Problem 15 - Beginning and Intermediate Algebra

Question

Solve each equation. Check the solutions. \(\frac{2}{x+1}+\frac{3}{x+2}=\frac{7}{2}\)

Step-by-Step Solution

Answer

The valid solution is \(x = 0\).

Step 1 - Find a common denominator

Identify the least common denominator (LCD) for the fractions. The denominators are \(x+1\), \(x+2\), and \2\. So, the least common denominator is \[ 2(x+1)(x+2) \].

Step 2 - Multiply each term by the LCD

Multiply all terms by the LCD \[ 2(x+1)(x+2) \] to eliminate the fractions:\[2(x+1)(x+2) \times \frac{2}{x+1} + 2(x+1)(x+2) \times \frac{3}{x+2} = 2(x+1)(x+2) \times \frac{7}{2}\ \Rightarrow 2 \times 2(x + 2) + 2 \times 3(x + 1) = 7(x + 1)(x + 2)\ \Rightarrow 4(x + 2) + 6(x + 1) = 7(x^2 + 3x + 2) \]

Step 3 - Simplify and combine like terms

Expand and combine like terms:\[4x + 8 + 6x + 6 = 7x^2 + 21x + 14\ \Rightarrow 10x + 14 = 7x^2 + 21x + 14 \]

Step 4 - Set up the quadratic equation

Move all terms to one side to set up the quadratic equation:\[0 = 7x^2 + 11x \]

Step 5 - Factor the quadratic equation

\[0 = x(7x + 11)\] Factor out \(x\):\[0 = x(7x + 11)\]

Step 6 - Solve for x

Set each factor equal to zero and solve:\[x = 0 \quad \text{or} \quad 7x + 11 = 0 \quad \text{which gives} \quad x = -\frac{11}{7}\]

Step 7 - Check the solutions

Substitute each solution back into the original equation to check:\1. For \(x = 0\): \ \ \[ \frac{2}{0+1} + \frac{3}{0+2} = \frac{2}{1} + \frac{3}{2} = 2 + \frac{3}{2} = \frac{7}{2} \] \ \ \is valid.\2. For \(x = -\frac{11}{7}\): Plugging in results in undefined fractions due to zero denominators. So, this solution is invalid.

Key Concepts

Common Denominator

When solving equations with fractions, finding a common denominator is essential. A common denominator allows you to combine or compare fractions easily. In this exercise, we have the denominators: \(x+1\), \(x+2\), and \(2\). The least common denominator (LCD) needs to include all these individual parts so that each fraction can be rewritten with this common base.

To find the LCD, take the product of each unique denominator component, resulting in \[ 2(x+1)(x+2) \]. By multiplying every term in the equation by the LCD, you effectively eliminate the fractions, making it easier to solve the equation. This technique simplifies complex rational equations and is a powerful tool for students to master.

Quadratic Equation

A quadratic equation is any equation that takes the form \( ax^2 + bx + c = 0 \), where \(a\), \(b\), and \(c\) are constants. In this exercise, after eliminating the fractions and combining like terms, we end up with:\[ 7x^2 + 11x = 0 \].

This is a classic quadratic equation form. Remember, solving a quadratic equation usually involves three main strategies:

• Factoring the quadratic expression
• Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \)
• Completing the square

Recognizing these different methods and choosing the most appropriate one based on the context is essential for efficiently solving such equations.

Factoring

Factoring involves breaking down a complex equation or expression into products of simpler factors. For the quadratic equation you arrived at: \[ 7x^2 + 11x = 0 \], it's apparent that both terms share a common factor of \(x\).

Thus, you can factor out \(x\) to get: \[ x(7x + 11) = 0 \]. This is a vital step because it translates the quadratic equation into a form that is straightforward to solve.

Next, set each factor equal to zero: \[ x = 0 \] and \[ 7x + 11 = 0 \]. Solving \( 7x + 11 = 0 \) gives \( x = -\frac{11}{7} \). Checking these solutions in the original equation will validate or invalidate them. Factoring thus remains a key skill in algebra, enabling students to parse and simplify complex expressions systematically.