- Combine the numerators of the fractions since they share a common denominator: \(\frac{x^2+6x+2+x+10}{5x+15} = \frac{x^2+7x+12}{5x+15}\).
- Factor both the numerator (\(x^2+7x+12 = (x+3)(x+4)\)) and the denominator (\(5x+15 = 5(x+3)\)).
- Cancel the common factor \((x+3)\) to simplify the expression to \(\frac{x+4}{5}\).
- The simplified result is \(\boxed{\frac{x+4}{5}}\).
Explanation
- Problem Analysis
The objective is to simplify the expression \(\frac{x^2+6x+2}{5x+15} + \frac{x+10}{5x+15}\). We observe that both fractions share the same denominator, \(5x+15\). To perform the addition, we can combine the numerators over this common denominator, provided that the expression is well-defined. The denominator \(5x+15\) can be factored as \(5(x+3)\), so the expression is defined for all real numbers \(x\) except \(x = -3\).
- Combining and Simplifying the Numerator
To combine the fractions, we write the sum as a single fraction over the common denominator:
\[ \frac{(x^2+6x+2) + (x+10)}{5x+15} \]
Combining the like terms in the numerator, we have \(x^2 + (6x+x) + (2+10) = x^2 + 7x + 12\). Thus, the expression becomes:
\[ \frac{x^2 + 7x + 12}{5x + 15} \]
- Factoring the Terms
Next, we factor the numerator and the denominator. The numerator \(x^2 + 7x + 12\) can be factored by finding two numbers that multiply to 12 and add to 7, which are 3 and 4. Therefore, \(x^2 + 7x + 12 = (x+3)(x+4)\). The denominator \(5x + 15\) can be factored by extracting the common factor 5, giving \(5(x+3)\). Now, substitute these back into the fraction:
\[ \frac{(x+3)(x+4)}{5(x+3)} \]
- Final Simplification
Finally, we cancel the common factor \((x+3)\) from the numerator and the denominator. Since we have the condition \(x
eq -3\), the division by \((x+3)\) is valid. This leaves us with:
\[ \frac{x+4}{5} \]
Thus, the simplified form of the expression is \(\frac{x+4}{5}\).
Examples
In real life, adding algebraic expressions like this often appears in resource allocation or engineering problems. For example, if you are distributing a budget for two projects that share the same overhead costs, you might find that the total cost is represented by a sum of two rational functions. Simplifying this expression allows you to easily calculate the total per-unit cost based on the variable \(x\), which could represent the number of participants or production units, making it much easier to determine how the total expense scales as you change the scale of your operations.