Solve $-2(8x-4)<2x+5$: Step-by-Step Solution

by Luna Greco 45 views

Hey guys! Let's dive into solving this inequality problem together. Inequalities might seem a bit tricky at first, but once you get the hang of the steps, they're totally manageable. This article will walk you through each step to find the solution to the inequality βˆ’2(8xβˆ’4)<2x+5-2(8x - 4) < 2x + 5. We'll break it down, so even if you're just starting with algebra, you'll be able to follow along. So, grab your pencils, and let's get started!

Understanding Inequalities

Before we jump into the problem, let's quickly recap what inequalities are. Unlike equations, which show when two expressions are equal, inequalities show when one expression is greater than, less than, greater than or equal to, or less than or equal to another expression. The symbols we use are:

  • < (less than)
  • > (greater than)
  • ≀ (less than or equal to)
  • β‰₯ (greater than or equal to)

The main difference when solving inequalities compared to equations is that when you multiply or divide both sides by a negative number, you need to flip the inequality sign. Keep this in mind as we work through the problem!

Step-by-Step Solution

Now, let’s tackle the inequality βˆ’2(8xβˆ’4)<2x+5-2(8x - 4) < 2x + 5 step by step. Our goal is to isolate x on one side of the inequality to find the solution.

Step 1: Distribute the -2

The first thing we need to do is get rid of the parentheses. We do this by distributing the -2 across the terms inside the parentheses:

βˆ’2βˆ—(8x)+(βˆ’2)βˆ—(βˆ’4)<2x+5-2 * (8x) + (-2) * (-4) < 2x + 5

This simplifies to:

βˆ’16x+8<2x+5-16x + 8 < 2x + 5

So far, so good! We've removed the parentheses and now have a more straightforward inequality to work with.

Step 2: Move the x Terms to One Side

Next, we want to get all the terms with x on one side of the inequality. To do this, let's add 16x to both sides. This will eliminate the -16x on the left side:

βˆ’16x+8+16x<2x+5+16x-16x + 8 + 16x < 2x + 5 + 16x

This simplifies to:

8<18x+58 < 18x + 5

Now, we have all our x terms on the right side, which is a step closer to isolating x.

Step 3: Move the Constant Terms to the Other Side

We need to isolate the x term further, so let's move the constant term (+5) from the right side to the left side. We do this by subtracting 5 from both sides:

8βˆ’5<18x+5βˆ’58 - 5 < 18x + 5 - 5

This simplifies to:

3<18x3 < 18x

We're getting closer! Now, it's just a matter of dividing to get x by itself.

Step 4: Isolate x

To get x by itself, we need to divide both sides of the inequality by 18:

rac{3}{18} < rac{18x}{18}

This simplifies to:

rac{1}{6} < x

Or, we can write it as:

x > rac{1}{6}

And there we have it! We've solved the inequality.

The Solution

The solution to the inequality βˆ’2(8xβˆ’4)<2x+5-2(8x - 4) < 2x + 5 is x > rac{1}{6}. This means that any value of x greater than 1/6 will satisfy the original inequality. Awesome job! You've just tackled an inequality problem like a pro. Remember, the key is to follow the steps carefully and keep track of the inequality sign, especially when multiplying or dividing by a negative number.

Checking the Solution

To make sure we've got the correct solution, it's always a good idea to check our work. We can do this by picking a value for x that is greater than 16\frac{1}{6} and plugging it back into the original inequality. If the inequality holds true, we know our solution is correct.

Let's choose a value for x that's greater than 16\frac{1}{6}. A simple choice is x=1x = 1. Now, let's substitute this value into the original inequality:

βˆ’2(8(1)βˆ’4)<2(1)+5-2(8(1) - 4) < 2(1) + 5

Simplify the expression inside the parentheses:

βˆ’2(8βˆ’4)<2+5-2(8 - 4) < 2 + 5

βˆ’2(4)<7-2(4) < 7

Multiply:

βˆ’8<7-8 < 7

This statement is true! Since -8 is indeed less than 7, our solution x>16x > \frac{1}{6} is correct. This step is crucial because it confirms that we haven't made any mistakes along the way. Checking our solution gives us confidence in our answer and ensures accuracy.

Now, let's think about why this check works. When we solve an inequality, we're essentially finding a range of values that make the inequality true. By choosing a value within that range and substituting it back into the original inequality, we're verifying that the range we found is indeed correct. If the inequality holds true for our chosen value, it's highly likely that our entire solution set is accurate.

However, what if we had made a mistake and our solution was incorrect? In that case, when we substitute a value from our supposed solution set back into the original inequality, it would not hold true. This is why checking our solution is such a valuable step – it allows us to catch errors and correct them before moving on.

In addition to choosing a value within our solution set, it can also be helpful to choose a value outside our solution set and see if the inequality does not hold true. For example, in this case, we could choose a value less than 16\frac{1}{6}, such as x=0x = 0, and substitute it into the original inequality:

βˆ’2(8(0)βˆ’4)<2(0)+5-2(8(0) - 4) < 2(0) + 5

βˆ’2(βˆ’4)<5-2(-4) < 5

8<58 < 5

This statement is false, as 8 is not less than 5. This further confirms that our solution set x>16x > \frac{1}{6} is correct, as values outside this range do not satisfy the inequality.

By consistently checking our solutions, we develop a strong habit of accuracy and attention to detail. This skill is invaluable not only in mathematics but also in many other areas of life. So, remember, always check your solutions whenever possible to ensure you're on the right track!

Common Mistakes to Avoid

Solving inequalities can sometimes be a bit tricky, and there are a few common mistakes that students often make. Being aware of these pitfalls can help you avoid them and ensure you get the correct solution. Let's go over some of the most frequent errors:

  1. Forgetting to Flip the Inequality Sign: This is probably the most common mistake. Remember, when you multiply or divide both sides of an inequality by a negative number, you must flip the direction of the inequality sign. For example, if you have -x < 5, and you multiply both sides by -1, the correct next step is x > -5, not x < -5. Many students forget this crucial step, leading to an incorrect solution.

  2. Incorrectly Distributing Negative Signs: When you have a negative number outside parentheses, like in our original problem, it's essential to distribute the negative sign correctly. Make sure you multiply each term inside the parentheses by the negative number. For instance, βˆ’2(8xβˆ’4)-2(8x - 4) becomes βˆ’16x+8-16x + 8, not βˆ’16xβˆ’8-16x - 8. Pay close attention to the signs when distributing.

  3. Combining Unlike Terms: This is a common mistake in algebra in general, not just with inequalities. Make sure you only combine like terms. For example, you can combine 2x and 16x to get 18x, but you cannot combine 18x with a constant term like 5. Mixing up unlike terms will lead to incorrect simplifications and an incorrect solution.

  4. Arithmetic Errors: Simple arithmetic mistakes can throw off your entire solution. Whether it's adding, subtracting, multiplying, or dividing, double-check your calculations to ensure accuracy. It's easy to make a small mistake, especially when working quickly, so take your time and be meticulous.

  5. Not Checking the Solution: As we discussed earlier, checking your solution is a crucial step in solving inequalities (and equations). Plugging your solution back into the original inequality helps you verify that you haven't made any mistakes. If the inequality doesn't hold true when you substitute your solution, you know you need to go back and find your error.

  6. Misunderstanding the Inequality Symbols: Make sure you understand what each inequality symbol means. < means