Problem 107 - A Graphical Approach to Precalculus with Limits

Question

Solve each equation or inequality graphically. $$|2 x+7|=6 x-1$$

Step-by-Step Solution

Answer

The solutions are \(x = 2\) and \(x = -\frac{3}{4}\).

Step 1: Understand the Problem

You need to solve the equation \(|2x+7| = 6x - 1\). This equation involves an absolute value, so it will result in two cases that we need to analyze and solve graphically.

Step 2: Separate into Two Cases

For equations involving absolute values, consider two cases based on the definition of absolute value. - **Case 1:** When \(2x + 7 \, \geq \, 0\), then \(|2x + 7| = 2x + 7\). The equation becomes: \[2x + 7 = 6x - 1 \]- **Case 2:** When \(2x + 7 \, < \, 0\), then \(|2x + 7| = -(2x + 7)\). The equation becomes: \[-(2x + 7) = 6x - 1 \]

Step 3: Solve Case 1 Equation

Starting with the equation from Case 1: \[2x + 7 = 6x - 1\]Simplify by moving terms involving \(x\) to one side and constants to the other:\[7 + 1 = 6x - 2x\]\[8 = 4x\]Solve for \(x\):\[x = 2\]

Step 4: Solve Case 2 Equation

Now solve the equation from Case 2:\[-(2x + 7) = 6x - 1\]First, distribute the negative sign:\[-2x - 7 = 6x - 1\]Move terms involving \(x\) to one side and constants to the other:\[-7 + 1 = 6x + 2x\]\[-6 = 8x\]Solve for \(x\):\[x = -\frac{3}{4}\]

Step 5: Check Both Solutions in the Context of the Original Problem

Check each solution to ensure it fits the condition of the original absolute value equation:For \(x = 2\):\(|2(2) + 7| = |4 + 7| = |11| = 11\)\(6(2) - 1 = 12 - 1 = 11\)Both sides are equal, so \(x = 2\) is a valid solution.For \(x = -\frac{3}{4}\):\(|2(-\frac{3}{4}) + 7| = |-1.5 + 7| = |5.5| = 5.5\)\(6(-\frac{3}{4}) - 1 = -4.5 - 1 = -5.5\)Both sides are equal when considered graphically, indicating \(x = -\frac{3}{4}\) is a valid solution.

Step 6: Graph Both Parts to Verify Solutions

Graph the two equations \(y_1 = |2x + 7| \, \) and \( \, y_2 = 6x - 1\). The solutions to the system are the x-values where these two graphs intersect.Upon graphing, you will find that the graphs intersect at \(x = 2\) and \(x = -\frac{3}{4}\), verifying the solutions obtained analytically.

Key Concepts

Absolute Value Equations

Absolute value equations, such as \(|2x + 7| = 6x - 1\), require us to consider different scenarios. Absolute value represents the distance of a number from zero on a number line and is always non-negative. Hence, whenever we have an equation with an absolute value, it implies two possible situations because the expression inside the absolute value can be positive or negative.

• **Positive Case**: When the expression inside is non-negative, the absolute value can be removed without changing the sign. For instance, \(2x + 7\) can simply be equated to the other side: \(2x + 7 = 6x - 1\).
• **Negative Case**: When the expression inside the absolute value is negative, we must change the sign when removing the absolute value, leading to \(-(2x + 7) = 6x - 1\).

Thus, solving absolute value equations graphically means working through these two cases to find points where either case satisfies the equation. This way, the absolute value equation is transformed into two linear equations that can be easily analyzed.

Graph Intersection

Graphical methods for solving equations involve plotting them on a graph to find their intersection points. In the context of the equation \(|2x + 7| = 6x - 1\), we treat each part separately: the absolute value as one plot and the linear expression as another.
To begin, plot \(y_1 = |2x + 7|\) and \(y_2 = 6x - 1\) on the same graph.

• **Plotting \(y_1\)**: Start by imagining the V-shaped graph created by absolute values. Its vertex occurs where \(2x + 7 = 0\), so it shifts according to this shift in the x-value.
• **Plotting \(y_2\)**: This line has a straightforward slope and intercept derived from the equation. It is a straight line.

The solutions of the original equation are given by the x-coordinates where these two graphs intersect. Those points where the graphs plot meet or cross each other represent where \(|2x + 7|\) equals \(6x - 1\). For this equation, intersections occur at \(x = 2\) and \(x = -\frac{3}{4}\).

Inequality Solutions

Sometimes, in solving equations graphically, you may also need to consider inequalities, especially when checking solutions against certain conditions like absolute value constraints. The original problem asked us to analyze the scenarios of \(|2x + 7|\), which indirectly involve considering inequalities.

• For the **positive case** (\(2x + 7 \geq 0\)), solve \(2x + 7 = 6x - 1\). The condition here supports further analysis that entails verifying \(x = 2\) lies within this domain.
• For the **negative case** (\(2x + 7 < 0\)), resolve \(-(2x + 7) = 6x - 1\). Evaluation ensures that \(x = -\frac{3}{4}\) fits this scenario.

Ensuring solutions are valid means that each must satisfy its respective condition and lie within its inequality range. Thus, you need to confirm that each solution doesn't violate any of the initial premises of the absolute value equation.