Problem 42 - Beginning and Intermediate Algebra

Question

Solve the following equations containing two absolute values. $$|4-11 r|=|5 r+3|$$

Step-by-Step Solution

Answer

The solutions for the equation \(|4-11r|=|5r+3|\) are \(r = \frac{1}{16}\) or \(r = \frac{7}{6}\).

Step 1: Case 1: Both absolute values are positive

If both the absolute values are positive, we can write the equation as: \(4-11r = 5r+3\)

Step 1: Isolate the variable "r"

To solve for "r," we need to isolate it on one side of the equation. Let's move all "r" terms to one side and constant terms to the other side: \(11r + 5r = 4-3\)

Step 2: Collect like terms

Now, combine the like terms: \(16r =1\)

Step 3: Solve for "r"

Divide both sides by 16: \(r=\frac{1}{16}\) This is the solution for the first case.

Step 5: Case 2: One absolute value is positive, and the other is negative

Now let's consider the case where the first absolute value is positive, and the second one is negative. We can write the equation as: \(4-11r = -(5r+3)\)

Step 1: Distribute the negative sign

Distribute the negative sign on the right-hand side of the equation: \(4-11r = -5r-3\)

Step 2: Isolate the variable "r"

Move all the "r" terms to one side and the constant terms to the other side: \(11r - 5r = 4+3\)

Step 3: Collect like terms

Combine the like terms: \(6r = 7\)

Step 4: Solve for "r"

Divide both sides by 6: \(r=\frac{7}{6}\) This is the solution for the second case.

Step 5: Combine the solutions

Now we have two possible solutions for "r," which are \(r = \frac{1}{16}\) and \(r = \frac{7}{6}\). It's important to check that these solutions satisfy the original equation, and, in this case, they do. So, we can write our final answer as: \(r = \frac{1}{16}\) or \(r = \frac{7}{6}\)

Key Concepts

Absolute Values

Absolute values can be a tricky concept at first, but it's really all about focusing on distance from zero. In mathematics, the absolute value of a number, represented as \(|a|\), is the non-negative value of \(a\) without regard to its sign. Hence:

• \(|5| = 5\)
• \(|-5| = 5\)

This property becomes particularly useful in algebraic equations because it allows us to evaluate expressions while considering both positive and negative possibilities for any term within an absolute value. In the example given, \(|4-11r|\) and \(|5r+3|\) must be equal in value, though not necessarily in immediate sign, prompting us to consider multiple cases during solution.

Algebraic Equations

Algebraic equations are statements of equality that involve variables and numbers. Solving them often requires isolating the variable to one side to uncover its value. When dealing with absolute values, equations can become somewhat nuanced since the value inside the absolute brackets may have multiple potential scenarios. For example:

• The entire expression might be positive, thus unchanged.
• The expression might be negative, hence need transforming by multiplication with \(-1\).

In solving the equation \(|4-11r| = |5r+3|\), we consider both cases: when their expressions are positive and when one expression is negative, to ensure comprehensive examination. This results in different, yet correct, solutions showing algebra's capability in handling varied conditions.

Intermediate Algebra

Intermediate algebra builds on basic algebraic principles by introducing more complex equation types, such as those involving absolute values. It encourages deeper understanding of concepts like variable isolation and managing multiple potential outcomes. By honing skills in:

• Handling multiple cases in equations.
• Simplifying expressions.
• Applying properties of equality efficiently.

Students learn to address diverse challenges. The exercise provided, involving \(|4-11r| = |5r+3|\), demands a level of comfort with such advanced algebraic techniques. It includes solving for more than one possible solution, testing conceptual understanding and competence in the subject. This prepares students for even more sophisticated topics further in mathematics.