Problem 16 - College Algebra with Modeling and Visualization

Question

Find all real solutions. Check your results. $$ \frac{1}{x+2}+\frac{1}{x+3}=\frac{2}{x^{2}+5 x+6} $$

Step-by-Step Solution

Answer

The real solution is \(x = -\frac{3}{2}\).

Step 1: Simplify the Denominator

First, observe that the denominator on the right side of the equation, \(x^2 + 5x + 6\), can be factored. This expression factors into \((x+2)(x+3)\).

Step 2: Rewrite the Equation

After factoring, the equation can be rewritten as: \[ \frac{1}{x+2} + \frac{1}{x+3} = \frac{2}{(x+2)(x+3)} \] This shows the equivalence of denominators on both sides of the equation.

Step 3: Apply Common Denominator

To combine the left side, note that the common denominator is \((x+2)(x+3)\): \[ \frac{x+3}{(x+2)(x+3)} + \frac{x+2}{(x+2)(x+3)} \].

Step 4: Combine the Fractions

Combine the fractions on the left side:\[ \frac{(x+3) + (x+2)}{(x+2)(x+3)} = \frac{2x + 5}{(x+2)(x+3)} \].

Step 5: Equating Numerators

Now that the denominators are the same, set the numerators equal:\[ 2x + 5 = 2 \].

Step 6: Solve for x

Subtract 5 from both sides:\[ 2x = -3 \]. Then divide by 2:\[ x = -\frac{3}{2} \].

Step 7: Check for Extraneous Solutions

Substitute \(x = -\frac{3}{2}\) into the original equation to verify it does not make any term undefined and satisfies the equation.

Step 8: Verify the Solution

Since \(-\frac{3}{2}\) does not make any denominator zero, substitute back to check equality:Compute \(\frac{1}{-\frac{3}{2} + 2} + \frac{1}{-\frac{3}{2} + 3} = \frac{1}{\frac{1}{2}} + \frac{1}{\frac{9}{6}} = 2 + \frac{2}{3} \), calculate the right side, it simplifies to \(2 + \frac{2}{3}\), verifying the solution is correct.

Key Concepts

Factoring

Factoring is a powerful mathematical technique that allows us to simplify expressions and solve equations. In our exercise, the expression in the denominator of the original equation is a quadratic polynomial: \( x^2 + 5x + 6 \). To factor this, we need to find two numbers that multiply to 6 (the constant term) and add up to 5 (the linear coefficient). These numbers are 2 and 3.Hence, \( x^2 + 5x + 6 \) can be factored into \( (x+2)(x+3) \). This transformation is crucial because it reveals common factors on the left and right-hand side of the equation. By identifying and using these factors, we can simplify the equation further by ensuring the denominators are the same, making it easier to solve. Factoring turns a seemingly complex quadratic expression into a pair of linear expressions that are much simpler to handle.

Common Denominator

A common denominator is essential when working with equations that involve adding or subtracting fractions. In this problem, once we've factored \( x^2 + 5x + 6 \) into \( (x+2)(x+3) \), we use this as a common denominator for both sides of the equation.On the left-hand side, we have two separate fractions: \( \frac{1}{x+2} \) and \( \frac{1}{x+3} \). To combine them, we adjust these fractions so they each have the common denominator of \( (x+2)(x+3) \). We do this by:

• Multiplying \( \frac{1}{x+2} \) by \( \frac{x+3}{x+3} \) to get \( \frac{x+3}{(x+2)(x+3)} \)
• Multiplying \( \frac{1}{x+3} \) by \( \frac{x+2}{x+2} \) to get \( \frac{x+2}{(x+2)(x+3)} \)

With the denominators now the same, the fractions can be added together, combining their numerators over this common denominator. This step is key to simplifying the equation and proceeding toward finding a solution.

Extraneous Solutions

Extraneous solutions are solutions that result from the process of solving the equation but do not actually satisfy the original equation. This often happens when dealing with fractions or radical equations.In our problem, after finding a potential solution, \( x = -\frac{3}{2} \), it's vital to check that this value doesn't make any part of the original fractions undefined. The denominators \( x+2 \) and \( x+3 \) should never be zero; otherwise, the fractions would be undefined. For the potential solution, plugging \( x = -\frac{3}{2} \) into these expressions:

• \( -\frac{3}{2} + 2 = \frac{1}{2} \)
• \( -\frac{3}{2} + 3 = \frac{3}{2} \)

Neither results in zero, indicating \( x = -\frac{3}{2} \) is not extraneous. Finally, verify by substituting \( x = -\frac{3}{2} \) back into the original equation to ensure the equality holds, confirming that this solution is indeed valid.