Problem 40 - Algebra for College Students

Question

For Problems \(1-44\), solve each equation. $$ \frac{n}{n+1}+\frac{1}{2}=\frac{-2}{n+2} $$

Step-by-Step Solution

Answer

The solution is \(n = -3\).

Step 1: Identify the LCD

The first step is to identify the least common denominator (LCD) for all the fractions in the equation. The denominators in this problem are \(n+1\), \(2\), and \(n+2\). The LCD for these denominators is \((n+1)(n+2)\).

Step 2: Multiply through by the LCD

Multiply each term of the equation by the common denominator \((n+1)(n+2)\) to eliminate the fractions:\[(n+1)(n+2) \cdot \frac{n}{n+1} + (n+1)(n+2) \cdot \frac{1}{2} = (n+1)(n+2) \cdot \frac{-2}{n+2}\]

Step 3: Simplify each term

Simplify each fraction by canceling the common terms:- \((n+1)(n+2) \cdot \frac{n}{n+1} = n(n+2) = n^2 + 2n\)- \((n+1)(n+2) \cdot \frac{1}{2} = \frac{1}{2}(n^2+3n+2)\)- \((n+1)(n+2) \cdot \frac{-2}{n+2} = -2(n+1)\)

Step 4: Expand and collect like terms

Expand the terms:\[n^2 + 2n + \frac{1}{2}n^2 + \frac{3}{2}n + 1 = -2n - 2\]Combine like terms:\[\frac{3}{2}n^2 + \frac{7}{2}n + 1 = -2n - 2\]

Step 5: Move all terms to one side

Rearrange all terms to form a standard quadratic equation:\[\frac{3}{2}n^2 + \frac{7}{2}n + 1 + 2n + 2 = 0\]Combine like terms: \[\frac{3}{2}n^2 + \frac{11}{2}n + 3 = 0\]

Step 6: Simplify the quadratic equation

Multiply through by 2 to clear the fractions:\[3n^2 + 11n + 6 = 0\]

Step 7: Factor the quadratic equation

Factor the quadratic equation:\[3n^2 + 11n + 6 = (3n + 2)(n + 3) = 0\]

Step 8: Solve for n

Set each factor equal to zero and solve for \(n\):- From \(3n + 2 = 0\), \(n = -\frac{2}{3}\)- From \(n + 3 = 0\), \(n = -3\)

Step 9: Verify solutions in the original equation

Substitute \(n = -\frac{2}{3}\) and \(n = -3\) back into the original equation to ensure they are valid solutions. Substitute to check: Only \(n = -3\) is valid since substitution retains equation integrity.

Key Concepts

Understanding the Least Common Denominator in Rational Equations

When solving rational equations such as \(\frac{n}{n+1}+\frac{1}{2}=\frac{-2}{n+2}\), dealing with fractions can get a bit tricky. The neat trick here is to find the least common denominator (LCD) for the fractions involved. This helps in getting rid of the fractions and eases the path towards finding a solution.

• The denominators present in the problem are \(n+1\), \(2\), and \(n+2\).
• To solve the equation efficiently, we need to identify the LCD, which is the smallest expression that each denominator can divide into without leaving a remainder.
• In this example, the LCD is \((n+1)(n+2)\). This common factor allows us to simplify the equation by multiplying each term by it, effectively eliminating the fractions.

Once the fractions are gone, we're left with a polynomial equation that's much simpler and more pleasant to solve. Understanding how to find and use the LCD is pivotal in resolving equations involving fractions.

Quadratic Factoring: A Method to Solve Polynomial Equations

After clearing fractions using the LCD and simplifying the equation, you often end up with a quadratic equation. In the given equation, we obtained \(3n^2 + 11n + 6 = 0\). Quadratic factoring is a method used to find the roots, or solutions, of this equation.

• First, we look for two numbers that multiply to give the product of the quadratic term coefficient and the constant term (\(3 \times 6 = 18\)), and also add up to the linear coefficient (\(11\)).
• For \(3n^2 + 11n + 6 = 0\), these numbers are \(2\) and \(9\), leading us to rewrite the middle term (11n) as \(9n + 2n\).
• Then we factored these terms, allowing us to express the quadratic as the product of two binomials \((3n + 2)(n + 3) = 0\).

Factoring transforms the seemingly complex equation into easily solvable linear equations. Remember, not all quadratics are easily factorable. Sometimes, you might need the quadratic formula or a different method if simple integers don't work out.

Verify Solutions to Ensure Validity

In mathematics, verifying the solutions is always a crucial final step. It ensures that the solutions derived are indeed correct and applicable to the problem statement.

• Begin by substituting the obtained solutions back into the original equation provided in the exercise.
• When substituting \(n = -\frac{2}{3}\) and \(n = -3\) into \(\frac{n}{n+1}+\frac{1}{2}=\frac{-2}{n+2}\), only \(n = -3\) maintains the equality.
• This check is particularly important in rational equations where certain values can lead to undefined expressions when substituting back into denominators.

By following these verification steps, you assure the integrity and correctness of your solution, promoting a thorough understanding and avoiding potential pitfalls with extraneous solutions that do not fit the original equation.