Exercise 2.5 Chapter 2 Solutions Detailed Holt Algebra 1 Practice Workbook 1st Edition Page 13 Problem 7 Answer

Given: a/b=c

To Find: Solve the equation for b.

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

Given equation can be written as

A/b=c

⇒a=bc

⇒b=a/c

Required equation is b=a/c

Holt Algebra 1 Chapter 2 Page 13 Problem 8 Answer

Given: h−4/j=k

To Find: Solve the equation for j.

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

Given equation can be written as

h−4/j=k

⇒h−4=kj

⇒j=h−4/k

Required equation is j=h−4/k

Page 13 Problem 9 Answer

Given: c=5p+215

To Find: Solve the equation for p.

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

The given equation can be written as

c=5p+215

⇒c−215=5p

⇒p=c−215/5

⇒p=c/5−43

​The required equation is p=c/5−43

Holt Algebra 1 Chapter 2 Page 13 Problem 10 Answer

Given: c=5p+215 where c=300

To Find: Solve the equation for p

Method Used: Put all the values in p=c−215/5

Given equation can be written asp=c−215/5

p=c−215/5

⇒p=300−215/5

⇒p=85/5

⇒p=17

​Required value is p=17

Student Edition Chapter 2 Exercise 2.5 Holt Algebra 1 Homework Workbook Solutions Guide Page 13 Problem 11 Answer

Given: A=1/2bh

To Find: Solve the equation for b.

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

Given equation can be written as

A=1/2bh

⇒2A=bh

⇒b=2A/h

Required equation is b=2A/h

Holt Algebra 1 Chapter 2 Page 13 Problem 12 Answer

Given: A=1/2bh, area is 192 mm² and height is 12 mm

To Find: Solve the value of base

Method Used: put all the values in b=2Ah

Given equation can be written as b=2A/h

Putting all the values as

b=2A/h

⇒b=2×192/12

⇒b=2×16

⇒b=32 mm

​Required equation is b=32 mm

Holt Algebra 1 Homework and Practice Workbook 1st Edition Chapter 2

Holt Algebra 1 Homework And Practice Workbook Chapter 2 Exercise 2.4 Solutions Page 12 problem 1 Answer

Given: 3d+8=2d−17

To Find: Solve the equation.

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

Simplifying and solving the equation as

3d+8=2d−17

⇒3d−2d=−17−8

⇒d=−25

​Required solution is d=−25

Read and Learn More Holt Algebra 1 Homework and Practice Workbook Solutions

Holt Algebra 1 Homework And Practice Workbook Chapter 2 Exercise 2.4 Solutions Page 12 Problem 2 Answer

Given: 2n−7=5n−10

To Find: Solve the equation.

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

Simplifying and solving the equation as

2n−7=5n−10

⇒5n−2n=−7+10

⇒3n=3

⇒n=1

​Required solution is n=1

Chapter 2 Exercise 2.4 Answers Holt Algebra 1 Practice Workbook 1st Edition Page 12 Problem 3 Answer

Page 12 Problem 4 Answer

Given: −t+5=t−19

To Find: Solve the equation.

Method Used: Simplification using addition, subtraction, division, or multiplication of integers.

Simplifying and solving the equation as

−t+5=t−19

⇒t+t=5+19

⇒2t=24

⇒t=12

​The required solution is t=12

Step-By-Step Solutions For Exercise 2.4 Chapter 2 Holt Algebra 1 Homework Workbook Page 12 Problem 5 Answer

Given: 15x−10=−9x+2

To Find: Solve the equation.

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

Simplifying and solving the equation as

15x−10=−9x+2

⇒15x+9x=10+2

⇒24x=12

⇒x=1/2

The required solution is x=1/2

Exercise 2.4 Solutions For Chapter 2 Holt Algebra 1 Homework And Practice Workbook Page 12 Problem 6 Answer

Given: 4n+6−2n=2(n+3)

To Find: Solve the equation.

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

Simplifying and solving the equation as

4n+6−2n=2(n+3)

⇒4n+6−2n=2n+6

⇒4n−2n−2n=6−6

⇒0=0

​The equation will always be true but the solution is not present.

Page 12 problem 7 Answer

Given: 6m−8=2+9m−1

To Find: Solve the equation.

