Solving Equations With Absolute Values And Verifying Solutions With Domain Of Definition

Hey guys! Let's dive into solving equations that involve absolute values. These can seem tricky at first, but with a systematic approach and a little understanding of what absolute value means, you'll be solving them like a pro in no time. We're also going to make sure we verify our solutions by checking them against the domain of definition (ODZ) to ensure they're valid. So, buckle up, and let's get started!

Understanding Absolute Value

Before we jump into the equations, let's quickly recap what absolute value actually means. The absolute value of a number is its distance from zero on the number line. This means it's always non-negative. For example, the absolute value of 5 (written as |5|) is 5, and the absolute value of -5 (written as |-5|) is also 5. This concept is crucial because when we're dealing with equations involving absolute values, we need to consider both the positive and negative possibilities of the expressions inside the absolute value symbols.

When you're faced with absolute value equations, remember that absolute value represents the distance from zero. This key concept helps in understanding why we need to consider both positive and negative scenarios for the expression inside the absolute value bars. For instance, in the equation |x| = 3, x can be either 3 or -3 because both numbers are three units away from zero. Similarly, when dealing with more complex expressions, such as |2x - 1| = 5, you need to consider two separate cases: 2x - 1 = 5 and 2x - 1 = -5. Each case will lead to a different solution, and it’s important to verify both against the original equation. The complexity arises when multiple absolute value expressions are involved, as seen in our first problem. In such cases, you need to identify the critical points where the expressions inside the absolute value change sign. These critical points will help you divide the number line into intervals, and you will need to solve the equation separately for each interval. Remember, the goal is to systematically eliminate the absolute value signs by considering the appropriate sign for the expression inside them in each interval. This process can be a bit tedious, but it ensures you're accounting for all possible solutions. Keep in mind that the domain of definition (ODZ) is particularly important when dealing with equations that might have restrictions, such as square roots or fractions. However, in the case of absolute value equations, the main focus of the ODZ check is to ensure that your solutions satisfy the initial conditions of the equation, especially when dealing with piecewise functions implicitly defined by the absolute value expressions. By carefully considering these aspects, you can confidently tackle absolute value equations and ensure the accuracy of your solutions.

Equation 1: |2x-3|-|3x-4|=|5x-7|

This equation involves multiple absolute values, which means we'll need to consider different cases based on the intervals where the expressions inside the absolute values change signs.

Step 1: Find the Critical Points

First, let's find the critical points where the expressions inside the absolute values equal zero:

• 2x - 3 = 0 => x = 3/2
• 3x - 4 = 0 => x = 4/3
• 5x - 7 = 0 => x = 7/5

These critical points (3/2, 4/3, and 7/5) divide the number line into four intervals:

1. x < 4/3
2. 4/3 ≤ x < 3/2
3. 3/2 ≤ x < 7/5
4. x ≥ 7/5

Step 2: Solve for Each Interval

Now, we'll solve the equation for each interval, considering the signs of the expressions inside the absolute values:

Interval 1: x < 4/3

In this interval, all three expressions (2x-3), (3x-4), and (5x-7) are negative. So, the equation becomes:

-(2x-3) - (-(3x-4)) = -(5x-7)
-2x + 3 + 3x - 4 = -5x + 7
x - 1 = -5x + 7
6x = 8
x = 4/3

However, x = 4/3 is not within the interval x < 4/3, so there's no solution in this interval.

Interval 2: 4/3 ≤ x < 3/2

In this interval, (2x-3) is negative, (3x-4) is non-negative, and (5x-7) is negative. The equation becomes:

-(2x-3) - (3x-4) = -(5x-7)
-2x + 3 - 3x + 4 = -5x + 7
-5x + 7 = -5x + 7

This equation is always true, meaning all x in the interval 4/3 ≤ x < 3/2 are solutions.

Interval 3: 3/2 ≤ x < 7/5

In this interval, (2x-3) is non-negative, (3x-4) is positive, and (5x-7) is negative. The equation becomes:

(2x-3) - (3x-4) = -(5x-7)
2x - 3 - 3x + 4 = -5x + 7
-x + 1 = -5x + 7
4x = 6
x = 3/2

x = 3/2 is within the interval 3/2 ≤ x < 7/5, so it's a solution.

