Solving Absolute Value Equations Find X In 4|x-5| = 10
Introduction: Understanding Absolute Value Equations
Absolute value equations, like the one presented—(f) 4|x-5| = 10—are a fascinating area of mathematics that often puzzles students. To truly grasp these equations, it's crucial to first understand the concept of absolute value itself. The absolute value of a number represents its distance from zero on the number line, irrespective of direction. This means that absolute value is always non-negative. For instance, the absolute value of 5, written as |5|, is 5, and the absolute value of -5, written as |-5|, is also 5. This fundamental property is what makes solving absolute value equations unique and sometimes tricky.
The equation at hand, 4|x-5| = 10, is a classic example of an absolute value problem. It asks us to find all values of x that make the equation true. The presence of the absolute value bars around x-5 indicates that we are interested in the distance between x and 5. The equation essentially states that four times this distance is equal to 10. Before we dive into the step-by-step solution, it’s vital to appreciate that absolute value equations typically have two possible solutions, stemming from the dual nature of distance – a number can be a certain distance away from zero in either the positive or the negative direction. This is a core concept that we will repeatedly use throughout the solution process. Ignoring this duality can lead to incomplete or incorrect answers, which is why a thorough understanding of the underlying principles is so important. As we proceed, we will not only find the solutions but also discuss the rationale behind each step, ensuring a comprehensive understanding of the methods involved in tackling such equations.
Step 1: Isolating the Absolute Value
In the equation 4|x-5| = 10, our initial goal is to isolate the absolute value expression, which in this case is |x-5|. Isolating the absolute value means getting it all by itself on one side of the equation. This is a critical first step because it sets the stage for splitting the equation into two separate cases, which we'll discuss in the next section. To isolate |x-5|, we need to undo the multiplication by 4 that is currently affecting it. We accomplish this by applying the inverse operation, which is division. Dividing both sides of the equation by 4 maintains the balance of the equation, ensuring that we're performing a valid algebraic manipulation. This step is rooted in the fundamental principle that what you do to one side of an equation, you must also do to the other side to preserve equality. It's a bedrock principle in algebra and is used extensively in solving various types of equations. By dividing both sides by 4, we transform the equation from 4|x-5| = 10 into a simpler form, which directly highlights the absolute value expression we need to address. The resulting equation becomes |x-5| = 10/4, which simplifies further to |x-5| = 5/2. Now that the absolute value expression is isolated, we can clearly see that the distance between x and 5 must be equal to 5/2. This sets us up perfectly for the next crucial step in solving absolute value equations: splitting the equation into two cases based on the possible positive and negative values within the absolute value.
Step 2: Splitting the Equation into Two Cases
Once the absolute value expression is isolated, as we achieved with |x-5| = 5/2, the next critical step is to split the equation into two distinct cases. This is necessary because the absolute value of a quantity can be the same whether the quantity itself is positive or negative. In other words, if |A| = B, then A could be either B or -B. This duality is the essence of why absolute value equations require a bifurcated approach. Applying this principle to our equation, |x-5| = 5/2, we recognize that the expression inside the absolute value, x-5, could be either 5/2 or -5/2. This gives rise to our two cases:
- Case 1: x - 5 = 5/2 This case considers the scenario where the expression x-5 is positive or zero and is equal to the positive value of 5/2. It essentially says that x is 5/2 units away from 5 in the positive direction.
- Case 2: x - 5 = -5/2 This case addresses the situation where the expression x-5 is negative but its absolute value is still 5/2. It means that x is 5/2 units away from 5 in the negative direction.
By splitting the equation into these two cases, we ensure that we account for all possible values of x that could satisfy the original absolute value equation. Failing to consider both cases is a common mistake that leads to incomplete solutions. Each of these cases now represents a simple linear equation that can be solved independently. The next step involves solving each of these equations to find the potential values of x. This methodical approach guarantees that we capture the complete solution set for the absolute value equation.
Step 3: Solving Case 1 (x - 5 = 5/2)
Having established the two cases arising from our absolute value equation, we now focus on solving Case 1, which is x - 5 = 5/2. This equation is a straightforward linear equation, and our goal is to isolate x on one side to determine its value. To do this, we need to undo the subtraction of 5 from x. The inverse operation of subtraction is addition, so we will add 5 to both sides of the equation. Adding the same value to both sides ensures that the equation remains balanced, adhering to a fundamental principle of algebraic manipulation. When we add 5 to both sides, the equation transforms as follows:
x - 5 + 5 = 5/2 + 5
On the left side, the -5 and +5 cancel each other out, leaving just x. On the right side, we need to add 5/2 and 5. To do this, we first express 5 as a fraction with a denominator of 2, which gives us 10/2. Now we can add the two fractions:
5/2 + 10/2 = 15/2
Thus, the equation simplifies to:
x = 15/2
This tells us that one possible solution for x is 15/2, or 7.5 in decimal form. This value represents a number that, when substituted into the original absolute value equation, will make the equation true, provided that the first case holds. However, to have a complete solution, we must also consider the second case. Solving each case independently and systematically is crucial in handling absolute value equations, ensuring that we capture all possible solutions. Now, we proceed to solve the second case to find the other potential value of x.
