Problem 39 - Algebra for College Students

Question

For Problems \(1-44\), solve each equation. $$ \frac{3 n}{n-1}-\frac{1}{3}=\frac{-40}{3 n-18} $$

Step-by-Step Solution

Answer

The solutions are approximately \(n \approx 1.446\) and \(n \approx -15.446\).

Step 1: Identify Common Denominators

To combine the fractions, find a common denominator. The denominators are \(n-1\), 3, and \(3n-18\). Notice that \(3n-18 = 3(n-6)\). So, the common denominator is \(3(n-1)(n-6)\).

Step 2: Multiply Through by the Common Denominator

Eliminate the fractions by multiplying every term by the common denominator, \(3(n-1)(n-6)\):\[3(n-1)(n-6) \cdot \frac{3n}{n-1} - 3(n-1)(n-6) \cdot \frac{1}{3} = 3(n-1)(n-6) \cdot \frac{-40}{3(n-6)}\].This simplifies to:\[3n(n-6) - (n-1)(n-6) = -40(n-1)\].

Step 3: Expand and Simplify

Expand each term:- \(3n(n-6) = 3n^2 - 18n\)- \((n-1)(n-6) = n^2 - 7n + 6\)Substitute these to get:\[3n^2 - 18n - n^2 + 7n - 6 = -40n + 40\].Combine like terms to simplify the equation:\[2n^2 - 11n - 6 = -40n + 40\].

Step 4: Rearrange into a Quadratic Equation

Bring all terms to one side to set the equation to zero:\[2n^2 - 11n - 6 + 40n - 40 = 0\],which simplifies to:\[2n^2 + 29n - 46 = 0\].

Step 5: Solve the Quadratic Equation

Use the quadratic formula, \(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), with \(a = 2\), \(b = 29\), and \(c = -46\):\[n = \frac{-29 \pm \sqrt{29^2 - 4 \cdot 2 \cdot (-46)}}{2 \cdot 2}\].Calculate the discriminant:\[29^2 = 841, \quad 4 \cdot 2 \cdot (-46) = -368\],\[841 + 368 = 1209\].Find the square root and solve:\[n = \frac{-29 \pm \sqrt{1209}}{4}\].

Step 6: Approximate and Verify Solutions

Calculate \(\sqrt{1209} \approx 34.785\). Find approximate solutions:\[n_1 = \frac{-29 + 34.785}{4} \approx 1.446\],\[n_2 = \frac{-29 - 34.785}{4} \approx -15.446\].Verify that both solutions satisfy the original equation. Check for undefined values like \(n = 1\) or \(n = 6\), and ensure neither is the solution, especially since the solutions don't equate to these values.

Key Concepts

Common Denominators

When dealing with fractions in equations, finding a common denominator is essential for simplifying and solving the problem. A common denominator allows you to combine or compare fractions effectively. In the given problem, the denominators were \( n-1 \), 3, and \( 3n-18 \). Notice that \( 3n-18 \) can be factored into \( 3(n-6) \). This helps in identifying a least common denominator, which is \( 3(n-1)(n-6) \).

Understanding how to find common denominators is crucial. Here's how to approach it:

• Factor the denominators, if possible, to find common factors.
• Make sure the common denominator includes all distinct factors from each fraction.
• Multiply each term in the equation by the common denominator to clear the fractions.

Establishing a combined denominator streamlines the equation and prepares it for solving.

Fractions

Manipulating fractions is an integral part of solving equations, especially when they're mixed with variables. Let's break down how fractions were handled in this exercise:

When you see fractions, start by trying to eliminate them. By multiplying each fraction by the common denominator found in the previous section, you can simplify the equation. Here's what to do:

• Convert every fraction to a form where it shares a common denominator with the others.
• Multiply every term by this common denominator to eliminate the fractions.
  For example, multiplying \( \frac{3n}{n-1} \) by the common denominator \( 3(n-1)(n-6) \) simplifies the term to \( 3n(n-6) \).

Removing fractions in this way simplifies the equation into polynomial form, making it easier to focus on solving for the variable.

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations, which are equations in the form of \( ax^2 + bx + c = 0 \). In this problem, the equation was boiled down to a quadratic form: \( 2n^2 + 29n - 46 = 0 \).

Here's the formula itself, a reliable way to find the solution(s) for any quadratic equation:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In this equation:

• \( a \) is the coefficient of \( n^2 \)
• \( b \) is the coefficient of \( n \)
• \( c \) is the constant term

The discriminant \( b^2 - 4ac \) determines the nature of the roots:

• If it's positive, you'll find two distinct real roots.
• If zero, you'll have exactly one real root (a repeated root).
• If negative, the roots are complex and not real.

For this problem, substituting \( a = 2 \), \( b = 29 \), and \( c = -46 \) into the formula provided the roots of \( n \). Learning this formula makes solving quadratics straightforward, particularly when factoring isn't viable.

Roots of Equations

Roots of an equation are the solutions that satisfy the equation, meaning that if you substitute them back into the equation, they should uphold the equality. In quadratic equations, the number of roots can vary but never exceeds two.

Think of roots as the points where the corresponding quadratic function crosses the x-axis on a graph. In the solved equation, the roots were calculated using the quadratic formula, which yielded:

• \( n_1 \approx 1.446 \)
• \( n_2 \approx -15.446 \)

To verify, plug these values back into the original equation to ensure they satisfy it. Note that certain roots might be excluded if they cause division by zero in the original fractions.

Understanding roots is vital, as they provide critical points that define the function's behavior and solutions.