Interval 4: x ≥ 7/5

In this interval, all three expressions are non-negative. The equation becomes:

(2x-3) - (3x-4) = (5x-7)
2x - 3 - 3x + 4 = 5x - 7
-x + 1 = 5x - 7
6x = 8
x = 4/3

However, x = 4/3 is not within the interval x ≥ 7/5, so there's no solution in this interval.

The approach to solving equations with multiple absolute values, as we've seen, involves breaking down the problem into intervals based on the critical points where the expressions inside the absolute value signs change their signs. This method is crucial because it allows us to eliminate the absolute value signs by considering the sign of the expression within each interval. The initial step is to identify these critical points by setting each expression inside the absolute value to zero and solving for x. These points act as boundaries, dividing the number line into distinct intervals where the equation's form changes. For each interval, you need to rewrite the equation without absolute value signs, taking into account whether the expressions inside the absolute value are positive or negative. Remember, if an expression (like 2x - 3) is negative in a given interval, then |2x - 3| is replaced with -(2x - 3). Conversely, if the expression is positive, |2x - 3| is simply replaced with (2x - 3). After rewriting the equation for each interval, you solve it algebraically. The key step is verifying whether the solutions obtained fall within the interval for which they were derived. Solutions that do not belong to their respective intervals must be discarded, as they do not satisfy the original equation under the conditions of that interval. This process needs to be meticulously repeated for each interval to ensure all potential solutions are found and that no extraneous solutions are included. When checking the domain of definition (ODZ) for these types of equations, the primary focus is to confirm that the solutions fit within the intervals they were calculated for. Unlike equations involving fractions or square roots, absolute value equations generally don't have restrictions that would exclude solutions due to undefined operations. Therefore, the ODZ check here is more about verifying the consistency of the solutions with the piecewise definition of the absolute value function. By systematically working through each interval and rigorously checking your solutions, you can confidently solve equations with multiple absolute values.

Step 3: Combine the Solutions

Combining the solutions from all intervals, we get:

• 4/3 ≤ x < 3/2
• x = 3/2

This can be written as 4/3 ≤ x ≤ 3/2.

Step 4: Verification (ODZ)

Since we're dealing with absolute values, there are no restrictions on the domain. However, we need to verify that our solutions satisfy the original equation. The solutions we found are within the intervals we considered, so they are valid.

Solution for Equation 1: 4/3 ≤ x ≤ 3/2

Equation 2: |2x-5|=4-7x

This equation has one absolute value, making it a bit simpler than the first one. We'll still need to consider two cases.

Step 1: Consider Both Cases

We need to consider two cases:

• Case 1: 2x - 5 = 4 - 7x
• Case 2: -(2x - 5) = 4 - 7x

Step 2: Solve Each Case

Case 1: 2x - 5 = 4 - 7x

2x - 5 = 4 - 7x
9x = 9
x = 1

Case 2: -(2x - 5) = 4 - 7x

-2x + 5 = 4 - 7x
5x = -1
x = -1/5

Step 3: Verification (ODZ)

Again, there are no restrictions on the domain due to the absolute value. However, we need to make sure our solutions satisfy the original equation. We need to check if the right-hand side (4-7x) is non-negative because the absolute value is always non-negative.

• For x = 1: 4 - 7(1) = -3 (Negative). This solution is extraneous.
• For x = -1/5: 4 - 7(-1/5) = 4 + 7/5 = 27/5 (Positive). This solution is valid.

When approaching an equation with a single absolute value, the strategy revolves around understanding the fundamental property of absolute value: it turns any expression into its non-negative counterpart. This is why the first crucial step is to split the problem into two separate cases. The first case considers the scenario where the expression inside the absolute value bars is non-negative, meaning the absolute value does not change its sign. For example, if you have |2x - 5|, the first case assumes that 2x - 5 is greater than or equal to zero, and you solve the equation 2x - 5 = the other side of the equation. The second case deals with the scenario where the expression inside the absolute value is negative. In this situation, the absolute value changes the sign of the expression, so you solve -(2x - 5) = the other side of the equation. After solving both cases, you obtain potential solutions. However, the process doesn't end there. It's imperative to verify these solutions by substituting them back into the original equation. This step is crucial because the process of splitting the equation into cases can sometimes introduce extraneous solutions—solutions that satisfy the transformed equations but not the original one. The verification step helps to filter out these extraneous solutions. Checking the domain of definition (ODZ) is particularly important in equations where the absolute value is set equal to an expression that could potentially be negative, as the absolute value itself is always non-negative. In such cases, you must ensure that the expression on the other side of the equation is also non-negative for the solution to be valid. For instance, in the equation |2x - 5| = 4 - 7x, you need to check whether 4 - 7x is non-negative for your solutions. If a solution makes 4 - 7x negative, it is an extraneous solution and must be discarded. By meticulously working through each case, solving for the potential solutions, and then rigorously verifying these solutions against the original equation and the ODZ, you can accurately solve equations involving a single absolute value.

