Solving 3|2x-1| + 5x ≤ 5|x+5| + 2: A Math Discussion
Hey guys! Today, we're diving deep into solving the inequality 3|2x-1| + 5x ≤ 5|x+5| + 2. This type of problem might seem a bit intimidating at first glance, with those absolute value signs staring back at you. But don't worry, we're going to break it down step by step, making it super easy to understand. So, grab your pencils and let's get started!
Understanding Absolute Values
Before we jump into the main problem, let's quickly recap what absolute value means. The absolute value of a number is its distance from zero. So, |x| is x if x is positive or zero, and it's -x if x is negative. This simple concept is crucial for tackling inequalities with absolute values.
When dealing with absolute values in inequalities, the key is to consider different cases based on the expressions inside the absolute value symbols. In our case, we have two absolute value expressions: |2x-1| and |x+5|. We need to figure out when these expressions are positive, negative, or zero. This will help us break down the problem into manageable parts.
For |2x-1|, the expression inside becomes zero when 2x-1 = 0, which means x = 1/2. So, we have two cases here: one where x is less than 1/2 and another where x is greater than or equal to 1/2. Similarly, for |x+5|, the expression becomes zero when x+5 = 0, meaning x = -5. This gives us two more cases: x less than -5 and x greater than or equal to -5.
By considering these critical points (x = 1/2 and x = -5), we can divide the number line into intervals and analyze the inequality in each interval separately. This is a standard technique for solving inequalities involving absolute values, and it’s important to get comfortable with this approach. Remember, the goal is to eliminate the absolute value signs by considering the different scenarios they create.
Breaking Down the Problem into Cases
To solve the inequality 3|2x-1| + 5x ≤ 5|x+5| + 2, we need to consider the critical points where the expressions inside the absolute value signs change their signs. These points are x = 1/2 (from 2x - 1 = 0) and x = -5 (from x + 5 = 0). These points divide the number line into three intervals:
- x < -5
- -5 ≤ x < 1/2
- x ≥ 1/2
We'll analyze the inequality in each of these intervals separately. This approach allows us to eliminate the absolute value signs by determining the sign of the expressions inside them for each interval. Remember, this is a crucial step in solving inequalities with absolute values because it transforms the problem into simpler algebraic inequalities.
Case 1: x < -5
When x < -5, both (2x - 1) and (x + 5) are negative. Therefore, |2x - 1| = -(2x - 1) and |x + 5| = -(x + 5). Substituting these into the original inequality gives us:
3(-(2x - 1)) + 5x ≤ 5(-(x + 5)) + 2
Simplifying this inequality, we get:
-6x + 3 + 5x ≤ -5x - 25 + 2
-x + 3 ≤ -5x - 23
4x ≤ -26
x ≤ -26/4
x ≤ -13/2
So, in this case, we have x ≤ -13/2. Since we are considering x < -5, the solution for this interval is x ≤ -13/2, which means x is less than or equal to -6.5. This is the first piece of our solution, and it’s important to remember that this is only valid for the interval x < -5. We’ll combine these solutions at the end to get the overall solution set.
Case 2: -5 ≤ x < 1/2
In this interval, (2x - 1) is negative, but (x + 5) is non-negative (positive or zero). So, |2x - 1| = -(2x - 1) and |x + 5| = (x + 5). Substituting these into the original inequality:
3(-(2x - 1)) + 5x ≤ 5(x + 5) + 2
Simplifying, we get:
-6x + 3 + 5x ≤ 5x + 25 + 2
-x + 3 ≤ 5x + 27
-6x ≤ 24
x ≥ -4
Here, we have x ≥ -4. Considering our interval -5 ≤ x < 1/2, the solution for this case is -4 ≤ x < 1/2. This means that any value of x in this range satisfies the inequality for this specific interval. Remember, we're piecing together the solution by looking at each interval separately, so it’s crucial to keep track of the interval boundaries.
Case 3: x ≥ 1/2
When x ≥ 1/2, both (2x - 1) and (x + 5) are non-negative. Thus, |2x - 1| = (2x - 1) and |x + 5| = (x + 5). Substituting these into the original inequality:
3(2x - 1) + 5x ≤ 5(x + 5) + 2
Simplifying, we get:
6x - 3 + 5x ≤ 5x + 25 + 2
11x - 3 ≤ 5x + 27
6x ≤ 30
x ≤ 5
In this case, we have x ≤ 5. Since we are considering x ≥ 1/2, the solution for this interval is 1/2 ≤ x ≤ 5. This is the final piece of the puzzle, and now we need to combine all the pieces to get the complete solution.
Combining the Solutions
Now, let's combine the solutions from all three cases:
- Case 1: x < -5, solution is x ≤ -13/2
- Case 2: -5 ≤ x < 1/2, solution is -4 ≤ x < 1/2
- Case 3: x ≥ 1/2, solution is 1/2 ≤ x ≤ 5
Combining these, we get the overall solution: x ≤ -13/2 or -4 ≤ x ≤ 5. This means that any value of x within these ranges will satisfy the original inequality. It's always a good idea to check some values within these ranges in the original inequality to ensure the solution is correct.
To express this in interval notation, the solution is (-∞, -13/2] ∪ [-4, 5]. This notation clearly shows the range of values that satisfy the inequality, including the endpoints.
Final Thoughts
Solving inequalities with absolute values can seem tricky, but by breaking the problem into cases based on the intervals created by the critical points, it becomes much more manageable. Always remember to consider the cases where the expressions inside the absolute value signs change signs. And don't forget to combine the solutions from each case to get the final solution set!
I hope this detailed explanation helps you guys understand how to tackle these types of problems. Keep practicing, and you'll become a pro in no time! If you have any questions or want to discuss more math problems, feel free to drop a comment below. Happy solving!