Problem 10 - The Calculus with Analytic Geometry

Question

In Exercises 1 through 10, solve for \(x\). $$ \left|\frac{3 x+8}{2 x-3}\right|=4 $$

Step-by-Step Solution

Answer

The solutions for \(x\) are \(5\) and \(\frac{4}{11}\).

Step 1: Set-Up the Equation

Begin by understanding that the absolute value equation \ \ \ \ \ \ \ \ \(\left|\frac{3 x+8}{2 x-3}\right|=4\) implies two separate equations. This is because the expression inside the absolute value could be either positive or negative.

Step 2: Remove Absolute Value for Case 1

For the first case, assume \ \ \ \( \frac{3 x+8}{2 x-3}=4\). Proceed to isolate \ \(x\ \) by first eliminating the fraction:

Step 3: Solve for x in Case 1

Multiply both sides of the equation by \ \(2 x - 3\ \) to obtain: \ \( 3 x + 8 = 4(2 x - 3)\ \). Next, distribute and simplify: \ \( 3x + 8 = 8x - 12 \ \ Rightarrow \ -4x = -20 \ \Rightarrow \ x = 5 \ \).

Step 4: Remove Absolute Value for Case 2

For the second case, assume \ \( \frac{3 x+8}{2 x-3}=-4 \ \). Again, eliminate the fraction by multiplying both sides by \ \(2 x - 3\ \):

Step 5: Solve for x in Case 2

Multiply both sides by \ \(2 x - 3\ \) to obtain: \ \( 3 x + 8 = -4(2 x - 3) \ \). Distribute and simplify the equation: \ \( 3 x + 8 = -8 x + 12 \ \Rightarrow \ 11 x = 4 \ \Rightarrow \ x = \frac{4}{11} \ \).

Step 6: Consolidate the Solutions

Combine the two solutions obtained from both cases. The values of \ \( x \ \) that satisfy the original equation are \ \( x = 5\ \ \) and \ \( x = \frac{4}{11}\ \ \).

Key Concepts

solving absolute value equations

Solving absolute value equations involves finding the values of the variable that make the equation true. Absolute value equations tend to have two possible solutions because the absolute value of a number represents its distance from zero, irrespective of the direction on a number line. This means that an equation like \(\left\| \frac{3x + 8}{2x - 3} \right\| = 4\) translates into two separate linear equations, representing the two scenarios where the expression can be either positive or negative. By setting up and solving these two cases, you effectively explore all possible values of the variable.

case analysis in equations

Case analysis in equations involves breaking down an absolute value equation into two distinct equations, one for each possible case of the variable's expression. For our example \(\left\| \frac{3x + 8}{2x - 3} \right\| = 4\), this means solving for both \( \frac{3x + 8}{2x - 3} = 4 \) and \( \frac{3x + 8}{2x - 3} = -4 \). Each equation is solved individually:
* In the first case: \( \frac{3x + 8}{2x - 3} = 4 \), multiply both sides by the denominator \(2x - 3\) to clear the fraction, getting \(3x + 8 = 4(2x - 3)\). Simplifying leads to solving for \(x = 5\).
* In the second case: \( \frac{3x + 8}{2x - 3} = -4 \), again multiply by \(2x - 3\), resulting in \(3x + 8 = -4(2x - 3)\). Simplifying this leads to solving for \(x = \frac{4}{11}\).
By employing case analysis, you are able to find all possible solutions to the absolute value equation.

isolating variables in algebra

Isolating variables in algebra typically involves rearranging the equation so that the variable you're solving for stands alone on one side of the equation. This process often requires several algebraic manipulation steps, including:
* Adding or subtracting terms on both sides to eliminate constants or coefficients.
* Multiplying or dividing both sides by the same number to isolate the variable.
For instance, in the case of solving \( \frac{3x + 8}{2x - 3} = 4 \), you first clear the fraction by multiplying both sides by \(2x - 3\). Then, you distribute and combine like terms which leads to: \(3x + 8 = 8x - 12\), which simplifies to \(-4x = -20\), thus \(x = 5\).
Similarly, in solving \( \frac{3x + 8}{2x - 3} = -4 \), the same process of multiplication, distribution, and combining like terms will eventually isolate \(x\) as \(x = \frac{4}{11}\). By practicing isolation techniques, you streamline the process of solving for variables effectively across various algebraic equations.