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