Solution for Equation 2: x = -1/5

Equation 3: |x+1|+|5-x|=2

This equation, like the first one, involves multiple absolute values. We'll follow a similar process of finding critical points and considering intervals.

Step 1: Find the Critical Points

• x + 1 = 0 => x = -1
• 5 - x = 0 => x = 5

These critical points divide the number line into three intervals:

1. x < -1
2. -1 ≤ x < 5
3. x ≥ 5

Step 2: Solve for Each Interval

Interval 1: x < -1

In this interval, (x+1) is negative and (5-x) is positive. The equation becomes:

-(x+1) + (5-x) = 2
-x - 1 + 5 - x = 2
-2x + 4 = 2
-2x = -2
x = 1

However, x = 1 is not within the interval x < -1, so there's no solution in this interval.

Interval 2: -1 ≤ x < 5

In this interval, (x+1) is non-negative and (5-x) is positive. The equation becomes:

(x+1) + (5-x) = 2
x + 1 + 5 - x = 2
6 = 2

This equation is never true, so there's no solution in this interval.

Interval 3: x ≥ 5

In this interval, (x+1) is positive and (5-x) is negative. The equation becomes:

(x+1) - (5-x) = 2
x + 1 - 5 + x = 2
2x - 4 = 2
2x = 6
x = 3

However, x = 3 is not within the interval x ≥ 5, so there's no solution in this interval.

When dealing with equations involving multiple absolute values, the approach we use is similar to that of the first equation but can sometimes lead to unexpected outcomes, as we see in this case. The fundamental strategy remains the same: identify the critical points, divide the number line into intervals, and solve the equation separately for each interval. However, in some instances, the algebraic manipulation might lead to contradictions, indicating that there are no solutions within certain intervals or even no solutions at all for the entire equation. This is a crucial aspect of solving absolute value equations – being able to recognize when an equation has no solution. In this specific scenario, after setting up the cases based on the intervals defined by the critical points, we encounter equations that simplify to contradictions. For example, in the second interval (-1 ≤ x < 5), the equation simplifies to 6 = 2, which is a clear falsehood. This tells us definitively that there are no values of x within this interval that can satisfy the original equation. Similarly, if solving the equation in another interval leads to a solution that does not fall within that interval, we conclude that there is no solution for that case either. The absence of solutions can stem from the specific constants and coefficients in the equation, which, when combined with the piecewise nature of the absolute value function, might create conditions that are mathematically impossible to fulfill. The domain of definition (ODZ) plays a subtle but important role here. While absolute value equations themselves do not impose direct restrictions on the domain in the way that, say, square roots or fractions do, the ODZ check helps ensure that the solutions we find are consistent with the conditions under which we derived them. In other words, it re-emphasizes the importance of verifying that the solutions belong to the intervals they were intended for. In cases where contradictions arise, the ODZ check further validates that no solutions exist, as any purported solution would necessarily violate the conditions defined by the intervals. Therefore, when solving equations with multiple absolute values, it's essential to be alert for contradictions and to rigorously check that any solutions obtained are not only algebraically correct but also logically consistent within the context of the original equation and its interval-based cases. This careful approach helps in accurately determining the solution set, which can sometimes be an empty set.

Step 3: Combine the Solutions

There are no solutions in any of the intervals.

Step 4: Verification (ODZ)

Since there are no solutions, there's nothing to verify.

Solution for Equation 3: No Solution

Conclusion

So, there you have it! We've tackled three equations with absolute values, making sure to verify our solutions using the domain of definition. Remember, the key to solving these equations is to break them down into cases based on the critical points and to always verify your solutions. Keep practicing, and you'll become a master of absolute value equations in no time! Let me know if you guys have any other math challenges you'd like to explore!