Problem 67 - A Graphical Approach to College Algebra

Question

An equation of the form \(|f(x)|=|g(x)|\) is given. (a) Solve the equation analytically and support the solution graphically. (b) Solve \(|f(x)|>|g(x)|\) (c) Solve \(|f(x)|<|g(x)|\) $$|3 x+1|=|2 x-7|$$

Step-by-Step Solution

Answer

(a) Solutions: \(x = -8\) and \(x = \frac{6}{5}\). (b) \(-8 < x < \frac{6}{5}\). (c) \(x < -8\) or \(x > \frac{6}{5}\).

Step 1: Identify cases of the equation

To solve the equation \(|3x + 1| = |2x - 7|\), consider the possible cases based on the properties of absolute values. These cases are where both expressions are equal and positive, both are equal and negative, or one is positive and the other negative.

Step 2: Case 1: Both expressions equal and positive

In this case, solve the equation \(3x + 1 = 2x - 7\). Subtract \(2x\) from both sides to get \(x + 1 = -7\). Then, subtract 1 from both sides to find \(x = -8\).

Step 3: Case 2: Both expressions equal and negative

Here, solve \(3x + 1 = -(2x - 7)\). This simplifies to \(3x + 1 = -2x + 7\). Add \(2x\) to both sides: \(5x + 1 = 7\). Then, subtract 1 from both sides: \(5x = 6\). Divide by 5: \(x = \frac{6}{5}\).

Step 4: Case 3: One expression positive, one negative

Two more cases arise here. First, solve \(3x + 1 = -2x + 7\). Proceed as before to determine solutions. Then, solve \(-3x - 1 = 2x - 7\). If calculations repeat prior results, disregard as these yield existing solutions.

Step 5: Verify solutions

Checking both \(x = -8\) and \(x = \frac{6}{5}\) in the original absolute value equation shows they satisfy the equation. Thus, these are valid solutions.

Step 6: Solution to inequality \/|f(x)| > |g(x)|\/

Consider the inequality \(|3x + 1| > |2x - 7|\). Analyze cases: \(3x + 1 > 2x - 7\) or \(3x + 1 < -2x + 7\). First, solve \(3x + 1 > 2x - 7\), yielding \(x > -8\). Solving \(3x + 1 < -2x + 7\) results in \(x < \frac{6}{5}\). Combine to find \(-8 < x < \frac{6}{5}\).

Step 7: Solution to inequality \/|f(x)| < |g(x)|\/

Analyze the inequality \(|3x + 1| < |2x - 7|\) using remaining inequalities analyzed: \(3x + 1 < 2x - 7\) or \(3x + 1 > -2x + 7\). Solving \(3x + 1 < 2x - 7\) gives \(x < -8\). Solving \(3x + 1 > -2x + 7\) gives \(x > \frac{6}{5}\). Combine to form the solution: \(x < -8 \text{ or } x > \frac{6}{5}\).

Key Concepts

Absolute Value Equations

When we work with absolute value equations, we delve into expressions that show the magnitude of a number, irrespective of its sign. Imagine you have an equation like \(|3x + 1| = |2x - 7|\). This means solving for values of \(x\) where both sides are equal after considering any signs of the expressions. To tackle these equations, it’s crucial to consider the cases that arise due to the properties of absolute value, namely:

• Both expressions are positive.
• Both expressions are negative.
• One expression is positive while the other is negative.

Let's delve into each scenario using our equation.
First up, both sides might just be straight-up positive. Here, simply equate the expressions without the absolute value: \(3x + 1 = 2x - 7\) and solve to find \(x = -8\).
Next, consider the situation where both are negative. This implies \(3x + 1 = -(2x - 7)\), which after solving yields \(x = \frac{6}{5}\).
Finally, what if one’s positive and the other’s negative? Picture solving \(3x + 1 = -(2x - 7)\) and \(-3x - 1 = 2x - 7\), leading back to our previous solutions. Essentially, stick to valid \(x\) values confirmed by substituting back into the original equation and verify both satisfy your equation.

Inequalities

Working with inequalities, especially those involving absolute values, requires pinpointing not just potential solutions but ranges. Consider \(|3x + 1| > |2x - 7|\), an inequality setting the stage for a comparison of magnitudes.
Here, break it down into manageable parts. Start with scenarios like:

• \(|3x + 1| > |2x - 7|\): Could mean \(3x + 1 > 2x - 7\) or \(3x + 1 < -2x + 7\).

The first piece to solve is \(3x + 1 > 2x - 7\). Solve this like a regular inequality, which yields \(x > -8\).
Next, handle \(3x + 1 < -2x + 7\). Solve carefully and you’ll find \(x < \frac{6}{5}\).
Combine these solutions, capturing the entire range: \(-8 < x < \frac{6}{5}\). This isolates \(x\) values satisfying the inequality and guides our understanding of what regions maximize \(|3x + 1|\). Separately investigating \(|3x + 1| < |2x - 7|\) directs us to the complementary intervals, concluded as \(x < -8\) or \(x > \frac{6}{5}\). Thus, we map inclusive and exclusive zones satisfying complex comparative statements.

Graphical Solutions

For visual learners, graphs provide an approachable way to solve absolute value equations and inequalities. Let’s sketch our scenario involving \(|3x+1| = |2x-7|\) and see it unfold on a coordinate plane.
Graph each expression \(|3x + 1|\) and \(|2x - 7|\), and these appear as 'V' shapes reflecting dependencies on different \(x\). Where these graphs intersect offer solutions to the absolute value equation since they depict points where expressions are equal in magnitude.
For inequalities, focus on zones bounded by these 'V' shapes:

• Regions where \(|3x + 1|\) is above \(|2x - 7|\) - it indicates \(|3x + 1| > |2x - 7|\).
• Conversely, places where \(|3x + 1|\) sits below \(|2x - 7|\) inform \(|3x + 1| < |2x - 7|\).

These graphical insights allow you to easily observe intervals as opposed to strictly analytical methods. Meshing the imagery with calculations lets you bridge between theory and practice, paving an intuitive path through the algebraic maze of absolute values and inequalities.