Problem 86 - A Graphical Approach to Precalculus with Limits

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 \(|f(x)|<|g(x)|\). $$|6 x+9|=|6 x-3|$$

Step-by-Step Solution

Answer

(a) \(x = -\frac{1}{2}\). (b) \(x < -\frac{1}{2}\). (c) \(x > -\frac{1}{2}\).

Step 1: Understand the Equation

The given equation is \(|6x+9|=|6x-3|\). This equation implies that the absolute values of two linear expressions are equal.

Step 2: Remove Absolute Values

For \(|f(x)|=|g(x)|\), consider two cases: 1. \(6x+9 = 6x-3\)2. \(6x+9 = -(6x-3)\).Let's solve each case.

Step 3: Case 1: Simplify Same Side Equation

Solve the equation \(6x+9 = 6x-3\). Subtract \(6x\) from both sides: \(9 = -3\). This gives a contradiction which means there is no solution from this case.

Step 4: Case 2: Simplify Opposite Side Equation

Solve the equation \(6x+9 = - (6x-3)\). Simplify the right side to get: \(6x + 9 = -6x + 3\).Add \(6x\) to both sides: \(12x + 9 = 3\). Subtract \(9\) from both sides: \(12x = -6\). Divide by \(12\): \(x = -\frac{1}{2}\).

Step 5: Graphical Representation

Graph the functions \(f(x) = |6x + 9|\) and \(g(x) = |6x - 3|\). The graph will show the two functions intersecting at \(x = -\frac{1}{2}\), confirming the solution derived analytically.

Step 6: Inequality for Greater Than

Solve \(|6x + 9| > |6x - 3|\). Use the same cases but alter the inequality:1. \(6x + 9 > 6x - 3\) leads to \(9 > -3\), which is always true for all \(x\).2. \(6x + 9 > - (6x - 3)\) with the solution \(x < -\frac{1}{2}\). Combine these results meaning the solution is \(x < -\frac{1}{2}\).

Step 7: Inequality for Less Than

Solve \(|6x + 9| < |6x - 3|\). Change cases:1. \(6x + 9 < 6x - 3\) yields \(9 < -3\), which is never true.2. \(6x + 9 < - (6x - 3)\) yields \(x > -\frac{1}{2}\).

Key Concepts

Graphical Solution

Visualizing mathematical equations helps us understand the concept better, especially when dealing with absolute value equations. When you graph the functions related to the equation \(|6x+9|=|6x-3|\), you plot two V-shaped graphs on the coordinate plane. Each graph represents an absolute value function:

• One for \(|6x+9|\), which opens upwards and shifts leftwards.
• Another for \(|6x-3|\), which also opens upwards but shifts rightwards.

Where these two graphs intersect represents the solution to the equation. As noted in the solution, they intersect at \(x = -\frac{1}{2}\). This point of intersection visually confirms the analytical solution, because both expressions are equal there. Graphs provide insight into how the values of linear expressions are distance-wise equal on a number line, depicted as intersections on the graph.

Analytical Solution

To solve \(|6x+9|=|6x-3|\), we initially need to consider the nature of absolute values. The two main cases derived from this equation are:

• First, assuming \(6x+9 = 6x-3\), leading to a contradiction (9 = -3), thus showing no solution exists in this scenario.
• Second, equating one side to the negative of the other, so \(6x+9 = -(6x-3)\). This simplifies to \(12x = -6\), giving the solution \(x = -\frac{1}{2}\).

Through these cases, we learn that absolute value equations can have complex logical outcomes (like contradictions or valid results) depending on the sign and shift of expressions. Analytically, this method helps us dissect what makes the expressions equal after considering their absolute distances.

Inequality Solutions

Solving inequalities involving absolute values, such as \(|6x+9|>|6x-3|\) and \(|6x+9|<|6x-3|\), requires a similar case-based approach:

• For \(|6x+9|>|6x-3|\), check when \(6x+9 > 6x-3\) (always true), and \(6x+9 > -(6x-3)\) (true when \(x < -\frac{1}{2}\)). The latter constraint \(x < -\frac{1}{2}\) governs the solution.
• For \(|6x+9|<|6x-3|\), compare \(6x+9 < 6x-3\) (never true) and \(6x+9 < -(6x-3)\) (establishes \(x > -\frac{1}{2}\)). Therefore, solution involves values greater than \(-\frac{1}{2}\).

Recognizing which inequality holds under specific conditions reveals the range of solutions. It demonstrates how inequalities evaluate differing distances between expressions via absolute values.

Linear Expressions

Linear expressions involved in absolute value equations like \(6x+9\) and \(6x-3\) denote simple algebraic equations that vary linearly when graphed. These expressions show direct relationships between the variables with the coefficient (slope) and constant (intercept):

• \(6x+9\) can be interpreted as a line that rises steeply with a starting point above the origin on the y-axis.
• \(6x-3\) similarly climbs but starts three units below the x-axis at the y-intercept.

These linear expressions allow us to visualize absolute value equations. They form straight lines that adopt a V-shape when transformed via absolute values. Understanding linear equations and their properties—such as slope (alignment) and y-intercept (starting point)—is vital for graphing and interpreting absolute value problems profoundly. They lay the groundwork for creating the functions involved in these equations.