Problem 49 - Intermediate Algebra

Question

Solve each equation. See Example 5. $$ |10 x|=|x-18| $$

Step-by-Step Solution

Answer

The solutions are \(x = -2\) and \(x = \frac{18}{11}\).

Step 1: Understand the Absolute Values

The equation \(|10x| = |x-18|\) contains two absolute values. An absolute value \(|a|\) represents the distance of \(a\) from 0, making it always non-negative. Therefore, this equation states that the distance of \(10x\) from 0 is the same as the distance of \(x - 18\) from 0.

Step 2: Set up Possible Equations

For two expressions \(|A| = |B|\), there are two possible cases. Case 1: \(A = B\) and Case 2: \(A = -B\). Here, we have two equations:1. \(10x = x - 18\)2. \(10x = -(x - 18)\)

Step 3: Solve Case 1: \(10x = x - 18\)

Rearrange the equation \(10x = x - 18\): Subtract \(x\) from both sides:\(10x - x = -18\)Simplify:\(9x = -18\)Divide both sides by 9:\(x = -2\)

Step 4: Solve Case 2: \(10x = -(x - 18)\)

Simplify the second case:Distribute the negative sign:\(10x = -x + 18\)Add \(x\) to both sides:\(10x + x = 18\)Simplify:\(11x = 18\)Divide both sides by 11:\(x = \frac{18}{11}\)

Step 5: Verify Solutions

Verify each solution by substituting back into the original equation.For \(x = -2\):\(|10(-2)| = |-2 - 18| \implies 20 = 20\) which is true.For \(x = \frac{18}{11}\):Calculate \(|10 \times \frac{18}{11}| \approx 16.36\) and \(|\frac{18}{11} - 18| \approx 16.36\) which are approximately equal. Thus, \(x = \frac{18}{11}\) is correct.

Key Concepts

Solving Linear Equations

Solving linear equations is an essential skill in algebra. It involves finding the value of the variable that makes the equation true. Linear equations are often in the form \(Ax + B = C\), where \(x\) is the variable, and \(A\), \(B\), and \(C\) are constants. To solve them, we use techniques such as:

• Adding or subtracting terms from both sides to isolate the variable.
• Multiplying or dividing both sides to solve for the variable.

When dealing with equations involving absolute values, like \(|10x| = |x-18|\), we need to consider two cases: when both expressions are equal and when one is the negative of the other.

Distance from Zero

The concept of distance from zero is what absolute values represent. For any real number \(a\), the absolute value \(|a|\) is the non-negative distance on the number line from \(a\) to zero. So, absolute value equations like \(|10x| = |x-18|\) indicate that both expressions are equidistant from zero. To understand better, let's consider:

• If \(a = b\), both \(a\) and \(b\) are at the same distance from zero, potentially in opposite directions.
• If \(a = -b\), they mirror each other across zero.

This interpretation allows us to set up equations that consider both situations. Keeping in mind the non-negative nature of absolute values helps simplify many seemingly complex problems.

Verification of Solutions

Once solutions to an equation are found, it's crucial to verify them. Verification ensures that the solutions satisfy the original equation. For absolute value equations, like our example, check each solution by substituting back into the initial equation \(|10x| = |x-18|\). For instance:- If \(x = -2\), substitute back: \(|10(-2)| = |-2 - 18|\), resulting in \(20 = 20\), confirming this solution.- Similarly, for \(x = \frac{18}{11}\), check \(|10 \times \frac{18}{11}| = |\frac{18}{11} - 18|\). Both sides should give approximately the same result, confirming this solution as well.This step is vital to ensure accuracy and to understand the behavior of the solutions.

Intermediate Algebra Concepts

Intermediate algebra often involves working with more complex equations and expressions, including those with absolute values. Such problems require an understanding of fundamental concepts like the properties of absolute values and manipulation of algebraic expressions. Key algebra techniques include:

• Breaking down complex expressions into simpler components.
• Applying properties of equality to maintain balance on both sides of an equation.
• Understanding how operations affect absolute values in the context of their distance-related nature.

These strategies help in dealing with seemingly tough algebra problems. As seen in our example with \(|10x| = |x-18|\), using these intermediate concepts systematically lets us solve and confidently verify solutions to real-world problems.