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

Simplifying and solving the equation as

6m−8=2+9m−1

⇒9m−6m=−8−2+1

⇒3m=−9

⇒m=−3​

Required solution is m=−3

Examples Of Problems From Exercise 2.4 Chapter 2 In Holt Algebra 1 Workbook Page 12 Problem 8 Answer

Given: −v+5+6v=1+5v+3

To Find: Solve the equation.

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

Simplifying and solving the equation as

−v+5+6v=1+5v+3

⇒5v+5=5v+4

​This equation is wrong.

the solution is not present as the equation is wrong.

Page 12 Problem 9 Answer

Given: 2(3b−4)=8b−11

To Find: Solve the equation.

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

Simplifying and solving the equation as

2(3b−4)=8b−11

⇒6b−8=8b−11

⇒2b=3

⇒b=3/2

Required solution is b=3/2

Common Core Exercise 2.4 Chapter 2 Solutions Detailed Holt Algebra 1 Practice Workbook 1st Edition Page 12 Problem 10 Answer

Given: 2(3b−4)=8b−11

To Find: Solve the equation.

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

Simplifying and solving the equation as

2(3b−4)=8b−11

⇒6b−8=8b−11

⇒2b=3

⇒b=3/2

Required solution is b=3/2

Page 12 problem 11 Answer

Given:  One company offers a starting salary of 28000 with a raise of 3000 and other company offers 36000 with a raise of 2000.

To Find:  After how many years would Janine’s salary be the same with both companies?

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

Let  us consider x represent the number of years then

At the first company, after x years Jennie’s salary will be: 28,000+3,000x

At the second company, after x years Jennie’s salary will be: 36,000+2,000x

Equating both values as

28,000+3,000x=36,000+2,000x

⇒1,000x=36,000−28,000

⇒x=8

​Required number of years is x=8

Student Edition Chapter 2 Exercise 2.4 Holt Algebra 1 Homework Workbook Solutions Guide Page 12 Problem 12 Answer

Given:  One company offers a starting salary of 28000 with a raise of 3000 and other company offers 36,000 with a raise of 2000 .

To Find:  After how many years would Janine’s salary be the same with both companies?

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

As we have calculated that after 8 years, salaries are same.

Putting this in any one of the salaries as

28,000+3,000∗x=28,000+3,000∗8

⇒28,000+3,000∗8=52,000.

​After 8 years the salary will be: 28,000+3,000∗8=52,000.

Page 12 Problem 13 Answer

Given:  Xian and his cousin both collect stamps. Xian has 56 stamps, and his cousin has 80 stamps.

Both have recently joined different stamp-collecting clubs.

Xian’s club will send him 12 new stamps per month, and his cousin’s club will send him 8 new stamps per month.

To Find:   After how many months will Xian and his cousin have the same number of stamps?

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

Let us consider the number of months be x

According to the question56+12x=80+8x

56+12x=80+8x

⇒12x−8x=80−56

⇒4x=24

⇒x=6

​After 6 many months will Xian and his cousin have the same number of stamps.

Step-By-Step Answers For Exercise 2.4 Chapter 2 In Holt Algebra 1 Homework And Practice Workbook Page 12 Problem 14 Answer

Given:  Xian and his cousin both collect stamps. Xian has 56 stamps, and his cousin has 80 stamps.

Both have recently joined different stamp-collecting clubs.

Xian’s club will send him 12 new stamps per month, and his cousin’s club will send him 8 new stamps per month.

To Find:  How many stamps will that be?

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

As we have calculated, After 6 many months will Xian and his cousin have the same number of stamps.

Then number of stamps is

56+12∗x=56+12∗6

⇒56+12∗6=128

80+8x=80+8∗4

⇒80+8∗4=128

​After six months number of stamps is 128

Holt Algebra 1 Homework and Practice Workbook 1st Edition Chapter 2

Holt Algebra 1 Homework And Practice Workbook Chapter 2 Exercise 2.3 Solutions 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.

Read and Learn More Holt Algebra 1 Homework and Practice Workbook Solutions

Holt Algebra 1 Homework And Practice Workbook Chapter 2 Exercise 2.3 Solutions 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.

Chapter 2 Exercise 2.3 Answers Holt Algebra 1 Practice Workbook 1st Edition 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.

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.

Step-By-Step Solutions For Exercise 2.3 Chapter 2 Holt Algebra 1 Homework Workbook 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.

