Problem 17 - Algebra 2 and Trigonometry

Question

In \(3-20,\) solve each equation and check. $$ 1=\frac{5}{x+3}+\frac{5}{(x+2)(x+3)} $$

Step-by-Step Solution

Answer

The solution to the equation is \(x=3\).

Step 1: Eliminate the denominator

To solve the equation \(1 = \frac{5}{x+3} + \frac{5}{(x+2)(x+3)}\), first eliminate the denominators by finding the common denominator. The common denominator here is \((x+2)(x+3)\). Multiply every term of the equation by this common denominator:\[ (x+2)(x+3) imes 1 = (x+2)(x+3) \times \frac{5}{x+3} + (x+2)(x+3) \times \frac{5}{(x+2)(x+3)} \]This simplifies to: \[ (x+2)(x+3) = 5(x+2) + 5 \]

Step 2: Expand and simplify

Now, expand both sides of the equation:\[ x^2 + 5x + 6 = 5x + 10 + 5 \]This simplifies to:\[ x^2 + 5x + 6 = 5x + 15 \]

Step 3: Isolate the quadratic

To isolate the quadratic term, subtract \(5x + 15\) from both sides:\[ x^2 + 5x + 6 - 5x - 15 = 0 \]This simplifies to:\[ x^2 - 9 = 0 \]

Step 4: Solve for x

Now solve the equation \(x^2 - 9 = 0 \) by using the difference of squares formula:\[ (x-3)(x+3) = 0 \]Set each factor equal to zero:\[ x-3=0 \quad \text{or} \quad x+3=0 \]Solving these gives:\[ x = 3 \quad \text{or} \quad x = -3 \]

Step 5: Verify the solutions

Substitute \(x=3\) into the original equation to check:\[ 1 = \frac{5}{3+3} + \frac{5}{(3+2)(3+3)} \]Which simplifies to:\[ 1 = \frac{5}{6} + \frac{5}{30} \]Simplifying the right-hand side:\[ 1 = \frac{5}{6} + \frac{1}{6} \]\[ 1 = 1 \]Now check \(x=-3\):This substitution will lead to division by zero in the first term, so it's not valid.Thus, \(x=3\) is the valid solution.

Key Concepts

Common Denominator

When solving rational equations, one of the first steps is to eliminate fractions by finding a common denominator. Let's break this down. A common denominator is a shared multiple of the denominators in your equation. This allows you to rewrite the equation without fractions by multiplying through by this common denominator.
In our example, we deal with two terms: \(\frac{5}{x+3}\) and \(\frac{5}{(x+2)(x+3)}\). Here, the denominators are \((x+3)\) and \((x+2)(x+3)\). The least common denominator for these would be \((x+2)(x+3)\), covering all parts of the denominators involved.
We then multiply every term in the equation by this common denominator, ensuring every numerator gets multiplied by \((x+2)(x+3)\). This step transforms the equation, eliminating fractions and allowing us to proceed with solving it.

Quadratic Equations

Quadratic equations are polynomial equations of degree two and often come in the form \(ax^2 + bx + c = 0\). Solving these equations is a key skill, and they can emerge from more complex algebra problems, much like in our exercise.
After using a common denominator to simplify the rational expression, we expanded and simplified the equation to reach \(x^2 + 5x + 6 = 5x + 15\). Further simplification led us to \(x^2 - 9 = 0\).
This is a classic example of a quadratic equation where \(a = 1\), \(b = 0\), and \(c = -9\). Recognizing this form is crucial for using algebraic methods to find solutions, such as factoring, completing the square, or using the quadratic formula.

Difference of Squares

Difference of squares is a special factoring technique used for expressions like \(x^2 - a^2\), which can be rewritten as \((x-a)(x+a)\). This formula is handy in simplifying polynomials and solving equations.
In our problem, we arrived at the expression \(x^2 - 9 = 0\), which fits the pattern of a difference of squares since it can be expressed as \(x^2 - 3^2\). This allows us to factor it into \((x-3)(x+3) = 0\).
Solving each factor separately gives the potential solutions \(x=3\) and \(x=-3\). Applying the difference of squares efficiently breaks down the problem and helps in finding neat solutions quickly.

Checking Solutions

Once solutions are determined, it is vital to validate them by substituting back into the original equation. This step checks for any extraneous solutions, often introduced when dealing with rational equations or when the original equation might have undefined points (such as division by zero).
For our example, substituting \(x = 3\) back into the original equation confirms the equality: \(1 = \frac{5}{3+3} + \frac{5}{(3+2)(3+3)}\), simplifying effectively to \(1 = 1\). This shows it is a valid solution.
On the other hand, substituting \(x = -3\) results in division by zero in one of the terms, which shows that \(x = -3\) is not a valid solution. Hence, checking prevents incorrect answers from being accepted and ensures consistency in problem-solving.