Problem 16 - Algebra and Trigonometry

Question

Solve the equation. $$ \left|\frac{3}{5} x+2\right|-\frac{1}{2}=4 $$

Step-by-Step Solution

Answer

The solutions are \( x = \frac{25}{6} \) and \( x = -\frac{65}{6} \).

Step 1: Understand the Absolute Value Equation

The equation given is \( \left|\frac{3}{5}x + 2\right| - \frac{1}{2} = 4 \). The goal is to isolate the absolute value expression and solve for \( x \).

Step 2: Isolate the Absolute Value

Add \( \frac{1}{2} \) to both sides of the equation to isolate the absolute value term: \( \left| \frac{3}{5}x + 2 \right| = 4 + \frac{1}{2} \), which simplifies to \( \left| \frac{3}{5}x + 2 \right| = 4.5 \).

Step 3: Set Up Two Equations

The absolute value equation \( \left| a \right| = b \) can be split into two separate equations: 1. \( \frac{3}{5}x + 2 = 4.5 \)2. \( \frac{3}{5}x + 2 = -4.5 \)

Step 4: Solve the First Equation

For \( \frac{3}{5}x + 2 = 4.5 \), subtract 2 from both sides: \( \frac{3}{5}x = 2.5 \). Multiply both sides by \( \frac{5}{3} \) to solve for \( x \): \[ x = \frac{5}{3} \times 2.5 \] \[ x = \frac{25}{6} \].

Step 5: Solve the Second Equation

For \( \frac{3}{5}x + 2 = -4.5 \), subtract 2 from both sides: \( \frac{3}{5}x = -6.5 \). Multiply both sides by \( \frac{5}{3} \) to solve for \( x \): \[ x = \frac{5}{3} \times (-6.5) \] \[ x = -\frac{65}{6} \].

Step 6: Verify Solutions

Check both solutions by substituting back into the original equation:For \( x = \frac{25}{6} \): \( \left| \frac{3}{5} \times \frac{25}{6} + 2 \right| - \frac{1}{2} = 4 \)For \( x = -\frac{65}{6} \): \( \left| \frac{3}{5} \times (-\frac{65}{6}) + 2 \right| - \frac{1}{2} = 4 \) Both hold true, so the solutions are verified.

Key Concepts

Solving Equations

Absolute value equations involve expressions enclosed within the absolute value bars, like \( \left| \frac{3}{5}x + 2 \right| \). These equations can be a bit tricky due to their dual nature. They represent two possible cases for solving.

The absolute value of a number is the number's distance from zero, meaning it is always nonnegative. When solving an absolute value equation, you will often need to set up two separate equations because the expression inside the absolute value can be positive or negative.

To solve such an equation:

• Begin by isolating the absolute value expression on one side of the equation.
• Once isolated, split the equation into two separate scenarios.
• Solve each equation independently to find potential solutions for \( x \).

These steps are crucial to handle the dual nature of absolute values, where the expression can equal either the positive or negative value of the other side.

Isolation of Variables

Isolation of variables involves rearranging an equation to get a particular variable by itself on one side, and everything else on the other. In the context of absolute value equations, you often begin with isolating the absolute value.

For instance, consider the equation \( \left| \frac{3}{5}x + 2 \right| - \frac{1}{2} = 4 \).
To isolate the absolute value:

• Add \( \frac{1}{2} \) to both sides to remove any constant outside the absolute value.
• The equation becomes \( \left| \frac{3}{5}x + 2 \right| = 4.5 \).

Now, the absolute value is isolated, and you can proceed by solving the equation, allowing the exploration of both the positive and negative cases. By isolating the absolute value, you simplify the solving process significantly.

Verifying Solutions

Verifying solutions ensures that the solutions you find actually satisfy the original equation. This step is often overlooked but is very important, especially with absolute value equations, which can sometimes yield extraneous solutions.

To verify the solutions:

• Substitute each tentative solution back into the original equation.
• Check whether both sides of the equation equal each other after substitution.

For example, with solutions \( x = \frac{25}{6} \) and \( x = -\frac{65}{6} \), plug these back into the initial equation:

\( \left| \frac{3}{5} \times \frac{25}{6} + 2 \right| - \frac{1}{2} = 4 \)

Both should simplify to the truth of the equation. If one solution does not satisfy the equation, it must be discarded. This verification step confirms the accuracy of your solutions and ensures that all found solutions are indeed correct.