Exercise 2.3 Solutions For Chapter 2 Holt Algebra 1 Homework And Practice Workbook 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.

Step-By-Step Answers For Exercise 2.3 Chapter 2 In Holt Algebra 1 Homework And Practice Workbook 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.

Examples Of Problems From Exercise 2.3 Chapter 2 In Holt Algebra 1 Workbook 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.

Common Core Exercise 2.3 Chapter 2 Solutions Detailed Holt Algebra 1 Practice Workbook 1st Edition 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∘.

Student Edition Chapter 2 Exercise 2.3 Holt Algebra 1 Homework Workbook Solutions Guide 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.

Holt Algebra 1 Homework and Practice Workbook 1st Edition Chapter 2

Holt Algebra 1 Homework And Practice Workbook Chapter 2 Exercise 2.2 Solutions Page 10 Problem 1 Answer

Given,d/8=6

To find : the value of d.

Solve the above equation and find the value of d.

Consider the given equation.

d/8=6

Cross multiplying, we get

d=8×6

d=48

Now, check our answer

48/8=6

6=6​

Hence, the value of d is 48.

Read and Learn More Holt Algebra 1 Homework and Practice Workbook Solutions

Holt Algebra 1 Homework And Practice Workbook Chapter 2 Exercise 2.2 Solutions Page 10 Problem 2 Answer

Given,−5=n/2

To find : the value of n.

Solve the above equation and find the value of n.

Consider the given equation.

−5=n/2

Cross multiplying, we get

−5×2=n

n=−10

Now, check our answer

−5=−10/2

−5=−5

​Hence, the value of n is −10.

Chapter 2 Exercise 2.2 Answers Holt Algebra 1 Practice Workbook 1st Edition Page 10 Problem 3 Answer

​Now, check my answer

−(−24)/2=12/24

2=12

12=12

​Hence, the value of t is −24.

Page 10 Problem 4 Answer

Given,−40=−4x

To find : the value of x

Solve the above equation and find the value of x.

Consider the given equation.

−40=−4x

On dividing the above equation by −4, we get

−40−4=−4x

−4/10=x

x=10

Now, check my answer

−40=−4(10)

−40=−40

​Hence, the value of x is 10.

Step-By-Step Solutions For Exercise 2.2 Chapter 2 Holt Algebra 1 Homework Workbook Page 10 Problem 5 Answer

Given,2r/3=16

To find : the value of r

Solve the above equation and find the value of r.

Consider the given equation.

2r/3=16

On solving the above equation, we get

2r=3×16

2r=48

r=48/2

r=24

​Now, check my answer

2(24)/3=16/48

3=16

16=16

​Hence, the value of  r is 24.

Page 10 Problem 6 Answer

Given,−49=7y

To find : the value of y

Solve the above equation and find the value of y.

Consider the given equation.

−49=7y

On dividing the above equation by 7, we get

−49/7=7y

7−7=y

y=−7

​Now, check our answer

−49=7(−7)

−49=−49

​Hence, the value of y is −7.

Exercise 2.2 Solutions For Chapter 2 Holt Algebra 1 Homework And Practice Workbook Page 10 Problem 7 Answer

Given,−15=−3n/5

To find : the value of n.

Solve the above equation and find the value of n.

Consider the given equation.

−15=−3n/5

On solving the above equation, we get

−15×5=−3n

−3n=−75

n=−75/−3

n=25

Now, check our answer

−15=−3(25)/5

−15=−15

​Hence, the value of n is 25.

Step-By-Step Answers For Exercise 2.2 Chapter 2 In Holt Algebra 1 Homework And Practice Workbook Page 10 Problem 8 Answer

Given,v−3=−6

To find : the value of v.

Solve the above equation and find the value of v.

Consider the given equation.

v−3=−6

On solving the above equation, we get

v=−6×−3

v=18

Now, check my answer

18−3=−6

−6=−6

​Hence, the value of v is 18.

Examples Of Problems From Exercise 2.2 Chapter 2 In Holt Algebra 1 Workbook Page 10 Problem 9 Answer

Given,

2.8=b/4

To find: the value of b

Solve the above equation and find the value of b.

Consider the given equation.

2.8=b/4

Cross multiplying, we get

b=2.8×4

b=11.2

Now, check our answer

2.8=11.2/4

2.8=2.8

​Hence, the value of b is 11.2.

