Question

as part of their algebra homework, a student was simplifying a rational expression. for the simplification problem below, the student’s work is shown. is the work shown for this problem correct? if not, identify and correct the error.
\\(\frac{x}{x - 5} + \frac{3(x - 5)}{x + 1}\\)
\\(\frac{x}{1} + \frac{3}{x + 1}\\)
\\(\frac{x(x + 1)}{x + 1} + \frac{3}{x + 1}\\)
\\(\frac{x^2 + x + 3}{x + 1}\\)

Explanation:

Step1: Analyze the first step

The original expression is $\frac{x}{x - 5}+\frac{3(x - 5)}{x + 1}$? Wait, no, looking at the student's first step: Wait, the first line (student's work) seems to have a typo? Wait, maybe the original problem is $\frac{x}{1}+\frac{3}{x + 1}$? Wait, no, the first step the student wrote is $\frac{x}{x - 5}+\frac{3(x - 5)}{x + 1}$? Wait, no, the image shows the student's work: first line: $\frac{x}{x - 5}+\frac{3(x - 5)}{x + 1}$? No, maybe the original problem is adding two fractions, say $\frac{x}{1}+\frac{3}{x + 1}$, but the student's first step is wrong. Wait, no, let's re - examine. Wait, the student's work: first expression: $\frac{x}{x - 5}+\frac{3(x - 5)}{x + 1}$? No, maybe the correct original problem is $\frac{x}{1}+\frac{3}{x + 1}$, and the student tried to get a common denominator. Wait, no, the second step: $\frac{x(x + 1)}{(x - 5)(x + 1)}+\frac{3(x - 5)}{(x - 5)(x + 1)}$? Wait, no, the student's second step is $\frac{x(x + 1)}{x + 1}+\frac{3}{x + 1}$? Wait, the image is a bit unclear, but let's assume the problem is adding $\frac{x}{1}+\frac{3}{x + 1}$. The correct way to add is to get a common denominator of $x + 1$. So $\frac{x(x + 1)}{x + 1}+\frac{3}{x + 1}=\frac{x^{2}+x + 3}{x + 1}$. But if the student's first step was wrong in the denominator. Wait, maybe the original problem was $\frac{x}{x - 5}+\frac{3}{x + 1}$, then the common denominator is $(x - 5)(x + 1)$. The student's mistake: in the first step, when getting the common denominator, the student might have multiplied the numerator and denominator incorrectly. Wait, looking at the student's work: first line: $\frac{x}{x - 5}+\frac{3(x - 5)}{x + 1}$? No, the second line: $\frac{x(x + 1)}{x + 1}+\frac{3}{x + 1}$. Ah, here's the error: the student should have multiplied the numerator and denominator of the first fraction by $(x + 1)$ and the second fraction by $(x - 5)$ to get a common denominator of $(x - 5)(x + 1)$, but the student multiplied the first fraction's denominator by $(x + 1)$ but the second fraction's denominator is still $x + 1$, and the numerator of the second fraction is not multiplied by $(x - 5)$. So the error is in the common - denominator step. The correct common denominator for $\frac{x}{x - 5}$ and $\frac{3}{x + 1}$ is $(x - 5)(x + 1)$. So $\frac{x}{x - 5}=\frac{x(x + 1)}{(x - 5)(x + 1)}$ and $\frac{3}{x + 1}=\frac{3(x - 5)}{(x - 5)(x + 1)}$. Then adding them gives $\frac{x(x + 1)+3(x - 5)}{(x - 5)(x + 1)}=\frac{x^{2}+x+3x - 15}{(x - 5)(x + 1)}=\frac{x^{2}+4x - 15}{(x - 5)(x + 1)}=\frac{(x + 5)(x - 3)}{(x - 5)(x + 1)}$ (if we factor). But the student's work: second line is $\frac{x(x + 1)}{x + 1}+\frac{3}{x + 1}$, which has the wrong denominator (should be $(x - 5)(x + 1)$ instead of $x + 1$ for the first fraction's denominator). Then the third line: $\frac{x^{2}+x + 3}{x + 1}$, which is wrong because of the denominator error. So the student's work is incorrect. The error is in the common - denominator step; the student did not use the correct common denominator (should be $(x - 5)(x + 1)$ if the first fraction's denominator is $x - 5$) or used an incorrect multiplier for the numerator and denominator.

Step2: Correct the work

Let's assume the original problem is $\frac{x}{x - 5}+\frac{3}{x + 1}$.
1. Find the common denominator: The common denominator of $x - 5$ and $x + 1$ is $(x - 5)(x + 1)$.
2. Rewrite each fraction with the common denominator:
- For $\frac{x}{x - 5}$, multiply the numerator and denominator by $x + 1$: $\frac{x(x + 1)}{(x - 5)(x + 1)}=\frac{x^{2}+x}{(x - 5)(x + 1)}$.
- For $\frac{3}{x + 1}$, multiply the numerator and denominator by $x - 5$: $\frac{3(x - 5)}{(x - 5)(x + 1)}=\frac{3x-15}{(x - 5)(x + 1)}$.
3. Add the two fractions:
- $\frac{x^{2}+x+(3x - 15)}{(x - 5)(x + 1)}=\frac{x^{2}+4x - 15}{(x - 5)(x + 1)}$.
- We can factor the numerator: $x^{2}+4x - 15=(x + 5)(x - 3)$, so the fraction becomes $\frac{(x + 5)(x - 3)}{(x - 5)(x + 1)}$ (if we don't simplify further) or if the original problem was $\frac{x}{1}+\frac{3}{x + 1}$, then:
- Common denominator is $x + 1$.
- $\frac{x(x + 1)}{x + 1}+\frac{3}{x + 1}=\frac{x^{2}+x + 3}{x + 1}$.

Answer:

The student's work is incorrect. The error is in the process of finding a common denominator (incorrect multiplication of numerator and denominator). The correct simplification (assuming the problem is $\frac{x}{1}+\frac{3}{x + 1}$) is $\frac{x^{2}+x + 3}{x + 1}$; if the problem is $\frac{x}{x - 5}+\frac{3}{x + 1}$, the correct simplification is $\frac{x^{2}+4x - 15}{(x - 5)(x + 1)}$ (or factored form $\frac{(x + 5)(x - 3)}{(x - 5)(x + 1)}$).