Search results
Most tricky and tough algebra word problems are covered here. If you can solve these, you can probably solve any algebra problems. Teachers! Feel free to select from this list and give them to your students to see if they have mastered how to solve tough algebra problems. Find out below how you can print these problems. You can also purchase a ...
- Algebra Word Problems
Other algebra word you should know how to solve Example #1:...
- Solving Absolute Value Equations
After applying the definition to example #1, you will have...
- Member Login
Member Login. Username/Email. Password
- Pre-algebra Lessons
Goal: Learn to add, subtract, multiply and divide whole...
- Awards
You have done a wonderful thing for children of all ages...
- Trigonometry Lessons
The problems you can solve with trigonometry are many. A...
- Algebra Word Problems
- Example 1: Algebra Word Problems
- A Few Notes About This Problem
- Example 2 - Identifying A Constant
- Solution
- Notes For Example 2
- Example 3 - Equations with Variables on Both Sides
- Notes For Example 3
Linda was selling tickets for the school play. She sold 10 more adult tickets than childrentickets and she sold twice as many senior tickets as children tickets. 1. Let x represent the number of children's tickets sold. 2. Write an expression to represent the number of adult tickets sold. 3. Write an expression to represent the number of senior tic...
1. In this problem, the variable was defined foryou. Let x represent the number ofchildren’s tickets sold tells what x stands for in this problem. If this had not been done for you, you mighthave written it like this: Let x = the number of children’s tickets sold 2. For the first expression, I knew that 10 more adulttickets were sold. Since more me...
A cell phone company charges a monthly rate of $12.95 and $0.25 a minute per call. The bill for mminutes is $21.20. 1. Write an equation that models this situation. 2. How many minutes were charged on this bill?
A cell phone company charges a monthly rate of $12.95 and $0.25 a minute per call. The bill for mminutes is $21.20. 1. Write an equation that models this situation.
$12.95 is a monthly rate. Since this is a set fee for each month, I know that this is a constant. The rate does not change; therefore, it is not associated with a variable.$0.25 per minute per call requires a variable because the total amount will change based on the number of minutes. Therefore, we use the expression 0.25mYou must solve the equation to determine the value for m, which is the number of minutes charged.You have $60 and your sister has $120. You are saving $7 per week and your sister is saving $5 per week. How long will it be before you and your sister have the same amount of money? Write an equation and solve.
$60 and $120 are constants because this is the amount of money that they each have to begin with. This amount does not change.$7 per week and $5 per week are rates. They key word "per" in this situation means to multiply.The key word "same" in this problem means that I am going to set my two expressions equal to each other.When we set the two expressions equal, we now have an equation with variables on both sides.- Problem 1 A salesman sold twice as much pears in the afternoon than in the morning. If he sold 360 kilograms of pears that day, how many kilograms did he sell in the morning and how many in the afternoon?
- Problem 2 Mary, Peter, and Lucy were picking chestnuts. Mary picked twice as much chestnuts than Peter. Lucy picked 2 kg more than Peter. Together the three of them picked 26 kg of chestnuts.
- Problem 3. Sophia finished $\frac{2}{3}$ of a book. She calculated that she finished 90 more pages than she has yet to read. How long is her book?
- Problem 4. A farming field can be ploughed by 6 tractors in 4 days. When 6 tractors work together, each of them ploughs 120 hectares a day. If two of the tractors were moved to another field, then the remaining 4 tractors could plough the same field in 5 days.
Symbolab is the best calculator for solving a wide range of word problems, including age problems, distance problems, cost problems, investments problems, number problems, and percent problems. An age problem is a type of word problem in math that involves calculating the age of one or more people at a specific point in time.
- To solve word problems start by reading the problem carefully and understanding what it's asking. Try underlining or highlighting key information,...
- Word problems in math can be identified by the use of language that describes a situation or scenario. Word problems often use words and phrases wh...
- An age problem is a type of word problem in math that involves calculating the age of one or more people at a specific point in time. These problem...
Learn to solve word problems. This is a collection of word problem solvers that solve your problems and help you understand the solutions. All problems are customizable (meaning that you can change all parameters). We try to have a comprehensive collection of school algebra problems. The good news is that the steps to solve word problems are ...
Solving Word Problems. Your school is sponsoring a pancake dinner to raise money for a field trip. You estimate that 200 adults and 250 children will attend. Let x represent the cost of an adult ticket and y represent the cost of a child ticket. Write an equation that can be used to find what ticket prices to set in order to raise $3800.
People also ask
Can you solve tough algebra word problems?
How do you write an algebra word problem?
How do you solve word problems in math?
What is an example of an algebra word problem?
Are all school algebra problems customizable?
Is there a calculator that can solve word problems?
Dec 19, 2022 · 2. Solve an equation for one variable. If you have only one unknown in your word problem, isolate the variable in your equation and find which number it is equal to. Use the normal rules of algebra to isolate the variable. Remember that you need to keep the equation balanced.
- 77.9K