Page 10 Problem 10 Answer

Given, 3r /4=1/8

To find: the value of r

Solve the above equation and find the value of r.

Consider the given equation.

3r/4=1/8

On solving the above equation, we get

3r=4/8

3r=1/2

r=1/6

Now, check my answer

3(1/6)/4=1/8

1/2/4=1/8

1/8=1/8

Hence, the value of r is 1/6.

Common Core Exercise 2.2 Chapter 2 Solutions Detailed Holt Algebra 1 Practice Workbook 1st Edition Page 10 Problem 11 Answer

Given, The price of the package = $2.07

The first-class rate at the time = $0.23 per ounce

To find the weight of the package.

Apply the concept of linear equation. And solve the equation.

Given, The price of the package =$2.07

The first-class rate at the time =$0.23  per ounce

Let ′x′ be the weight of the package.

According to the question,

0.23x=2.07

x=2.07/0.23

x=9

​Hence, the weight of the package is 9 ounce.

Student Edition Chapter 2 Exercise 2.2 Holt Algebra 1 Homework Workbook Solutions Guide Page 10 Problem 12 Answer

Given, Lola spends $8 per at the movies.

To find : Lola’s weekly allowance.

Apply the concept of linear equation.

And solve the equation.

Given, Lola spends $8 per at the movies.

Let ′x′ be Lola’s weekly allowance.

According to the given question,

1/3

x=8

x=8×3

x=24

​Hence, Lola’s weekly allowance is $24.

Holt Algebra 1 Homework and Practice Workbook 1st Edition Chapter 2

Page 9 Problem 1 Answer

Given: g−7=14

To Solve each equation.

To solve the given expression, we will add the same thing on both the sides.

g−7=15

Add 7 on both the sides of the equation,

g−7+7=15+7

g=22

​After solving the given expression, the value of g is 22.

Read and Learn More Holt Algebra 1 Homework and Practice Workbook Solutions

Holt Algebra 1 Homework And Practice Workbook Chapter 2 Exercise 2.1 Solutions Page 9 Problem 2 Answer

Given: t+4=6

To Solve each equation.

To solve the given expression, we will subtract the same thing on both the sides.

t+4=6

Subtract 4 from both the sides,

t+4−4=6−4

t=2

​After solving the given expression, the value of t is 2.

Chapter 2 Exercise 2.1 Answers Holt Algebra 1 Practice Workbook 1st Edition Page 9 Problem 3 Answer

Page 9 Problem 4 Answer

Given: x+3.4=9.1

To Solve each equation.

To solve the given expression, we will subtract the same thing on both the sides.

x+3.4=9.1

Subtract 3.4 from both the sides,

x+3.4−3.4=9.1−3.4

x=5.7

After solving the given expression, the value of x is 5.7.

Step-By-Step Solutions For Exercise 2.1 Chapter 2 Holt Algebra 1 Homework Workbook Page 9 Problem 5 Answer

Given: n−3/8=1/8

To Solve each equation.

To solve the given expression, we will add the same thing on both the sides.

Given: n−3/8=1/8

To Solve each equation.

To solve the given expression, we will add the same thing on both the sides.

Simplify,

n−3/8+3/8=1/8+3/8

n=4/8

On simplifying 4/8 we get, 1/2.

The value of n is 1/2.

Page 9 Problem 6 Answer

Given: p−1/3=2/3

To Solve each equation.

To solve the given expression, we will add the same thing on both the sides.

p−1/3=2/3

Add 1/3 on both the sides,

p−1/3+1/3=2/3+1/3

Simplify,

p−1/3+1/3=2/3+1/3

p=3/3

On simplifying 3/3 we get 1.

The value of p is 1.

Exercise 2.1 Solutions For Chapter 2 Holt Algebra 1 Homework and Practice Workbook Page 9 Problem 7 Answer

Given: −6+k=32

To Solve each equation.

To solve the given expression, we will add the same thing on both the sides.

−6+k=32

Add 6 on both the sides.

−6+k=32

−6+6+k=32+6

k=38

​The value of k is 38.

Page 9 Problem 8 Answer

Given: 7=w+9.3

To Solve each equation.

To solve the given expression, we will subtract the same thing on both the sides.

7=w+9.3

Subtract 9.3 from both the sides,

7−9.3=w+9.3−9.3

−2.3=w

​The value of w is −2.3.

