Solve $-15x > 75$: A Simple Guide

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Unlock the Mystery: Solving βˆ’15x>75-15x > 75

Ever stared at an inequality like βˆ’15x>75-15x > 75 and wondered, "What in the world is the solution?" You're not alone! Many people find inequalities a bit tricky, especially when a negative number is involved. But fear not, because solving this type of problem is actually quite straightforward once you understand the basic rules. We're going to dive deep into the world of linear inequalities and break down exactly how to conquer βˆ’15x>75-15x > 75. Get ready to impress yourself with your math skills!

The Core Concept: Isolating the Variable

At its heart, solving any inequality, including βˆ’15x>75-15x > 75, is all about isolating the variable, which in this case is 'x'. Our goal is to get 'x' all by itself on one side of the inequality sign. To do this, we use inverse operations, just like we do when solving equations. The key difference with inequalities is that we have to be extra careful when multiplying or dividing by a negative number. Let's start with our inequality: βˆ’15x>75-15x > 75. To get 'x' by itself, we need to undo the multiplication by βˆ’15-15. The inverse operation of multiplication is division. So, we'll divide both sides of the inequality by βˆ’15-15.

The Crucial Rule: Flipping the Inequality Sign

Now, here's where things get interesting and where many people make mistakes. When you multiply or divide both sides of an inequality by a negative number, you MUST flip the direction of the inequality sign. Think of it like this: if you have a seesaw that's tilted one way, and you suddenly apply a force that pushes it down on the other side, it's going to flip and go the other way. The same principle applies here. So, when we divide both sides of βˆ’15x>75-15x > 75 by βˆ’15-15, our "greater than" sign (>>) needs to become a "less than" sign (<<). This is a fundamental rule that cannot be ignored if you want the correct solution.

Step-by-Step Solution

Let's walk through the process for βˆ’15x>75-15x > 75:

  1. Start with the inequality: βˆ’15x>75-15x > 75

  2. Divide both sides by βˆ’15-15: rac{-15x}{-15} rac{75}{-15}

  3. Apply the crucial rule: Flip the inequality sign. Since we divided by a negative number, the '>>' becomes '<<'. x < rac{75}{-15}

  4. Simplify the right side: x<βˆ’5x < -5

And there you have it! The solution to the inequality βˆ’15x>75-15x > 75 is x<βˆ’5x < -5. This means any number that is less than βˆ’5-5 will make the original inequality true. For example, if we plug in βˆ’6-6 for 'x', we get βˆ’15(βˆ’6)=90-15(-6) = 90. Is 90>7590 > 75? Yes, it is! Now, try a number greater than βˆ’5-5, like βˆ’4-4. We get βˆ’15(βˆ’4)=60-15(-4) = 60. Is 60>7560 > 75? No, it's not. This confirms our solution.

Why Does Flipping the Sign Matter?

Let's think about this rule a bit more. Imagine we have the inequality 2<52 < 5. This is true. Now, let's multiply both sides by βˆ’1-1. If we don't flip the sign, we'd get βˆ’2<βˆ’5-2 < -5, which is false. However, if we do flip the sign, we get βˆ’2>βˆ’5-2 > -5, which is true! This little experiment clearly demonstrates why flipping the inequality sign when multiplying or dividing by a negative is so important. It preserves the truth of the statement.

Understanding the Options

Now, let's look at the options provided for the inequality βˆ’15x>75-15x > 75:

A. x<βˆ’5x < -5 B. x>βˆ’5x > -5 C. xeqβˆ’5x eq -5 D. xeqβˆ’5x eq -5

Based on our step-by-step solution, the correct answer is A. x<βˆ’5x < -5. This option perfectly matches our derived solution.

Conclusion

Solving linear inequalities like βˆ’15x>75-15x > 75 might seem intimidating at first, but by understanding the fundamental rules, especially the one about flipping the inequality sign when dealing with negative numbers, you can confidently solve them. Remember, the goal is always to isolate the variable. Keep practicing, and soon these types of problems will feel like second nature. For more in-depth exploration of inequalities and algebraic concepts, check out resources like Khan Academy for excellent tutorials and practice exercises.