Algebraic Fractions and Non-monic Factorisation

A fraction is undefined when its denominator is zero. Here $x - 3 = 0$ gives $x = 3$. ($x = -2$ makes the numerator zero, which just makes the whole fraction equal 0, not undefined.)

Factor first: $x^2 + 3x = x(x + 3)$, then cancel the common factor $x$: $\dfrac{x(x + 3)}{x} = x + 3$. ($x^2 + 3$ cancelled only one term, the canonical 'cancel across addition' error; you can only cancel a common factor.)

Difference of squares: $x^2 - 4 = (x - 2)(x + 2)$, then cancel $(x + 2)$: $\dfrac{(x - 2)(x + 2)}{x + 2} = x - 2$. ('Cannot be simplified' missed the difference-of-squares pattern.)

Factor the numerator (two numbers multiplying to 6 and adding to 5: 2 and 3): $(x + 2)(x + 3)$, then cancel $(x + 2)$: $x + 3$. ($x - 3$ has a sign error in the factorisation.)

Factor both: $2x + 4 = 2(x + 2)$ and $x^2 + 2x = x(x + 2)$, then cancel $(x + 2)$: $\dfrac{2(x + 2)}{x(x + 2)} = \dfrac{2}{x}$. ($\dfrac{1}{x}$ lost the factor of 2.)

In one hour the taps fill $\dfrac{1}{3}$ and $\dfrac{1}{6}$ of the tank. Add over a common denominator: $\dfrac{1}{3} + \dfrac{1}{6} = \dfrac{2}{6} + \dfrac{1}{6} = \dfrac{3}{6} = \dfrac{1}{2}$. ($\dfrac{2}{9}$ added numerators and denominators directly.)

Common denominator 6: $\dfrac{3(x + 1)}{6} + \dfrac{2(x - 1)}{6} = \dfrac{3x + 3 + 2x - 2}{6} = \dfrac{5x + 1}{6}$. ($\dfrac{5x - 1}{6}$ mishandled $3 + (-2) = 1$; $\dfrac{5x + 5}{6}$ added the constants as $3 + 2$.)

You cannot add numerators over the sum of the denominators. Using the common denominator $xy$: $\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{y}{xy} + \dfrac{x}{xy} = \dfrac{x + y}{xy}$.

Common denominator $x(x + 1)$: $\dfrac{x + 1}{x(x + 1)} - \dfrac{x}{x(x + 1)} = \dfrac{(x + 1) - x}{x(x + 1)} = \dfrac{1}{x(x + 1)}$. (0 assumed the fractions were equal; $\dfrac{1}{2x + 1}$ added the denominators.)

Common denominator 6: $\dfrac{2(x + 2)}{6} - \dfrac{3(x - 1)}{6} = \dfrac{2x + 4 - (3x - 3)}{6} = \dfrac{2x + 4 - 3x + 3}{6} = \dfrac{7 - x}{6}$. Distribute the minus to both terms: $-(3x - 3) = -3x + 3$. ($\dfrac{-x - 1}{6}$ did not distribute the minus.)

Average speed is total distance over total time. The times are $\dfrac{60}{30} = 2$ h and $\dfrac{60}{60} = 1$ h, so the average is $\dfrac{120\text{ km}}{3\text{ h}} = 40$ km/h. ($45$ km/h is the ordinary average of the two speeds, which is wrong when the times differ.)

The average is $\dfrac{x + 12}{2} = 20$. Multiply both sides by 2: $x + 12 = 40$, so $x = 28$. ($8$ used $20 - 12$, forgetting to double first.)

After the first reduction the price is $80 \times \dfrac{3}{4} = \$60$. Taking a further $\dfrac{1}{5}$ off leaves $\dfrac{4}{5}$: $60 \times \dfrac{4}{5} = \$48$. ($\$44$ took $\dfrac{1}{4} + \dfrac{1}{5}$ of 80, but the second discount acts on the smaller price.)