Examples Of Problems From Exercise 2.1 Chapter 2 In Holt Algebra 1 Workbook Page 9 Problem 9 Answer

Given: y−57=−40

To Solve each equation.

To solve the given expression, we will add the same thing on both the sides.

y−57=−40

Add 57 on both the sides,

y−57+57=−40+57

y=17

​The value of y is 17.

Page 9 Problem 10 Answer

Given: −5.1+b=−7.1

To Solve each equation.

To solve the given expression, we will add the same thing on both the sides.

−5.1+b=−7.1

Add 5.1 on both the sides,

−5.1+5.1+b=−7.1+5.1

b=−2

​The value of b is −2.

Common Core Exercise 2.1 Chapter 2 Solutions Detailed Holt Algebra 1 Practice Workbook 1st Edition Page 9 Problem 11 Answer

Given: a+15=15

To Solve each equation.

To solve the given expression, we will subtract the same thing on both sides.

a+15=15

Subtract 15 from both the sides,

a+15−15=15−15

a=0

​The value of a is 0.

Page 9 Problem 12 Answer

To write and solve an equation to determine Marietta’s hourly wage before her raise.

Show that your answer is reasonable.

Assume hourly wage and solve the resulting equation.

Let x be her hourly wage before her raise. She was given a raise of $0.75 an hour, which brought her hourly wage to $12.25.

So, the equation, that represents this situation, will be x+0.75=12.25.

Solve for x to get

x=12.25−0.75

x =11.5

​So, her hourly wage before the raise was $11.5.

So, her hourly wage before the raise was  $11.5.Check if this is true or if your answer is reasonable, by plugging x=11.5 back in the said equation 11.5+0.75=12.25.

Step-By-Step Answers For Exercise 2.1 Chapter 2 In Holt Algebra 1 Homework And Practice Workbook Page 9 Problem 13 Answer

Given: Brad grew 41/4 inches this year and is now 567/8 inches tall.

To find Brad’s height at the start of the year.

Show that your answer is reasonable.

We will convert both the given mixed fractions into improper fractions and then we will subtract the value of height grown from the total height to get Brad’s height at the start of the year.

Brad grew 41/4 inches this year. We will convert the mixed fraction to an improper fraction.

We will multiply the denominator of the fraction with the whole number and add whatever there is in the numerator, it can now be written as,

41/4=(4×4)+1/4

On simplifying, we get: 41/4=17/4

Therefore, Brad grew 17/4 inches this year.

Currently Brad is 567/8 inches tall. We will convert the mixed fraction to an improper fraction.

We will multiply the denominator of the fraction with the whole number and add whatever there is in the numerator, it can now be written as,

567/8=(56×8)+7/8

On simplifying, we get:567/8=455/8

Therefore, Brad is 455/8 tall.

Now brads height at the start of the year will be the subtraction of the current height and the height grown.

It can be written as,

⇒455/8−17/4

On taking the lowest common multiple in the denominator, we get,

⇒455/8−34/8

On taking the denominator common, we get,

⇒455−34/8

On simplifying the terms in the numerator, we get,

⇒421/8

Brad’s height at the start of the year is 421/8.

To show the answer reasonable add Brad’s height at the start of the year in his grown height i.e.,

421/8+17/4=421+34/8

421+34/8=455/8

From Step 2 we can write 455/8 as 567/8 which is the current height of Brad’s.

Student Edition Chapter 2 Exercise 2.1 Holt Algebra 1 Homework Workbook Solutions Guide Page 9 Problem 14 Answer

To solve an equation to find Heather’s practice time.

Heather’s finishing time is 2.6 seconds less than her practice time, so ,

Finishing time=Practice time−2.6 seconds.

Let P= practice time.

58.4=P−2.6

Add 2.6 to both sides,61=P.

Heather’s practice time is 61 seconds.

Page 9 Problem 15 Answer

Given: Radius of Earth 6378.1 km

To determine the radius of Mars.

Let the radius of Mars be R

As it is given that the radius of Earth is 2981.1 Km bigger than the radius of Mars,

So,6378.1km=R+(2981.1km)

Subtract (2981.1 km) from both sides of the equation.

6378.1−2981.1=R+2981.1−2981.1

3397=R

​The Radius of Mars is 3397 km.

To show the answer reasonably add the radius of Mars in 2981.1 we will get the Radius of earth.