Teaching Algebraic Fractions - Free Printable

Let's solve each of these algebraic fraction problems step by step. The goal is to add or subtract the given algebraic fractions by finding a common denominator and simplifying.

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a) $\frac{1}{f} + \frac{2}{3f}$

- Both terms have $f$ in the denominator.
- The least common denominator (LCD) is $3f$.
- Rewrite $\frac{1}{f}$ as $\frac{3}{3f}$:
$$
\frac{1}{f} = \frac{3}{3f}
$$
- Now add:
$$
\frac{3}{3f} + \frac{2}{3f} = \frac{5}{3f}
$$

✔ Answer: $\boxed{\frac{5}{3f}}$

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b) $\frac{h-1}{2} - \frac{h-2}{5}$

- LCD of 2 and 5 is 10.
- Rewrite both fractions:
$$
\frac{h-1}{2} = \frac{5(h-1)}{10} = \frac{5h - 5}{10}
$$
$$
\frac{h-2}{5} = \frac{2(h-2)}{10} = \frac{2h - 4}{10}
$$
- Subtract:
$$
\frac{5h - 5}{10} - \frac{2h - 4}{10} = \frac{(5h - 5) - (2h - 4)}{10}
$$
$$
= \frac{5h - 5 - 2h + 4}{10} = \frac{3h - 1}{10}
$$

✔ Answer: $\boxed{\frac{3h - 1}{10}}$

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c) $\frac{4}{2d} - \frac{2r}{4d}$

- Simplify first:
$$
\frac{4}{2d} = \frac{2}{d}, \quad \frac{2r}{4d} = \frac{r}{2d}
$$
- Now: $\frac{2}{d} - \frac{r}{2d}$
- LCD is $2d$
- Rewrite $\frac{2}{d} = \frac{4}{2d}$
- So:
$$
\frac{4}{2d} - \frac{r}{2d} = \frac{4 - r}{2d}
$$

✔ Answer: $\boxed{\frac{4 - r}{2d}}$

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d) $\frac{2}{c} - \frac{y+2}{4}$

- LCD of $c$ and 4 is $4c$
- Rewrite:
$$
\frac{2}{c} = \frac{8}{4c}, \quad \frac{y+2}{4} = \frac{c(y+2)}{4c}
$$
- Subtract:
$$
\frac{8}{4c} - \frac{c(y+2)}{4c} = \frac{8 - c(y+2)}{4c}
$$
$$
= \frac{8 - cy - 2c}{4c}
$$

✔ Answer: $\boxed{\frac{8 - cy - 2c}{4c}}$

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e) $\frac{w}{x} - \frac{1}{y}$

- LCD of $x$ and $y$ is $xy$
- Rewrite:
$$
\frac{w}{x} = \frac{wy}{xy}, \quad \frac{1}{y} = \frac{x}{xy}
$$
- Subtract:
$$
\frac{wy}{xy} - \frac{x}{xy} = \frac{wy - x}{xy}
$$

✔ Answer: $\boxed{\frac{wy - x}{xy}}$

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f) $\frac{4}{x-y} + \frac{5}{x+y}$

- LCD is $(x-y)(x+y)$
- Rewrite:
$$
\frac{4}{x-y} = \frac{4(x+y)}{(x-y)(x+y)}, \quad \frac{5}{x+y} = \frac{5(x-y)}{(x-y)(x+y)}
$$
- Add:
$$
\frac{4(x+y) + 5(x-y)}{(x-y)(x+y)} = \frac{4x + 4y + 5x - 5y}{(x-y)(x+y)}
$$
$$
= \frac{9x - y}{(x-y)(x+y)}
$$

✔ Answer: $\boxed{\frac{9x - y}{(x-y)(x+y)}}$

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g) $\frac{2}{a+c} - \frac{5}{a-c}$

- LCD is $(a+c)(a-c)$
- Rewrite:
$$
\frac{2}{a+c} = \frac{2(a-c)}{(a+c)(a-c)}, \quad \frac{5}{a-c} = \frac{5(a+c)}{(a+c)(a-c)}
$$
- Subtract:
$$
\frac{2(a-c) - 5(a+c)}{(a+c)(a-c)} = \frac{2a - 2c - 5a - 5c}{(a+c)(a-c)}
$$
$$
= \frac{-3a - 7c}{(a+c)(a-c)}
$$

✔ Answer: $\boxed{\frac{-3a - 7c}{(a+c)(a-c)}}$

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h) $\frac{2}{w^2 - 1} + \frac{3}{w + 1}$

- Note: $w^2 - 1 = (w - 1)(w + 1)$, so LCD is $(w - 1)(w + 1)$
- Rewrite:
$$
\frac{2}{w^2 - 1} = \frac{2}{(w-1)(w+1)}
$$
$$
\frac{3}{w+1} = \frac{3(w-1)}{(w-1)(w+1)}
$$
- Add:
$$
\frac{2 + 3(w-1)}{(w-1)(w+1)} = \frac{2 + 3w - 3}{(w-1)(w+1)} = \frac{3w - 1}{(w-1)(w+1)}
$$

✔ Answer: $\boxed{\frac{3w - 1}{(w-1)(w+1)}}$

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✔ Final Answers:

a) $\boxed{\frac{5}{3f}}$
b) $\boxed{\frac{3h - 1}{10}}$
c) $\boxed{\frac{4 - r}{2d}}$
d) $\boxed{\frac{8 - cy - 2c}{4c}}$
e) $\boxed{\frac{wy - x}{xy}}$
f) $\boxed{\frac{9x - y}{(x-y)(x+y)}}$
g) $\boxed{\frac{-3a - 7c}{(a+c)(a-c)}}$
h) $\boxed{\frac{3w - 1}{(w-1)(w+1)}}$

Let me know if you'd like a visual explanation or further steps!