Problem 38 - Algebra 2

Question

Solve each equation. Check your solutions. \(3|p-5|=2 p\)

Step-by-Step Solution

Answer

The solutions are \(p = 15\) and \(p = 3\). Both satisfy the equation.

Step 1: Identify the Absolute Value Expression

The given equation is \(3|p-5|=2p\). Here, the absolute value expression is \(|p-5|\). We need to consider the two possible cases for the absolute value, as \(|a| = a\) if \(a \geq 0\) and \(|a| = -a\) if \(a < 0\).

Step 2: Case 1 - Positive Inner Expression

Assume \(p-5 \geq 0\). This gives us \(|p-5| = p-5\). Substitute into the equation:\[3(p-5) = 2p\]Distribute and simplify:\[3p - 15 = 2p\]Subtract \(2p\) from both sides:\[p - 15 = 0\]Add 15 to both sides:\[p = 15\]

Step 3: Check Solution for Case 1

Substitute \(p = 15\) back into the original equation:\[3|15-5| = 2 \times 15\]\[3 \times 10 = 30\]\[30 = 30\]The solution \(p = 15\) satisfies the original equation.

Step 4: Case 2 - Negative Inner Expression

Assume \(p-5 < 0\). This gives us \(|p-5| = -(p-5) = -p + 5\). Substitute into the equation:\[3(-p + 5) = 2p\]Distribute and simplify:\[-3p + 15 = 2p\]Add \(3p\) to both sides:\[15 = 5p\]Divide both sides by 5:\[p = 3\]

Step 5: Check Solution for Case 2

Substitute \(p = 3\) back into the original equation:\[3|3-5| = 2 \times 3\]\[3 \times 2 = 6\]\[6 = 6\]The solution \(p = 3\) satisfies the original equation.

Key Concepts

Case Analysis in Absolute Value Equations

To solve absolute value equations effectively, we utilize a process called case analysis. This method involves considering all potential scenarios for the expression within the absolute value brackets. In simpler terms, you are determining how the absolute value expression behaves depending on its sign.

Let’s apply case analysis to understanding the two situations:

• Positive Case: When the expression inside the absolute value is zero or positive, the absolute value does not alter it. For example, in the equation \(|p-5|\), assuming \(p-5 \geq 0\) means the absolute value can be replaced with \((p-5)\).
• Negative Case: When the expression inside the absolute value is negative, the absolute value acts to flip its sign. So if \(p-5 < 0\), you would consider \(-1(p-5)\) instead, which simplifies to \(-p + 5\).

This case analysis helps ensure that we explore all possible situations for the variable and lay the groundwork for the subsequent steps of solving the equation.

Algebraic Manipulation Techniques

Once the cases have been identified, the next step involves solving the resulting equations using algebraic manipulation. This technique is all about systematically isolating the variable to one side of the equation.

Let's break down the steps:

• Distribution: If there are parentheses involving the variable, distribute the factor outside. For instance, in \(3(p-5)\), you distribute the \(3\) to each term inside the parentheses, resulting in \(3p - 15\).
• Transposing Terms: This involves moving terms from one side to the other to isolate the variable. For instance, from \(3p - 15 = 2p\), you can subtract \(2p\) from both sides to result in \(p - 15 = 0\).
• Addition or Subtraction: Use these operations to further isolate the variable. Continuing from \(p - 15 = 0\), you simply add 15 to both sides to solve for \(p = 15\).
• Division or Multiplication: Final isolates are sometimes necessary with these operations. For example, solving \(15 = 5p\) would involve dividing both sides by \(5\) to get \(p = 3\).

These techniques require attention to detail and proper execution to avoid errors, ultimately providing the correct solutions.

Checking Solutions for Accuracy

After arriving at potential solutions through case analysis and algebraic manipulation, it's important to verify these solutions to confirm their validity. This process involves substituting the solutions back into the original equation and ensuring the equation holds true.

Let's examine how to check the solutions:

• Substitution: Take the found solution and plug it back into the original equation. This is crucial, as it confirms the solution’s validity without rerunning complex calculations. For instance, substituting \(p = 15\) back into \(3|p-5| = 2p\), verifies if both sides equal when the equation balances as expected.
• Comparison: After substituting, compare both sides of the equation to ensure they are equal. If they match, the solution is correct. For instance, \(30 = 30\) after substituting \(p = 15\), indicates correctness.

Without this step, there could be oversight resulting in incorrect conclusions. Verifying solutions is a critical aspect of problem solving, solidifying confidence in the correctness of the answers found.