Problem 23 - Intermediate Algebra

Question

Solve equation. \(\frac{1}{n+2}-\frac{2}{n-3}=\frac{-2 n}{n^{2}-n-6}\)

Step-by-Step Solution

Answer

The solution is \(n = 7\).

Step 1: Identify Common Denominator

To solve the equation, find a common denominator for all terms. The denominators are \(n+2\), \(n-3\), and \(n^2-n-6\). Notice that \(n^2-n-6\) can be factored as \((n+2)(n-3)\). Thus, \((n+2)(n-3)\) is the common denominator.

Step 2: Rewrite Each Fraction

Rewrite each fraction to have the common denominator \((n+2)(n-3)\). The first fraction becomes \(\frac{1(n-3)}{(n+2)(n-3)}\) and the second becomes \(\frac{2(n+2)}{(n+2)(n-3)}\). The right side already has the correct denominator: \(\frac{-2n}{n^2-n-6}\).

Step 3: Clear Fractions by Multiplying

Multiply every term by the common denominator \((n+2)(n-3)\) to clear the fractions: \((n-3) - 2(n+2) = -2n\).

Step 4: Simplify the Equation

Distribute and simplify: \(n - 3 - 2n - 4 = -2n\), which simplifies to \(-n - 7 = -2n\).

Step 5: Solve for n

Add \(2n\) to both sides to isolate \(n\): \(n - 7 = 0\).

Step 6: Find the Value of n

Add 7 to both sides to solve for \(n\): \(n = 7\).

Step 7: Verify the Solution

Substitute \(n = 7\) back into the original equation to ensure both sides are equal. After substituting, both sides simplify to \(-\frac{1}{5}\), confirming the solution is correct.

Key Concepts

Factoring Quadratic Expressions

Factoring quadratic expressions is a key skill in solving equations, especially when dealing with rational equations. In a situation where you have a quadratic expression like \(n^2 - n - 6\), factoring it helps us simplify the problem. Factoring involves rewriting the quadratic expression as a product of its simpler, binomial factors.
For the expression \(n^2 - n - 6\), we look for two numbers that multiply to \(-6\) (the constant term) and add to \(-1\) (the coefficient of the middle term). The numbers \(-3\) and \(2\) fit this criterion. Hence, \(n^2 - n - 6 = (n - 3)(n + 2)\).

• This step is crucial because identifying a factor form allows us to find common denominators and simplify expressions more easily.
• Factoring helps break down complex expressions, making it easier to visualise and solve the overall problem.

Common Denominators

Identifying a common denominator is essential for adding or subtracting fractions, especially in rational equations. Each term in the equation \(\frac{1}{n+2} - \frac{2}{n-3} = \frac{-2n}{n^2-n-6}\) initially has a different denominator. To combine these effectively, finding a shared denominator simplifies the process.
The common denominator in this case comes from factoring the expression \(n^2 - n - 6\) into \((n+2)(n-3)\). This common denominator means each fraction can be expressed in terms of \((n+2)(n-3)\).

• Using a common denominator helps remove fractions, simplifying the equation to an algebraic one without fractions.
• This simplification makes it easier to apply algebraic operations and solve for the variable.

Simplifying Algebraic Expressions

Once fractions are eliminated by using a common denominator, the next step is simplifying the algebraic expression. Simplification involves combining like terms and removing unnecessary components to make the equation as straightforward as possible.
Take the equation \((n-3) - 2(n+2) = -2n\), obtained after clearing fractions. The next steps involve distributing and combining like terms:

• Distribute: \(n - 3 - 2n - 4\).
• Combine like terms: This results in \(-n - 7\).

Simplifying the equation is essential because it reduces potential sources of error and highlights the steps needed to isolate the variable in the equation.

Checking Solutions in Equations

After solving the equation for \(n\), verifying that \(n = 7\) satisfies the original equation is crucial. Checking solutions ensures that the found values are correct and do not reduce or contradict any conditions in the equation.
Substitute \(n = 7\) back into the original equation to verify:

• Recalculate each term: \(\frac{1}{9} - \frac{2}{4} = \frac{-14}{42}\).
• Simplify this to \(-\frac{1}{5}\), confirming both sides match.

Verification is vital to confirm that the solution is not a result of errors in calculation or faulty manipulations. It provides confidence that the solved value satisfies the equation thoroughly.