Question

emma, steve, maria, and george are comparing the solution of a math assignment problem below.\\(\frac{1}{x - 1}-\frac{3}{x^2 + 2x - 3}\\)\
select the student who correctly subtracted the rational expressions.\
\
| emma: \\(\

$$\begin{aligned}\\frac{1}{x - 1}-\\frac{3}{(x - 1)(x + 3)}\\&=\\frac{1(x + 3)+3}{(x - 1)(x + 3)}\\&=\\frac{x + 6}{(x - 1)(x + 3)}\\end{aligned}$$

\\) | steve: \\(\

$$\begin{aligned}\\frac{1}{x - 1}-\\frac{3}{(x - 1)(x + 3)}\\&=\\frac{1(x + 3)-3}{(x - 1)(x + 3)}\\&=\\frac{x}{(x - 1)(x + 3)}\\end{aligned}$$

\\) |\
| maria: \\(\

$$\begin{aligned}\\frac{1}{x - 1}-\\frac{3}{(x - 1)(x + 3)}\\&=\\frac{-2}{x + 3}\\end{aligned}$$

\\) | george: \\(\

$$\begin{aligned}\\frac{1 - 3}{x - 1 - x^2 - 2x + 3}\\&=\\frac{-2}{-x^2 - x + 2}\\end{aligned}$$

\\) |\
\
\\(\bigcirc\\) a. emma\
\\(\bigcirc\\) b. george\
\\(\bigcirc\\) c. steve\
\\(\bigcirc\\) d. maria

Explanation:

Step1: Factor the denominator

First, factor $x^2+2x-3$. We find that $x^2+2x-3=(x-1)(x+3)$, so the original expression becomes:
$$\frac{1}{x-1} - \frac{3}{(x-1)(x+3)}$$

Step2: Get common denominator

Rewrite $\frac{1}{x-1}$ with the common denominator $(x-1)(x+3)$:
$$\frac{1(x+3)}{(x-1)(x+3)} - \frac{3}{(x-1)(x+3)}$$

Step3: Subtract numerators

Subtract the numerators over the common denominator:
$$\frac{(x+3)-3}{(x-1)(x+3)}$$

Step4: Simplify numerator

Simplify the numerator by combining like terms:
$$\frac{x+3-3}{(x-1)(x+3)} = \frac{x}{(x-1)(x+3)}$$

Answer:

C. Steve