Problem 48 - Algebra 2

Question

Solve each equation. Check each solution. $$ \frac{5}{x+2}=\frac{-1}{x^{2}+7 x+10}+\frac{3}{-x-5} $$

Step-by-Step Solution

Answer

The solution to the equation is \(x = -10\)

Step 1: Simplify the Fractional Equation

First, note that \(x^{2}+7 x+10\) can be factored to \((x+2)(x+5)\). Substitute this in the equation: \[\frac{5}{x+2}=-\frac{1}{(x+2)(x+5)}+\frac{3}{-x-5}\]Then, multiply the entire equation by \((x+2)(x+5)\) to remove the denominators, leaving:\[ 5(x + 5) = -1 + 3(x + 2)\]

Step 2: Solve for X

Now solve for x by first simplifying both sides: \[ 5x + 25 = -1 + 3x + 6\]Subtract \(3x\) from both sides:\[ 2x + 25 = -1 + 6\]Then subtract \(25\) from both sides: \[ 2x = -25 + 5\]Finally, divide by \(2\): \[ x = \frac{-20}{2}\]

Step 3: Check the Solution

To check the solution, substitute \(x = -10\) into the original equation and make sure that the left-hand side equals to the right-hand side.

Key Concepts

Factoring Quadratic Expressions

Factoring quadratic expressions is a foundational skill in algebra. It's often necessary when solving rational equations, especially when tackling denominators that involve polynomial expressions. To factor a quadratic expression like \(x^2 + 7x + 10\), you need to find two numbers that both add up to the middle coefficient (which is 7 in this case) and multiply to give the constant term (which is 10 here). The numbers 2 and 5 fit these criteria perfectly. Thus, this expression can be rewritten as \((x+2)(x+5)\).

Factoring converts a quadratic expression into a product of simpler linear expressions. This can simplify equations and make further computation easier. When these expressions are in the denominator of a fraction, factoring is often the first step in attempting to simplify or solve the equation. Pay close attention to signs while factoring, as a minus sign can completely change the extraction process.

Solving Linear Equations

Solving linear equations, once the denominators are eliminated, involves simplifying both sides until you isolate the variable. In the step-by-step solution, once we factored and multiplied everything by the common denominator \((x+2)(x+5)\), it resulted in a simpler equation: \(5(x + 5) = -1 + 3(x + 2)\). From here, algebraic manipulation takes over.

Here are the steps briefly outlined to isolate \(x\):

• Distribute constants into parentheses: \(5x + 25\) and \(3x + 6\).
• Combine like terms on both sides of the equation.
• Move all terms involving \(x\) to one side to easily isolate \(x\).
• Finally, simplify to find the value of \(x\), which in this case is \(-10\).

Solving linear equations is often straightforward once terms are simplified, emphasizing the importance of each step in the process. Carefully follow each step, checking algebraic manipulations to ensure accuracy.

Checking Solutions

After solving a rational equation, verifying the solution by substituting it back into the original equation is crucial. For our solution \(x = -10\), substitute \(-10\) back into the original equation to ensure the results balance.

• Substitute \(-10\) into the denominators of each term to check they don't become undefined. Since \(-10 + 2 = -8\) and \(-10 + 5 = -5\), there are no zero denominators, thus it's valid.
• Calculate the value of each fraction on both sides with \(x = -10\), ensuring each side simplifies to the same value.

Verifying solutions serves as a reliability check—an assurance that the calculations were performed correctly and that results hold true within the problem's context. This step is a good practice even in professional work, as it prevents errors and strengthens understanding.