Arithmetic with Polynomials and Rational Expressions: Operating with Rational Expressions (CCSS.A-APR.7)

Since the denominators are the same, add the numerators: $\frac{(2x+3)+(x-5)}{x-1}=\frac{3x-2}{x-1}$. No common factor cancels with $x-1$. Slip links: (A) dropped the constant terms, (B) incorrectly added denominators to make $2x-2$, (D) incorrectly multiplied or squared the common denominator.

Same denominator, so subtract numerators: $\frac{2x-(x-4)}{x+2}=\frac{2x-x+4}{x+2}=\frac{x+4}{x+2}$. Slip links: (B) forgot to distribute the minus, (C) incorrectly multiplied/squared the denominator, (D) incorrectly subtracted/added denominators.

Factor $x^2-4=(x-2)(x+2)$ and cancel $x+2$: $\frac{x+2}{(x-2)(x+2)}\cdot\frac{x+5}{x+2}=\frac{1}{x-2}\cdot\frac{x+5}{1}=\frac{x+5}{x-2}$. Slip links: (A) did not factor/cancel, (B) added numerators instead of multiplying, (C) dropped the remaining denominator after canceling.

Keep-change-flip, then factor and cancel: $\frac{x^2-9}{x+1}\cdot\frac{x+1}{x-3}=\frac{(x-3)(x+3)}{x+1}\cdot\frac{x+1}{x-3}=x+3$. Slip links: (A) forgot to flip and left an extra $x+1$ below, (C) divided numerators/denominators separately or forgot to flip, (D) canceled incorrectly to leave $\frac{x-3}{x+1}$.

Same denominator, so add numerators: $\frac{(x+4)+(2x-3)}{x^2+x-6}=\frac{3x+1}{x^2+x-6}$. The denominator factors as $(x+3)(x-2)$, but $3x+1$ shares no common factor, so no cancellation. Slip links: (B) incorrectly squared/multiplied the denominator, (C) dropped the denominator entirely, (D) incorrectly added denominators.

Keep the common denominator $(x+2)$ and add the numerators: $(2x+3)+(x-5)=3x-2$, so the result is $\frac{3x-2}{x+2}$. B adds denominators, giving $2x+4$ incorrectly. C drops the denominator entirely. D multiplies numerators instead of adding, a wrong operation for addition.

With like denominators, subtract numerators over the same denominator: $(x^2+4x+1)-(x+3)=x^2+3x-2$, so $\frac{x^2+3x-2}{x^2-4}$. The denominator factors as $(x-2)(x+2)$, but there is no common factor with the numerator, so stop. A adds the numerators instead of subtracting. B arises from incorrectly canceling a noncommon factor after factoring $x^2-4$, leaving only $(x-2)$. D drops the denominator.

Multiply and simplify by canceling the common factor $(x+1)$: $\frac{x+1}{x-2}\cdot\frac{x-3}{x+1}=\frac{\cancel{(x+1)}(x-3)}{(x-2)\cancel{(x+1)}}=\frac{x-3}{x-2}$. A inverts the result (wrong reciprocal idea). C fails to cancel the common factor $(x+1)$. D cancels the wrong factor, removing $(x-2)$ instead.

Division by a rational expression means multiply by its reciprocal: $\frac{x^2-9}{x+4}\cdot\frac{x-2}{x-3}$. Factor $x^2-9=(x-3)(x+3)$ and cancel $(x-3)$ to get $\frac{(x+3)(x-2)}{x+4}$. A does not flip the divisor and also drops a factor when multiplying. B cancels the wrong factor in the denominator (removing $x+4$ instead of $x-3$). C combines denominator terms incorrectly (turning $x+4$ into $x^2+4$).

Factor $x^2-1=(x-1)(x+1)$, then cancel the common $(x+1)$: $\frac{2x}{(x-1)(x+1)}\cdot\frac{x+1}{x+3}=\frac{2x,\cancel{(x+1)}}{(x-1),\cancel{(x+1)},(x+3)}=\frac{2x}{(x-1)(x+3)}$. B leaves the denominator unfactored, so the common factor is not canceled (unsimplified). C incorrectly cancels $x$ from $2x$ with $(x+1)$. D drops the $(x+1)$ factor from the numerator without justification, losing a factor.