Question

question #18
tyson is adding the following rational expressions: $\frac{x + 2}{x - 1}+\frac{3x}{x - 2}$. tysons work is shown.
did tyson make a mistake? if so, in which step was the mistake made? what was the mistake?
step 1: $lcd=(x - 1)(x - 2)$
step 2: $\frac{(x + 2)(x - 2)}{(x - 1)(x - 2)}+\frac{3x(x - 1)}{(x - 2)(x - 1)}$
step 3: $\frac{x^{2}-4}{(x - 1)(x - 2)}+\frac{3x^{2}-3x}{(x - 2)(x - 1)}$
step 4: $\frac{x^{2}-4 + 3x^{2}-3x}{(x - 1)(x - 2)}$
step 5: $\frac{x^{2}-x - 4}{(x - 1)(x - 2)}$
step 2; tyson multiplied each rational expression by the incorrect factor to create common denominators.
step 3; tyson simplified the numerators from step 2 incorrectly.
step 5; tyson simplified the numerators incorrectly when writing the sum in simplest form.
no errors were made. the work is correct.

Explanation:

Step1: Check Step 1

The least - common denominator (LCD) of $\frac{x + 2}{x-1}$ and $\frac{3x}{x - 2}$ is indeed $(x - 1)(x - 2)$. So step 1 is correct.

Step2: Check Step 2

To get a common denominator, we multiply $\frac{x + 2}{x-1}$ by $\frac{x - 2}{x - 2}$ and $\frac{3x}{x - 2}$ by $\frac{x - 1}{x - 1}$, resulting in $\frac{(x + 2)(x - 2)}{(x - 1)(x - 2)}+\frac{3x(x - 1)}{(x - 2)(x - 1)}$. Step 2 is correct.

Step3: Check Step 3

Expand the numerators: $(x + 2)(x - 2)=x^{2}-4$ and $3x(x - 1)=3x^{2}-3x$. So step 3 is correct.

Step4: Check Step 4

Combine the numerators over the common denominator: $\frac{x^{2}-4+3x^{2}-3x}{(x - 1)(x - 2)}=\frac{4x^{2}-3x - 4}{(x - 1)(x - 2)}$. But in step 5, Tyson combined $x^{2}$ and $3x^{2}$ incorrectly as $x^{2}-x - 4$.

Answer:

Step 5; Tyson simplified the numerators incorrectly when writing the sum in simplest form.