Problem 17 - Precalculus Student Solutions Manual 5th

Question

Solve each equation. $$\left|\frac{6 x+1}{x-1}\right|=3$$

Step-by-Step Solution

Answer

The solutions are \( x = -\frac{4}{3} \) and \( x = \frac{2}{9} \).

Step 1: Understand the Absolute Value Equation

The given equation is an absolute value equation: \ \ \ \ \[ \left|\frac{6 x+1}{x-1}\right|=3 \]. \ \ The absolute value equation \( |A| = B \) means that either \( A = B \) or \( A = -B \). Therefore, the equation \( \left|\frac{6 x+1}{x-1}\right| = 3 \) implies that \ \ \( \frac{6 x+1}{x-1} = 3 \) or \( \frac{6 x+1}{x-1} = -3 \).

Step 2: Solve the First Equation

Solve the first equation \( \frac{6 x+1}{x-1} = 3 \): \ \ \ \( 6x + 1 = 3(x-1) \) \ \ Expand and simplify: \ \ \( 6x + 1 = 3x - 3 \) \ \ Move all terms involving x to one side: \ \ \( 6x - 3x = -3 - 1 \) \ \ \( 3x = -4 \) \ \ Divide both sides by 3: \ \ \( x = -\frac{4}{3} \).

Step 3: Solve the Second Equation

Solve the second equation \( \frac{6 x+1}{x-1} = -3 \): \ \ \ \( 6x + 1 = -3(x-1) \) \ \ Expand and simplify: \ \ \( 6x + 1 = -3x + 3 \) \ \ Move all terms involving x to one side: \ \ \( 6x + 3x = 3 - 1 \) \ \ \( 9x = 2 \) \ \ Divide both sides by 9: \ \ \( x = \frac{2}{9} \).

Step 4: Verify the Solutions

Verify the solutions in the original equation to ensure they do not make the denominator zero. \ \ For \( x = -\frac{4}{3} \): \ \ The denominator is \( x - 1 = -\frac{4}{3} - 1 = -\frac{4}{3} - \frac{3}{3} = -\frac{7}{3} \), which is not zero. \ \ For \( x = \frac{2}{9} \): \ \ The denominator is \( x - 1 = \frac{2}{9} - 1 = \frac{2}{9} - \frac{9}{9} = -\frac{7}{9} \), which is also not zero. Therefore, both solutions are valid.

Key Concepts

Understanding Absolute Value

An absolute value represents the distance of a number from zero on the number line. It is always non-negative. For example, the absolute value of both -5 and 5 is 5, written as \(|-5| = 5\) and \(|5| = 5\). In equations involving absolute values, such as \(|A| = B\), the expression inside the absolute value symbol can be equal to B or the negative of B. This is critical since \(|A| = B\) is true when A is either equal to B or \(-B\). Applied to our problem, \(|\(\frac{6x+1}{x-1}\)| = 3\) translates into two separate equations: \((\frac{6x+1}{x-1} = 3)\) and \((\frac{6x+1}{x-1} = -3)\). We need to solve both to find all potential solutions.

Working with Rational Expressions

A rational expression is a fraction where both the numerator and the denominator are polynomials. In our problem, we deal with \(\frac{6x+1}{x-1}\), which has a polynomial numerator and denominator. To solve rational expressions effectively, follow these steps:

• Simplify the expression as much as possible.
• Identify and handle any values that make the denominator zero, as these are restrictions on the variable.
• Solve for the variable as usual.

Keeping these points in mind helps navigate the complexities of rational expressions. When solving the given equation, recognizing that the denominator cannot be zero is crucial. Therefore, we must ensure that our solutions do not fall into those restricted values.

Solving Absolute Value Equations Step by Step

To solve an absolute value equation such as \(|\(\frac{6x+1}{x-1}\)| = 3\), break it down into two separate equations, \((\frac{6x+1}{x-1} = 3)\) and \((\frac{6x+1}{x-1} = -3)\).
For the first equation, multiply both sides by \(x-1\) to clear the denominator:
\[6x + 1 = 3(x-1)\]
Simplify:
\[6x + 1 = 3x - 3\]
Combine like terms:
\[6x - 3x = -3 - 1\]
\[3x = -4\]
\[x = -\frac{4}{3}\]
For the second equation, follow similar steps:
\[6x + 1 = -3(x-1)\]
Simplify:
\[6x + 1 = -3x + 3\]
Combine like terms:
\[6x + 3x = 3 - 1\]
\[9x = 2\]
\[x = \frac{2}{9}\]
Thus, our potential solutions are \(-\frac{4}{3}\) and \(-\frac{2}{9}\).

Verification of Solutions

After solving the equations, it is essential to verify that the solutions are valid and do not make the denominator zero. Substituting the solutions back into the original equation will help determine their validity:

• For \(x = -\frac{4}{3}\): Substituting results in the denominator \(\frac{-4}{3}-1 = \frac{-4}{3} - \frac{3}{3} = \frac{-7}{3}\), which is non-zero.
• For \(x = \frac{2}{9}\): Substituting results in the denominator \(\frac{2}{9}-1 = \frac{2}{9} - \frac{9}{9} = \frac{-7}{9}\), which is also non-zero.

Since neither solution makes the denominator zero, both \(-\frac{4}{3}\) and \(\frac{2}{9}\) are valid solutions. Always verify your solutions to ensure they are applicable and do not violate any conditions originally set by the equation.