Algebraic fraction problems for simplification.

Let’s solve each problem one by one.

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Problem 20: $\frac{3}{7x} - \frac{1}{2x}$

We need to subtract these two fractions. To do that, we need a common denominator.

The denominators are $7x$ and $2x$. The least common denominator (LCD) is $14x$, because 14 is the least common multiple of 7 and 2, and both have $x$.

Now rewrite each fraction with denominator $14x$:

- $\frac{3}{7x} = \frac{3 \cdot 2}{7x \cdot 2} = \frac{6}{14x}$
- $\frac{1}{2x} = \frac{1 \cdot 7}{2x \cdot 7} = \frac{7}{14x}$

Now subtract:

$\frac{6}{14x} - \frac{7}{14x} = \frac{6 - 7}{14x} = \frac{-1}{14x}$

✔ Final Answer for #20: $-\frac{1}{14x}$

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Problem 21: $\frac{4}{6x} - \frac{5}{9x}$

First, simplify if possible. $\frac{4}{6x}$ can be simplified to $\frac{2}{3x}$, but let’s keep it as is for now to find LCD.

Denominators: $6x$ and $9x$

Find LCD of 6 and 9 → LCM is 18, so LCD is $18x$

Rewrite each fraction:

- $\frac{4}{6x} = \frac{4 \cdot 3}{6x \cdot 3} = \frac{12}{18x}$
- $\frac{5}{9x} = \frac{5 \cdot 2}{9x \cdot 2} = \frac{10}{18x}$

Subtract:

$\frac{12}{18x} - \frac{10}{18x} = \frac{2}{18x} = \frac{1}{9x}$

✔ Final Answer for #21: $\frac{1}{9x}$

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Problem 22: $\frac{x+1}{5} + \frac{x+2}{5}$

Same denominator! So just add numerators:

$(x + 1) + (x + 2) = x + 1 + x + 2 = 2x + 3$

So result is: $\frac{2x + 3}{5}$

✔ Final Answer for #22: $\frac{2x + 3}{5}$

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Problem 23: $\frac{x - 2}{6} + \frac{x + 3}{6}$

Again, same denominator. Add numerators:

$(x - 2) + (x + 3) = x - 2 + x + 3 = 2x + 1$

Result: $\frac{2x + 1}{6}$

✔ Final Answer for #23: $\frac{2x + 1}{6}$

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Final Answers:

20. $-\frac{1}{14x}$
21. $\frac{1}{9x}$
22. $\frac{2x + 3}{5}$
23. $\frac{2x + 1}{6}$