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Understanding Word Problems

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Teacher
Teacher

Today, weโ€™re going to learn how to solve word problems using algebra. Can anyone tell me what a word problem is?

Student 1
Student 1

Is it a math problem that is written in words instead of numbers?

Teacher
Teacher

Exactly! Word problems present a scenario and require us to translate that into a mathematical equation. For example, โ€˜three times a number plus five equals twenty.โ€™ Can someone express this in equation form?

Student 2
Student 2

It would be written as 3x + 5 = 20, right?

Teacher
Teacher

Absolutely! In this case, 'x' represents the unknown number we want to find. Now, letโ€™s discuss the steps to isolate x.

Solving the Equation

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Teacher
Teacher

To solve the equation 3x + 5 = 20, we need to isolate x. Whatโ€™s the first step?

Student 3
Student 3

I think we need to subtract 5 from both sides to get rid of the 5.

Teacher
Teacher

Correct! So, when we subtract 5, what do we have?

Student 4
Student 4

We get 3x = 15.

Teacher
Teacher

Great! Now, we need to divide both sides by 3. What do we get?

Student 1
Student 1

x = 5!

Teacher
Teacher

Fantastic! So, the number we were looking for is 5. Remember this process of isolating variables; itโ€™s key for solving word problems.

Real-World Application of Word Problems

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Teacher
Teacher

Now, letโ€™s think about why mastering word problems is important. Can anyone give me a real-life example that could be modeled with a word problem?

Student 2
Student 2

Maybe something like calculating the cost of tickets for a concert?

Teacher
Teacher

Yes! If a ticket costs $10 and you want to buy several tickets, you can represent the total cost with an equation. For instance, if you want to buy x tickets, the equation would be 10x.

Student 3
Student 3

And if I want to spend $100, then it would be 10x = 100?

Teacher
Teacher

Exactly! You can solve for x to find out how many tickets you can buy. This kind of setup and calculation is very common in everyday scenarios.

Introduction & Overview

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Quick Overview

This section illustrates how to solve word problems using algebraic equations.

Standard

The section focuses on the application of algebraic concepts to solve word problems, specifically focusing on the process of setting up and solving linear equations derived from the problem statement.

Detailed

In this section, we delve into the strategy of solving word problems using algebraic expressions and equations. Linear equations often represent word problems, enabling us to derive the unknown through systematic steps. Understanding the structure of a word problem helps in translating the words into mathematical statements, thereby allowing algebraic manipulation to find solutions. We explore a practical example: finding a number that satisfies the equation built from a narrative, emphasizing the importance of isolating variables and using inverse operations to achieve the correct answer.

Audio Book

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Understanding the Word Problem

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"If 3 times a number plus 5 equals 20, find the number"

Detailed Explanation

This statement presents a problem where we need to find an unknown number. The phrase '3 times a number' suggests that we multiply a variable, usually represented by 'x'. The equation is constructed as follows: 3x + 5 = 20. Here, '3x' represents three times the unknown number, and '+ 5' indicates we add 5 to it. We want to find the value of 'x' such that the equation is true.

Examples & Analogies

Think of it like a treasure hunt. You have a treasure chest (your number), and you know that if you multiply the number of treasures by 3 and then add 5 other items, you will have a total of 20 items in the chest. Your job is to figure out how many treasures you originally had!

Setting Up the Equation

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3x + 5 = 20

Detailed Explanation

In this equation, we are expressing the relationship given in the word problem mathematically. The left side, 3x + 5, must equal the right side, which is 20. We will use algebraic methods to manipulate this equation to find the value of 'x'.

Examples & Analogies

Imagine you're trying to balance a scale. On one side, you have 3 times the number of candies you have (plus 5 extra candies), and on the other side, you know you need to have exactly 20 candies to balance. You need to find out how many candies you originally had on your side.

Isolating the Variable

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First, subtract 5 from both sides: 3x + 5 - 5 = 20 - 5

Detailed Explanation

To isolate the variable 'x', you need to remove the '+ 5' by performing the same operation on both sides of the equation. Subtracting 5 from both sides gives us: 3x = 15. This process is known as maintaining the balance of the equationโ€”it ensures that what we do on one side must be done on the other to keep the equation true.

Examples & Analogies

Think of sharing a cake equally; if you take away the same amount from both sides, the portions remain equal. Here, by taking away 5 from both sides, you're simplifying the problem to find out how many times 'x' fits into the remaining total.

Finding the Value of x

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Now, divide both sides by 3: x = 15 / 3

Detailed Explanation

After simplifying the equation to 3x = 15, we can find the value of 'x' by dividing both sides by 3. This gives us x = 5. By dividing, we are figuring out how many times 3 fits into 15, thereby solving for 'x'.

Examples & Analogies

Imagine you have 15 slices of pizza, and you need to share them evenly among 3 friends (you included). Dividing the total number of slices by 3 gives you how much pizza each person gets; in this case, 5 slices each. Similarly, here we are dividing to find out the original amount.

Conclusion of the Problem

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The solution is x = 5

Detailed Explanation

At this point, we have solved the equation and found that the number we were looking for is 5. This means that if you take 3 times 5, you get 15, and when you add 5, you end up with 20, which confirms our solution.

Examples & Analogies

So, if you have found that you ate 5 pieces of chocolate out of a larger box, and your friend added 5 more pieces, it should feel complete when you look back and see that you ended up with a total of 20 pieces in the box!

Definitions & Key Concepts

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Key Concepts

  • Word Problems: Scenarios described in words that can be converted into algebraic expressions.

  • Linear Equations: Equations of the form ax + b = c, used to represent relationships.

  • Isolating Variables: The process of rearranging an equation to solve for a specific variable.

Examples & Real-Life Applications

See how the concepts apply in real-world scenarios to understand their practical implications.

Examples

  • If three times a number plus five equals twenty, the equation can be represented as 3x + 5 = 20.

  • To find how many tickets you can buy for $100 if each ticket costs $10, set up the equation 10x = 100.

Memory Aids

Use mnemonics, acronyms, or visual cues to help remember key information more easily.

๐ŸŽต Rhymes Time

  • When a word problem comes your way, set it up right and solve today!

๐Ÿ“– Fascinating Stories

  • Imagine a baker with x cookies, who sold 5. If he has 20 left, how many did he bake in the first place?

๐Ÿง  Other Memory Gems

  • W.R.S. - Word-Represents-Solve. First, recognize that itโ€™s a word problem, represent it as an equation, and then solve.

๐ŸŽฏ Super Acronyms

S.I.S.T. - Set Up the equation, Isolate the variable, Solve for the variable, and Thoroughly check your answer.

Flash Cards

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Glossary of Terms

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  • Term: Word Problem

    Definition:

    A mathematical problem expressed in words, requiring translation into an algebraic expression or equation.

  • Term: Variable

    Definition:

    A symbol (often a letter) used to represent an unknown value in mathematical equations.

  • Term: Linear Equation

    Definition:

    An equation that makes a straight line when graphed, often in the form ax + b = c.

  • Term: Isolate

    Definition:

    To manipulate an equation to solve for one variable, typically by getting the variable alone on one side.