Problem 49 - Elementary and Intermediate Algebra

Question

Solve each equation. \(|10 x|=|x-18|\)

Step-by-Step Solution

Answer

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

Step 1: Consider the definition of absolute value

An absolute value equation can split into two separate equations. We need to consider both scenarios where the expressions inside the absolute value signs are equal or opposite. For the given equation \[|10x| = |x - 18|\]We consider two cases: when both expressions are equal \[10x = x - 18\]and when they are opposite \[10x = -(x - 18)\].

Step 2: Solve the first equation

Let's solve the equation where the expressions are considered equal. Starting from \[10x = x - 18\], we rearrange the expression to find x:\[10x - x = -18\]\[9x = -18\]Divide both sides by 9:\[x = -2\]

Step 3: Solve the second equation

Now solve for the situation where the expressions are opposite: Starting from \[10x = -(x - 18)\], we first simplify:\[10x = -x + 18\]Combine like terms:\[10x + x = 18\]\[11x = 18\]Divide by 11:\[x = \frac{18}{11}\]

Step 4: Verify solutions within the context of absolute value

You need to check that both solutions satisfy the original absolute value equation.- For \(x = -2\): - \(|10(-2)| = |18(-2) + 18|\) - \(20 = 20\), which is true.- For \(x = \frac{18}{11}\): - \(10 \times \frac{18}{11} = |\frac{18}{11} - 18|\) - It can be verified that both sides are equal.

Key Concepts

Understanding Solving Equations

When it comes to solving equations, especially those involving absolute values, it's crucial to understand that an absolute value equation often leads to more than one solution. This is because absolute value measures distance from zero, which means that two different numbers can have the same absolute value.
In the context of the given exercise, \(|10x| = |x - 18|\), we interpret the absolute value signs by considering two possible scenarios:

• The expressions inside the absolute value bars are equal, implying \(10x = x - 18\).
• The expressions are opposite, which means \(10x = -(x - 18)\).

By tackling these scenarios individually, students can solve the corresponding equations to find potential solutions. This approach is a fundamental step in learning how to handle absolute value equations effectively.

The Role of Systems of Equations

Systems of equations are sets of equations with multiple variables. However, in the context of absolute value equations like the one in our exercise, solving the equation involves treating each "case" as a simple linear equation to derive a system of two separate equations.
In our particular example, the absolute value equation \(|10x| = |x - 18|\) results in two linear equations:

• First equation from equal expression: \(10x = x - 18\)
• Second equation from opposite expression: \(10x = -(x - 18)\)

Solving each case separately allows us to explore multiple possible solutions for \(x\). The expression of two cases broadens the concept of solving systems of equations, illustrating how different approaches deliver all potential solutions. This problem-solving strategy emphasizes exploring all possibilities inherent in absolute value scenarios.

Exploring Algebraic Manipulation

Algebraic manipulation is the process of rearranging and simplifying equations to make them easier to solve. For absolute value equations, this often includes:

• Expanding expressions and distributing signs correctly.
• Combining like terms to simplify the equation structure.
• Using inverse operations like addition and subtraction to isolate variables.

For the given equation, \(10x = x - 18\) necessitates combining like terms and simplifying as follows:
Start by subtracting \(x\) from both sides to get \(9x = -18\). Then divide by 9 to isolate \(x\), resulting in \(x = -2\).
Similarly, \(10x = -(x - 18)\) requires distributing the negative sign, leading to \(10x = -x + 18\). Combining like terms gives us \(11x = 18\), and dividing by 11 results in \(x = \frac{18}{11}\).
Thus, through algebraic manipulation, the solutions emerge incrementally, highlighting how change in approach affects resolution. Practicing these techniques makes solving such equations more intuitive and straightforward.