QUESTION IMAGE
Question
add the rational expressions to find the sum. \\(\frac{x^2 - 2}{3x - 2} + \frac{x + 5}{3x - 2}\\). (1 point) \\(\bigcirc \frac{x^2 + x + 7}{3x - 2}\\) \\(\bigcirc \frac{x^2 + x + 3}{6x - 4}\\) \\(\bigcirc \frac{x^2 + 3}{3x - 2}\\) \\(\bigcirc \frac{x^2 + x + 3}{3x - 2}\\)
Step1: Add the numerators (common denominator)
Since the denominators \(3x - 2\) are the same, we add the numerators: \((x^2 - 2)+(x + 5)\)
Step2: Simplify the numerator
Combine like terms: \(x^2 - 2 + x + 5=x^2+x + 3\)
Step3: Write the sum
The denominator remains \(3x - 2\), so the sum is \(\frac{x^2+x + 3}{3x - 2}\) (Wait, wait, let's check again. Wait, \(-2 + 5 = 3\)? Wait no, \(-2+5=3\)? Wait original numerators: \(x^2 - 2\) and \(x + 5\). So \(x^2 - 2+x + 5=x^2+x+( - 2 + 5)=x^2+x + 3\). Wait but let's check the options. Wait the fourth option is \(\frac{x^2 + x + 3}{3x - 2}\)? Wait the user's options: let's re-express. Wait the first option: \(\frac{x^2 + x + 7}{3x - 2}\), second: \(\frac{x^2 + x + 3}{6x - 4}\), third: \(\frac{x^2 + 3}{3x - 2}\), fourth: \(\frac{x^2 + x + 3}{3x - 2}\)? Wait maybe I miscalculated. Wait \(-2 + 5 = 3\), so numerator is \(x^2 + x + 3\), denominator \(3x - 2\). So the fourth option (if the fourth is \(\frac{x^2 + x + 3}{3x - 2}\)) is correct. Wait let's recheck the addition:
\(\frac{x^2 - 2}{3x - 2}+\frac{x + 5}{3x - 2}=\frac{(x^2 - 2)+(x + 5)}{3x - 2}=\frac{x^2 - 2 + x + 5}{3x - 2}=\frac{x^2+x + 3}{3x - 2}\)
Yes, so the correct option is the one with numerator \(x^2 + x + 3\) and denominator \(3x - 2\), which is the fourth option (assuming the fourth option is \(\frac{x^2 + x + 3}{3x - 2}\)). Wait looking at the user's options:
First: \(\frac{x^2 + x + 7}{3x - 2}\)
Second: \(\frac{x^2 + x + 3}{6x - 4}\) (which is \(2(3x - 2)\), so that's wrong, we don't add denominators)
Third: \(\frac{x^2 + 3}{3x - 2}\) (missing the \(x\) term)
Fourth: \(\frac{x^2 + x + 3}{3x - 2}\) (correct)
So the correct answer is the fourth option (let's check the original problem's options again, the fourth option is written as \(\frac{x^2 + x + 3}{3x - 2}\) (assuming the user's typo: maybe "3x - 2" not "3x - 2" with space? So the correct option is the one with numerator \(x^2 + x + 3\) and denominator \(3x - 2\), which is the fourth option.
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\(\boldsymbol{\frac{x^2 + x + 3}{3x - 2}}\) (corresponding to the option, e.g., if the fourth option is D. \(\frac{x^2 + x + 3}{3x - 2}\), then D. \(\frac{x^2 + x + 3}{3x - 2}\))