Problem 36 - Algebra for College Students

Question

For Problems \(1-44\), solve each equation. $$ -1+\frac{2 x}{x+3}=\frac{-4}{x+4} $$

Step-by-Step Solution

Answer

The solutions are \( x = 0 \) and \( x = -5 \).

Step 1: Eliminate the Fractions

To eliminate the fractions, identify the common denominator, which in this case is \((x + 3)(x + 4)\). Multiply every term in the equation by this common denominator to eliminate the fractional components.

Step 2: Distribute and Simplify

Distribute the common denominator across each term. This results in:\[-(x + 3)(x + 4) + 2x(x + 4) = -4(x + 3)\]. Now, expand and simplify each term.

Step 3: Expand the Equation

Expand each term:\[-(x^2 + 7x + 12) + 2x^2 + 8x = -4x - 12\]. This simplifies to:\(-x^2 - 7x - 12 + 2x^2 + 8x = -4x - 12\).

Step 4: Combine Like Terms

Combine like terms on both sides. This results in:\[x^2 + x - 12 = -4x - 12\].

Step 5: Rearrange the Equation

Rearrange the equation to bring all terms to one side, resulting in:\[x^2 + 5x = 0\].

Step 6: Factor the Quadratic Equation

Factor the quadratic equation. The equation \[x^2 + 5x = 0\] can be factored as \[x(x + 5) = 0\].

Step 7: Solve for the Variable x

Using the zero product property, set each factor to zero and solve for \(x\): 1. \[x = 0\] 2. \[x + 5 = 0\] which implies \[x = -5\].

Key Concepts

Factoring Quadratic Equations

Factoring quadratic equations can sometimes feel tricky, but with practice, it becomes an effective tool for solving problems. A quadratic equation is generally in the form of \(ax^2+bx+c=0\), where \(a\), \(b\), and \(c\) are constants. In this particular exercise, we arrived at the quadratic \(x^2 + 5x = 0\).
The goal of factoring is to write the equation as a product of its factors. For example, in this equation, notice how there is an \(x\) in both terms? This common factor of \(x\) can be factored out.

• You take \(x^2 + 5x\) and simplify it as \(x(x + 5)\).
• This means the two numbers whose product is zero are \(x\) and \(x + 5\).

Factoring is particularly useful because it converts a polynomial into simpler terms, preparing us to apply the zero product property in the solving process.

Common Denominators

When dealing with rational equations, fractions can make the solving process complicated. Identifying a common denominator enables us to work without fractions. The common denominator has to be a multiple of all the denominators in the equation.
In our problem, the equation \(-1 + \frac{2x}{x+3} = \frac{-4}{x+4}\) involves two fractions. Here, the denominators are \(x+3\) and \(x+4\).

• The common denominator is \((x + 3)(x + 4)\).
• Once determined, every term in the equation is multiplied by this common denominator.

This allows you to "clear" the fractions, transforming the equation into one with whole numbers. This step is crucial as it simplifies the process and makes subsequent steps like distribution and factoring more straightforward.

Zero Product Property

The zero product property is a simple yet powerful tool in algebra. It states that if the product of two expressions is zero, then at least one of the expressions must be zero. In other words, if \(a \cdot b = 0\), then either \(a = 0\) or \(b = 0\).
In our equation, after factoring \(x^2 + 5x = 0\) into \(x(x + 5) = 0\), we directly apply this property.

• Set each factor equal to zero: \(x = 0\) and \(x + 5 = 0\).
• Solving these gives \(x = 0\) and \(x = -5\).

Using the zero product property simplifies finding solutions from a potentially complex quadratic equation. Each factor represents a possible solution, which you evaluate separately.

Eliminating Fractions

Eliminating fractions is a key step in solving equations that involve fractions. The process involves getting rid of the fractions to make the equation easier to solve, typically by finding and multiplying by a common denominator.
Once you have multiplied every term by this common denominator, the fractions are "cleared" and you can proceed with solving the equation as if it contained only integers. In the original problem:

• We found the common denominator \((x + 3)(x + 4)\).
• We multiplied each term by this to eliminate fractions.

Doing so provided a clearer path forward, allowing us to focus on distribution, combining like terms, and eventually factoring out the equation. This step makes the equation much more manageable, ensuring clarity moving forward.