Step 4: Solving Case 2 (x - 5 = -5/2)
Having solved Case 1 and found one potential solution for x, we now turn our attention to Case 2, which is x - 5 = -5/2. This equation, similar to Case 1, is a linear equation, and our objective remains to isolate x to determine its value. The approach we take here mirrors that of Case 1: we need to undo the subtraction of 5 from x. As before, we accomplish this by adding 5 to both sides of the equation, maintaining the balance and integrity of the equation. Adding 5 to both sides gives us:
x - 5 + 5 = -5/2 + 5
On the left side, -5 and +5 cancel out, leaving x. On the right side, we need to add -5/2 and 5. To do this, we again express 5 as a fraction with a denominator of 2, which is 10/2. Now we can add the fractions:
-5/2 + 10/2 = 5/2
This simplifies our equation to:
x = 5/2
Thus, the second possible solution for x is 5/2, or 2.5 in decimal form. This value represents another number that, when substituted into the original absolute value equation, will make the equation true, provided that the conditions of the second case are met. With both cases now solved, we have two potential solutions for x: 15/2 and 5/2. The next crucial step is to verify these solutions to ensure they satisfy the original equation. This verification process is a vital safeguard against errors and ensures the accuracy of our solutions.
Step 5: Verifying the Solutions
After solving the two cases of an absolute value equation, it's essential to verify the solutions. This step is not just a formality; it's a critical part of the solution process that ensures the values we've found actually satisfy the original equation. By verifying, we can catch any potential errors made during the solving process, such as incorrect algebraic manipulations or misinterpretations of the absolute value properties. For our equation, 4|x-5| = 10, we found two potential solutions: x = 15/2 and x = 5/2. To verify these, we will substitute each value back into the original equation and check if the equation holds true.
Verification for x = 15/2:
Substitute x = 15/2 into the equation 4|x-5| = 10:
4|15/2 - 5| = 10
First, we need to evaluate the expression inside the absolute value. We rewrite 5 as 10/2 to have a common denominator:
15/2 - 10/2 = 5/2
Now, substitute this back into the equation:
4|5/2| = 10
The absolute value of 5/2 is 5/2, so we have:
4 * (5/2) = 10
Multiplying 4 by 5/2 gives us:
20/2 = 10
Which simplifies to:
10 = 10
Since the equation holds true, x = 15/2 is indeed a valid solution.
Verification for x = 5/2:
Now, we substitute x = 5/2 into the equation 4|x-5| = 10:
4|5/2 - 5| = 10
Again, we need to evaluate the expression inside the absolute value. We rewrite 5 as 10/2:
5/2 - 10/2 = -5/2
Substitute this back into the equation:
4|-5/2| = 10
The absolute value of -5/2 is 5/2, so we have:
4 * (5/2) = 10
Multiplying 4 by 5/2 gives us:
20/2 = 10
Which simplifies to:
10 = 10
Since the equation holds true, x = 5/2 is also a valid solution.
Both values, x = 15/2 and x = 5/2, satisfy the original equation. This verification process confirms the accuracy of our solution and reinforces the importance of this step in solving absolute value equations.
Final Answer: The Solutions to 4|x-5| = 10
Having meticulously worked through the process of solving the absolute value equation 4|x-5| = 10, we have arrived at the final and crucial step: presenting the solutions. We began by understanding the fundamental principles of absolute value, then methodically isolated the absolute value expression, split the equation into two distinct cases, solved each case independently, and rigorously verified our potential solutions. This comprehensive approach ensures that we have not only found the correct answers but also thoroughly understood the underlying concepts.
Our step-by-step solution revealed two values for x that satisfy the equation. In Case 1, we found that x = 15/2. In Case 2, we determined that x = 5/2. The verification process confirmed that both of these values indeed make the original equation true. Therefore, the solutions to the equation 4|x-5| = 10 are x = 15/2 and x = 5/2. These are the only two values of x that, when substituted into the equation, will result in a true statement.
These solutions can also be expressed in decimal form for easier interpretation. x = 15/2 is equivalent to x = 7.5, and x = 5/2 is equivalent to x = 2.5. Thus, the solutions can also be stated as x = 7.5 and x = 2.5. Regardless of the format, the key is that we have identified the two precise values that fulfill the conditions of the absolute value equation.
In summary, the solutions to the absolute value equation 4|x-5| = 10 are:
- x = 15/2 or 7.5
- x = 5/2 or 2.5
This concludes our comprehensive solution, providing a clear and verified answer to the problem.
