Fractions can be made much easier to work with when we look for structure in numerators and denominators.

One useful technique is to decompose denominators into factors and express numerators as sums of those factors. In this post we’ll apply that idea to evaluate the expression

$$\frac{11}{30}-\frac{13}{42}+\frac{15}{56},$$

using the given decompositions

$$30=5\cdot 6,\qquad 42=6\cdot 7,\qquad 56=7\cdot 8$$

and writing the numerators as sums aligned with those factors:

$$11=5+6,\qquad 13=6+7,\qquad 15=7+8.$$

Key algebraic rule

If a numerator equals a sum of the denominator’s factors, i.e.

$$\frac{a+b}{a\cdot b},$$

we can split it by linearity of fractions:

$$\frac{a+b}{a\cdot b}=\frac{a}{a\cdot b}+\frac{b}{a\cdot b} =\frac{1}{b}+\frac{1}{a}.$$

(We used $\dfrac{a}{a\cdot b}=\dfrac{1}{b}$​ and $\dfrac{b}{a\cdot b}=\dfrac{1}{a}$)

Apply the rule to each term

1. For $\dfrac{11}{30}$ with $11=5+6$ and $30=5\cdot6$:

$$\frac{11}{30}=\frac{5+6}{5\cdot6}=\frac{5}{5\cdot6}+\frac{6}{5\cdot6}=\frac{1}{6}+\frac{1}{5}.$$

2. For $\dfrac{13}{42}$ with $13=6+7$ and $42=6\cdot7$

$$\frac{13}{42}=\frac{6+7}{6\cdot7}=\frac{6}{6\cdot7}+\frac{7}{6\cdot7}=\frac{1}{7}+\frac{1}{6}.$$

3. For $\dfrac{15}{56}$ with $15=7+8$ and $56=7\cdot8$:

$$\frac{15}{56}=\frac{7+8}{7\cdot8}=\frac{7}{7\cdot8}+\frac{8}{7\cdot8}=\frac{1}{8}+\frac{1}{7}.$$

Rewrite the whole expression using these decompositions

$$\frac{11}{30}-\frac{13}{42}+\frac{15}{56} = \Big(\frac{1}{6}+\frac{1}{5}\Big) – \Big(\frac{1}{7}+\frac{1}{6}\Big) + \Big(\frac{1}{8}+\frac{1}{7}\Big).$$

Combine like terms (observe cancellations)
Group the terms and cancel where possible:

• The $\dfrac{1}{6}$ from the first term and the $-\dfrac{1}{6}$​ from the second term cancel.
• The $-\dfrac{1}{7}$​ from the second term and the $+\dfrac{1}{7}$ from the third term cancel.

What remains is:

$$\frac{1}{5} + \frac{1}{8}.$$

Add the remaining fractions
Find a common denominator for $\dfrac{1}{5}$ and $\dfrac{1}{8}$​. The least common denominator is $40$:

$$\frac{1}{5}=\frac{8}{40},\qquad \frac{1}{8}=\frac{5}{40}.$$

So

$$\frac{1}{5}+\frac{1}{8}=\frac{8}{40}+\frac{5}{40}=\frac{13}{40}.$$

Final result

$$\boxed{\frac{11}{30}-\frac{13}{42}+\frac{15}{56}=\frac{13}{40}.}$$

Conclusion

By decomposing denominators into factors and writing numerators as sums of those factors, the expression simplified through cancellation to just $\dfrac{1}{5}+\dfrac{1}{8}$, which equals $\dfrac{13}{40}$. This technique is particularly useful for telescoping-like cancellations and for seeing structure that makes arithmetic simpler.