Problem 14 - Algebra for College Students

Question

Find the solution set to each equation. $$\frac{x}{x-5}+\frac{5}{x}=\frac{11}{6}$$

Step-by-Step Solution

Answer

x = 15 or x = 2

Step 1: Identify the common denominator

To combine the fractions, identify the common denominator. The common denominator for \(\frac{x}{x-5}\), \(\frac{5}{x}\), and \(\frac{11}{6}\) is \(6x(x-5)\).

Step 2: Multiply each term by the common denominator

Multiply each term of the equation by \(6x(x-5)\) to eliminate the fractions: \[6x(x-5) \times \frac{x}{x-5} + 6x(x-5) \times \frac{5}{x} = 6x(x-5) \times \frac{11}{6}\].

Step 3: Simplify the equation

Simplify each term after multiplying: \[6x \times x + 6(x-5) \times 5 = 11x(x-5)\]. This simplifies to \[6x^2 + 30(x - 5) = 11x(x - 5)\].

Step 4: Distribute and combine like terms

Distribute and then combine like terms: \[6x^2 + 30x - 150 = 11x^2 - 55x\], which simplifies to \[0 = 5x^2 - 85x + 150\].

Step 5: Solve the quadratic equation

Solve the quadratic equation \(5x^2 - 85x + 150 = 0\) using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 5\), \(b = -85\), and \(c = 150\).

Step 6: Calculate the discriminant

First, calculate the discriminant: \[b^2 - 4ac = (-85)^2 - 4(5)(150) = 7225 - 3000 = 4225\].

Step 7: Compute the solutions

Compute the solutions using the quadratic formula: \[x = \frac{85 \pm \sqrt{4225}}{10} = \frac{85 \pm 65}{10}\]. Therefore, \[x = 15\] or \[x = 2\].

Step 8: Verify the solutions

Verify each solution by substituting back into the original equation. When \(x = 15\), the equation is true. When \(x = 2\), the equation is also true. Thus, both solutions are valid.

Key Concepts

Common Denominator

Understanding how to find a common denominator is key in solving rational equations. A common denominator is a shared multiple of the denominators in each fraction. It allows us to combine fractions into a single fraction, making equations easier to solve.

For the equation \(\frac{x}{x-5} + \frac{5}{x} = \frac{11}{6}\), the denominators are \(x-5\), \(x\), and \(6\). The common denominator must be a multiple of all these values. Here, it's \(6x(x-5)\). By multiplying each term by this common denominator, we eliminate the fractions, which simplifies the equation significantly.

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations of the form \(ax^2 + bx + c = 0\). The formula is:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \(a\), \(b\), and \(c\) represent coefficients from the quadratic equation. For example, in the equation \(5x^2 - 85x + 150 = 0\), we identify \(a = 5\), \(b = -85\), and \(c = 150\). Plug these values into the formula to find the solutions. It’s important because it provides a direct way to solve for \(x\), especially when factoring is complex or impossible.

Discriminant

The discriminant is part of the quadratic formula and determines the nature of the roots of the equation. It is calculated by the expression \(b^2 - 4ac\).

In our example, \(b = -85\), \(a = 5\), and \(c = 150\), so the discriminant is:

\[ (-85)^2 - 4(5)(150) = 7225 - 3000 = 4225 \]

The discriminant helps us understand the type of solutions:

• If the discriminant is positive, there are two real solutions.
• If it is zero, there is one real solution.
• If it is negative, there are no real solutions (only complex ones).

In this exercise, the discriminant is positive (\(4225\)), indicating two distinct real solutions.

Verification of Solutions

After you find potential solutions, it's important to verify them by substituting back into the original equation. This step ensures that the solutions are correct and applicable to the problem.

For the values \(x = 15\) and \(x = 2\), substitute them back:

• \(x = 15\): \[\frac{15}{15-5} + \frac{5}{15} = \frac{11}{6} \]
  This simplifies to: \[\frac{15}{10} + \frac{1}{3} = \frac{9}{6} + \frac{2}{6} = \frac{11}{6} \]
  
• \(x = 2\): \[\frac{2}{2-5} + \frac{5}{2} = \frac{11}{6} \]
  This simplifies to: \[-\frac{2}{3} + \frac{2.5}{3} = \frac{-2}{3} + \frac{3.5}{3} = \frac{1.5}{3} = \frac{11}{6} \]

Both values satisfy the equation, confirming them as valid solutions. Verification helps avoid errors and confirms the correctness of your solutions.