Solving Equations: Roots, Solutions, And Textbooks

Hey everyone! Let's dive into some math problems today. We're going to tackle equations, figure out how many solutions they have, and even solve a word problem about textbooks on shelves. Ready? Let's get started!

1. Determining the Number of Roots in Equations

Okay, so the first task is to figure out how many roots (or solutions) each equation has. This is a fundamental concept in algebra, and understanding it is crucial for solving more complex problems. Let's break down each equation.

1) 0x = 10

Understanding the Equation: In this equation, we're essentially asking, "What number, when multiplied by zero, equals 10?" Think about it for a moment. Zero multiplied by any number is always zero. It can never be 10.

The Solution: Therefore, this equation has no solution. There is no value of 'x' that can satisfy this equation. In mathematical terms, we say the solution set is empty.

Why This Matters: This type of equation highlights the importance of understanding the properties of zero in multiplication. Recognizing such equations early on can save you time and effort when solving problems.

2) 8x = 3

Understanding the Equation: Here, we want to find the value of 'x' that, when multiplied by 8, gives us 3.

The Solution: To solve for 'x', we can divide both sides of the equation by 8:

x = 3/8

So, x = 3/8 is the solution to this equation. This equation has one solution.

Why This Matters: This is a straightforward linear equation. These types of equations are very common, and being able to solve them quickly and accurately is a key skill in algebra. Knowing that a linear equation generally has one solution (unless it's a special case like the one above) helps you anticipate the answer.

2. Solving Equations

Now, let’s actually solve a couple of equations. This is where we find the specific value(s) of the variable that make the equation true. Solving equations is a core skill in algebra, and these examples will give you some practice.

1) -5x = 65

The Goal: Our aim is to isolate 'x' on one side of the equation. To do this, we need to get rid of the -5 that's multiplying 'x'.

The Steps: To isolate 'x', divide both sides of the equation by -5:

-5x / -5 = 65 / -5

x = -13

The Solution: Therefore, the solution to the equation is x = -13.

Checking Your Work: It’s always a good idea to check your answer. Substitute -13 back into the original equation:

-5 * (-13) = 65

65 = 65 (This is true, so our solution is correct!)

2) 0.3x - 2.1 = 0

The Goal: Again, we want to isolate 'x'. This time, we have a term being subtracted from '0.3x', so we'll need to address that first.

The Steps: First, add 2.1 to both sides of the equation:

0. 3x - 2.1 + 2.1 = 0 + 2.1

1. 3x = 2.1

Now, divide both sides by 0.3 to isolate 'x':

0.3x / 0.3 = 2.1 / 0.3

x = 7

The Solution: Therefore, the solution to the equation is x = 7.

Checking Your Work: Substitute 7 back into the original equation:

0. 3 * 7 - 2.1 = 0

1. 1 - 2.1 = 0

0 = 0 (This is true, so our solution is correct!)

3. Determining Equivalent Equations

Next, we need to determine if two equations are equivalent. Equivalent equations are equations that have the same solution set. In other words, if you solve both equations, you should get the same value(s) for 'x'.

The Equations: We have two equations:

1. 2x - 3 = x + 5

2. 2(x - 7) = x - 6

Solving the First Equation: Let's solve 2x - 3 = x + 5 for 'x'.

Subtract 'x' from both sides: 2x - x - 3 = x - x + 5 => x - 3 = 5

Add 3 to both sides: x - 3 + 3 = 5 + 3 => x = 8

Solving the Second Equation: Now, let's solve 2(x - 7) = x - 6 for 'x'.

Distribute the 2: 2x - 14 = x - 6

Subtract 'x' from both sides: 2x - x - 14 = x - x - 6 => x - 14 = -6

Add 14 to both sides: x - 14 + 14 = -6 + 14 => x = 8

The Conclusion: Both equations have the same solution, x = 8. Therefore, the equations are equivalent.

Why This Matters: Understanding equivalent equations is important because it allows you to manipulate equations into simpler forms without changing their solutions. This is a powerful tool in algebra.

4. Solving a Word Problem: Textbooks on Shelves

Alright, time for a word problem! Word problems are a great way to apply your algebraic skills to real-world scenarios. This one involves textbooks on shelves.

The Problem: On one shelf, there are three times more textbooks than on another shelf. We need to find out how many textbooks are on each shelf.

Setting up the Equations: Let's use variables to represent the unknowns:

Let 'x' be the number of textbooks on the second shelf.

Then, the number of textbooks on the first shelf is 3x (since it has three times more).

We are missing a crucial piece of information here! We need to know the total number of textbooks on both shelves to solve this problem. Let's assume the total number of textbooks on both shelves is, say, 24.

Creating the Equation: Now we can write an equation representing the total number of textbooks:

x + 3x = 24

Solving the Equation: Combine like terms:

4x = 24

Divide both sides by 4:

x = 6

Finding the Number of Textbooks on Each Shelf: Now that we know 'x', we can find the number of textbooks on each shelf:

Second shelf: x = 6 textbooks

First shelf: 3x = 3 * 6 = 18 textbooks

The Solution: Therefore, there are 6 textbooks on the second shelf and 18 textbooks on the first shelf.

Key Takeaway: The most important skill is to represent the sentence in math language. First, assign variables for unknowns, then build an equation.

Conclusion

So, there you have it! We've solved equations, determined the number of roots, checked for equivalence, and even tackled a word problem. Remember, math is all about practice, so keep working at it, and you'll get better and better. Keep practicing, guys! You've got this!