Problem 67 - Beginning and Intermediate Algebra

Question

Solve each equation. Check the solutions. \(2+\frac{5}{3 x-1}=\frac{-2}{(3 x-1)^{2}}\)

Step-by-Step Solution

Answer

The solutions are \(x = \frac{1}{6}\) and \(x = \frac{-1}{3}\). Both satisfy the original equation.

Step 1: Isolate the Fractions

First, isolate the fractions on one side of the equation. The given equation is:\[2 + \frac{5}{3x - 1} = \frac{-2}{(3x - 1)^2}\]Subtract 2 from both sides to isolate the fraction:\[\frac{5}{3x - 1} = \frac{-2}{(3x - 1)^2} - 2\]

Step 2: Find a Common Denominator

To combine the terms on the right-hand side, find a common denominator. The common denominator is \((3x - 1)^2\):\[\frac{5}{3x - 1} = \frac{-2 - 2(3x - 1)^2}{(3x - 1)^2}\]

Step 3: Simplify the Equation

Simplify the right-hand side:\[\frac{5}{3x - 1} = \frac{-2 - 2(3x - 1)^2}{(3x - 1)^2}\]Expand and simplify the numerator:\[\frac{5}{3x - 1} = \frac{-2 - 2(9x^2 - 6x + 1)}{(3x - 1)^2}\]\[\frac{5}{3x - 1} = \frac{-2 - 18x^2 + 12x - 2}{(3x - 1)^2}\]\[\frac{5}{3x - 1} = \frac{-18x^2 + 12x - 4}{(3x - 1)^2}\]

Step 4: Cross-Multiply to Solve for x

Cross-multiply to solve for \(x\):\[5(3x - 1) = -18x^2 + 12x - 4\]Expand and simplify the equation:\[15x - 5 = -18x^2 + 12x - 4\]Combine like terms to set the equation to zero:\[18x^2 + 3x - 1 = 0\]

Step 5: Solve the Quadratic Equation

Solve the quadratic equation using the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]For \(18x^2 + 3x - 1 = 0\), \(a = 18\), \(b = 3\), and \(c = -1\):\[x = \frac{-3 \pm \sqrt{3^2 - 4(18)(-1)}}{2(18)}\]Simplify under the square root:\[x = \frac{-3 \pm \sqrt{9 + 72}}{36}\]\[x = \frac{-3 \pm \sqrt{81}}{36}\]\[x = \frac{-3 \pm 9}{36}\]Therefore, the solutions are:\[x = \frac{6}{36} = \frac{1}{6}\] and \[x = \frac{-12}{36} = \frac{-1}{3}\]

Step 6: Check the Solutions

Substitute \(x = \frac{1}{6}\) into the original equation to check the solution:\[2 + \frac{5}{3(\frac{1}{6}) - 1} = \frac{-2}{(3(\frac{1}{6}) - 1)^2}\]\[2 + \frac{5}{\frac{1}{2} - 1} = \frac{-2}{(\frac{1}{2} - 1)^2}\]\[2 + \frac{5}{\frac{-1}{2}} = \frac{-2}{(\frac{-1}{2})^2}\]\[2 - 10 = -2 / \frac{1}{4}\]\[-8 = -8\] (True)Now substitute \(x = \frac{-1}{3}\):\[2 + \frac{5}{3(\frac{-1}{3}) - 1} = \frac{-2}{(3(\frac{-1}{3}) - 1)^2}\]\[2 + \frac{5}{-1 - 1} = \frac{-2}{(-1 - 1)^2}\]\[2 + \frac{5}{-2} = \frac{-2}{4}\]\[2 - 2.5 = - \frac{1}{2}\]\[-0.5 = -0.5\] (True)Both solutions are correct.

Key Concepts

quadratic equations

Quadratic equations are a core element of algebra. They take the form: \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. In our problem, we ended up with the quadratic equation \(18x^2 + 3x - 1 = 0\). To solve it, we used the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). This equation requires us to:

• Identify \(a\), \(b\), and \(c\) from the quadratic equation
• Plug these values into the formula
• Simplify the equation under the square root
• Calculate the potential solutions for \(x\)

For our equation with \(a = 18\), \(b = 3\), and \(c = -1\), we found: \(x = \frac{1}{6}\) and \(x = \frac{-1}{3}\). If you follow these steps carefully, solving quadratic equations becomes easier.

common denominators

When you're dealing with rational expressions, it’s essential to find a common denominator to combine fractions. In our exercise, we needed to combine the fractions on the right side of the equation, which required a common denominator: \((3x - 1)^2\).

• Rewrite each fraction with this common denominator
• Convert \(2\) to a fraction by multiplying it with the common denominator, leading to: \(\frac{-2 - 2(3x - 1)^2}{(3x - 1)^2}\).

Combining fractions becomes simple when we use a single common denominator. This step helps in simplifying and solving the entire equation more systematically.

cross-multiplication

Cross-multiplication is a method to solve equations that involve fractions. Once we simplified the given equation to have a single fraction on each side, we used cross-multiplication to eliminate the denominators:

• Multiply the numerator of one side by the denominator of the other side
• Set them equal to each other

For example, in our problem: \(\frac{5}{3x-1} = \frac{-18x^2+12x-4}{(3x-1)^2}\), we cross-multiplied to get: \(5(3x-1) = -18x^2+12x-4\). This simplification step allowed us to transition to a quadratic equation, which we then solved using the quadratic formula.

checking solutions

After finding potential solutions for \(x\), the final step was to check these solutions by substituting them back into the original equation. This ensures that our solutions are indeed correct. For each potential \(x\):

• Substitute it back into the original equation
• Simplify both sides to verify they are equal

For \(x = \frac{1}{6}\): \[2 + \frac{5}{3(\frac{1}{6}) - 1} = \frac{-2}{(3(\frac{1}{6}) - 1)^2}\]\[2 - 10 = -2 / \frac{1}{4}\]\[-8 = -8\] (True).Similarly, we checked \(x = \frac{-1}{3}\) and confirmed it was valid. This step is crucial in verifying our solutions are correct.