Problem 13 - Algebra 2

Question

Solve each equation or inequality. Check your solutions. $$ a+1=\frac{6}{a} $$

Step-by-Step Solution

Answer

The solutions are \( a = -3 \) and \( a = 2 \).

Step 1: Clear the Fraction

First, we want to eliminate the fraction by multiplying both sides of the equation by the denominator, \( a \). This gives us: \[ a(a + 1) = a \cdot \frac{6}{a} \]Simplifying the right side, we have:\[ a^2 + a = 6 \]

Step 2: Rearrange the Equation

We need to form a standard quadratic equation by setting the equation equal to zero:\[ a^2 + a - 6 = 0 \]

Step 3: Factor the Quadratic

We next factor the quadratic equation. We need two numbers that multiply to \(-6\) and add to \(1\). The numbers \(3\) and \(-2\) work:\[ (a + 3)(a - 2) = 0 \]

Step 4: Solve for the Values of 'a'

Now, set each factor equal to zero and solve for \(a\):\[ a + 3 = 0 \quad \Rightarrow \quad a = -3 \]\[ a - 2 = 0 \quad \Rightarrow \quad a = 2 \]

Step 5: Check the Solutions

Substitute each solution back into the original equation to check:For \(a = -3\):\[ -3 + 1 = \frac{6}{-3} \]Simplifying both sides gives:\[ -2 = -2 \quad \text{(True)} \]For \(a = 2\):\[ 2 + 1 = \frac{6}{2} \]Simplifying both sides gives:\[ 3 = 3 \quad \text{(True)} \]

Step 6: List the Solutions

Since both solutions satisfy the original equation, the final solutions are:\( a = -3 \) and \( a = 2 \).

Key Concepts

Factoring

Factoring is a fundamental skill when dealing with quadratic equations, such as in this exercise. Once we have a quadratic in the standard form, like \[ a^2 + a - 6 = 0 \]we aim to express it as the product of two linear expressions. Consider, in particular, the term's coefficient and the constant that require careful selection of two numbers. We need numbers that multiply to \(-6\) and add up to the \(1\) in front of the linear term. In this problem, the numbers \(3\) and \(-2\) are perfect as they meet both conditions:

• \(3 \times (-2) = -6\)
• \(3 + (-2) = 1\)

By rewriting the quadratic as \( (a + 3)(a - 2) \), we can effectively break down and solve it swiftly, making successful factoring crucial in solving the equation.

Checking Solutions

Once we solve the quadratic equation by factoring or another method, it's crucial to verify each solution directly against the original equation. For our equation \(a + 1 = \frac{6}{a}\), the checking process ensures that no solutions were introduced that don't actually satisfy the original condition. To check, substitute each solution back into the original equation:

• For \(a = -3\), substituting gives \(-3 + 1 = \frac{6}{-3}\), which simplifies to\(-2 = -2\).
• For \(a = 2\), substituting gives \(2 + 1 = \frac{6}{2}\), which simplifies to\(3 = 3\).

Both checks confirm the correctness of our solutions. This step helps avoid false positives, especially if extraneous solutions are inadvertently introduced.

Eliminating Fractions

Dealing with equations that have fractions can be tricky, and cleaning them up is often a good initial step. In the original problem\(a + 1 = \frac{6}{a}\), we eliminate the fraction to simplify our work. By multiplying every term by \(a\), we simultaneously clear the denominator and create a polynomial:

• This leads to the product \(a(a + 1) = a \cdot \frac{6}{a}\).
• After simplification, the equation becomes \(a^2 + a = 6\), removing the fractional expression makes the expression solvable through straightforward algebraic techniques.

This simplification also helps avoid errors that can arise from the presence of fractions by converting the entire equation into an easier form to handle.

Solving Equations

In solving equations, especially quadratics, organizing your approach systematically is key. Our exercise started by clearing fractions, which boiled it down to:\[ a^2 + a - 6 = 0 \].From there, solving the equation involved organized steps of:

• Factoring: We broke it into simpler terms of \((a + 3)(a - 2) = 0\).
• Isolating the variable: Solving each part, \(a + 3 = 0\) and \(a - 2 = 0\), gave potential solutions.
• Checking: Verification confirmed that both \(a = -3\) and \(a = 2\) satisfied the initial equation.

Whether it's factoring, using the quadratic formula, or any other method, the process needs clear steps. This ensures accuracy and covers each possible solution.