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Quiz 19
1.) Add $$\dfrac{5}{x} + \dfrac{2}{x^2}.$$
Solution: The least common denominator is $x^2$, so write the fractions using that common denominator and add: $$\dfrac{5}{x}+\dfrac{2}{x^2}=\dfrac{5x}{x^2}+\dfrac{2}{x^2}=\dfrac{5x+2}{x^2}.$$

2.) Subtract $$\dfrac{x}{(x+2)(x+1)} - \dfrac{2}{(x+3)(x+1)}.$$
Solution: The common denominator is $(x+1)(x+2)(x+3)$. Write both fractions in terms of this common denominator and subtract: $$\begin{array}{ll} \dfrac{x}{(x+2)(x+1)} - \dfrac{2}{(x+3)(x+1)} &= \dfrac{x(x+3)}{(x+1)(x+2)(x+3)} - \dfrac{2(x+2)}{(x+1)(x+2)(x+3)} \\ &= \dfrac{x(x+3) - 2(x+2)}{(x+1)(x+2)(x+3)} \\ &= \dfrac{x^2 + 3x - 2x -4}{(x+1)(x+2)(x+3)} \\ &=\dfrac{x^2 + x-4}{(x+1)(x+2)(x+3)} \end{array}$$

3.) Add $$\dfrac{1}{x^2+2x+1} + \dfrac{1}{x^2+3x+2}.$$
Solution: First notice that the first denominator factors as $$x^2+2x+1=(x+1)(x+1)$$ and the second denominator factors as $$x^2+3x+2=(x+1)(x+2).$$ Therefore the common denominator is $(x+1)(x+1)(x+2)=(x+1)^2(x+2)$. So rewrite both fractions in terms of that common denominator and add: $$\begin{array}{ll} \dfrac{1}{x^2+2x+1} + \dfrac{1}{x^2+3x+2} &= \dfrac{1}{(x+1)(x+1)} + \dfrac{1}{(x+1)(x+2)} \\ &= \dfrac{x+2}{(x+1)^2 (x+2)} + \dfrac{x+1}{(x+1)^2(x+2)} \\ &= \dfrac{2x+3}{(x+1)^2(x+2)}. \end{array}$$

4.) Simplify $$\dfrac{t+t^{-1}}{t^{-1}+2}.$$
Solution: First remove the negative exponents by writing $t^{-1}=\dfrac{1}{t}$ (this is what negative exponents mean!!) to get $$\dfrac{t+t^{-1}}{t^{-1}+2} = \dfrac{t+\frac{1}{t}}{\frac{1}{t}+2}.$$ The common denominators of the top and the bottom are both $t$. Get common denominators on top and bottom and perform the additions. After this use the rule $\dfrac{\frac{a}{b}}{\frac{c}{d}}=\dfrac{a}{b} \dfrac{d}{c}$: $$\begin{array}{ll} \dfrac{t+\frac{1}{t}}{\frac{1}{t}+2} &= \dfrac{\frac{t^2+1}{t}}{\frac{1+2t}{t}} \\ &=\dfrac{t^2+1}{t} \dfrac{t}{1+2t} \\ &=\dfrac{t^2+1}{1+2t}. \end{array}$$

5.) Simplify $$\dfrac{x-2+\frac{1}{x}}{\frac{1}{x+2}+x}.$$
Solution: The common denominator on the top is $x$ and the common denominator on the bottom is $x+2$. Find those, perform the sums, and then simplify using the rule $\dfrac{\frac{a}{b}}{\frac{c}{d}}=\dfrac{a}{b} \dfrac{d}{c}$: $$\begin{array}{ll} \dfrac{x-2+\frac{1}{x}}{\frac{1}{x+2}+x} &= \dfrac{\frac{x^2}{x}-\frac{2x}{x}+\frac{1}{x}}{\frac{1}{x+2} + \frac{x^2+2x}{x+2}} \\ &= \dfrac{\frac{x^2-2x+1}{x}}{\frac{x^2+2x+1}{x+2}} \\ &= \dfrac{x^2-2x+1}{x} \dfrac{x+2}{x^2+2x+1} \\ &= \dfrac{(x-1)^2(x+2)}{x(x+1)^2}. \end{array}$$