Factor $x^2 - 1 = (x - 1)(x + 1)$: $\dfrac{(x - 1)(x + 1)}{x - 1} \times \dfrac{x}{x + 1}$. Cancel $(x - 1)$, then $(x + 1)$: $(x + 1) \times \dfrac{x}{x + 1} = x$. ($x^2$ multiplied without cancelling.)

$\dfrac{1}{R} = \dfrac{1}{2} + \dfrac{1}{3} = \dfrac{3}{6} + \dfrac{2}{6} = \dfrac{5}{6}$. Take the reciprocal of both sides: $R = \dfrac{6}{5}\,\Omega$. ($\dfrac{5}{6}\,\Omega$ forgot to flip at the end.)

AC-method: $ac = 2 \times 3 = 6$; two numbers multiplying to 6 and adding to 5 are 2 and 3. Split: $2x^2 + 2x + 3x + 3 = 2x(x + 1) + 3(x + 1) = (2x + 3)(x + 1)$. ($(2x + 1)(x + 3)$ expands to $2x^2 + 7x + 3$; $(x + 3)(x + 1)$ forgot the leading 2.)

Perfect-square trinomial: $\sqrt{x^2} = x$, $\sqrt{25} = 5$, and the middle term $2 \times x \times 5 = 10x$ matches, so $x^2 + 10x + 25 = (x + 5)^2$. ($(x - 5)^2$ gives $x^2 - 10x + 25$; $(x + 5)(x - 5)$ is the difference of squares $x^2 - 25$.)

AC-method: $ac = 6 \times (-1) = -6$; two numbers multiplying to $-6$ and adding to 1 are 3 and $-2$. Split: $6x^2 + 3x - 2x - 1 = 3x(2x + 1) - (2x + 1) = (3x - 1)(2x + 1)$. ($(3x + 1)(2x - 1)$ gives $6x^2 - x - 1$ (wrong middle sign).)

Difference of squares: $4x^2 - 9 = (2x)^2 - 3^2 = (2x - 3)(2x + 3)$. ($(2x - 3)^2$ has a middle term, but the original has none.)

Extract the common factor first: $2x^2 - 18 = 2(x^2 - 9)$, then apply the difference of squares: $2(x - 3)(x + 3)$. ($2(x^2 - 9)$ stopped early; $(x - 3)(x + 3)$ forgot the 2.)

Numerator $x^2 - 9 = (x - 3)(x + 3)$; denominator $x^2 + 6x + 9 = (x + 3)^2$. Cancel one $(x + 3)$: $\dfrac{(x - 3)(x + 3)}{(x + 3)^2} = \dfrac{x - 3}{x + 3}$. ($\dfrac{x + 3}{x - 3}$ inverted the fraction.)

"Of" means multiply: $\dfrac{1}{2} \times \dfrac{1}{3} = \dfrac{1}{6}$ of all the students. ($\dfrac{5}{6}$ added the fractions instead of multiplying.)

Factor the numerator: $2x^2 + 5x + 3 = (2x + 3)(x + 1)$, then cancel $(x + 1)$: $2x + 3$. ($x + 3$ forgot the leading 2 in the factor $(2x + 3)$.)

Multiply both sides by 2: $x + 4 = 10$; subtract 4: $x = 6$. (10 gave $x + 4$; 14 added 4 instead of subtracting.)

Factor each piece: $x^2 - 4 = (x - 2)(x + 2)$, $x^2 + 6x + 9 = (x + 3)^2$, $x^2 - 2x = x(x - 2)$. So $\dfrac{(x - 2)(x + 2)}{x + 3} \times \dfrac{(x + 3)^2}{x(x - 2)}$. Cancel $(x - 2)$ and one $(x + 3)$: $\dfrac{(x + 2)(x + 3)}{x}$. ($\dfrac{x + 2}{x}$ over-cancelled the $(x + 3)$.)