How do you simplify \[ - \dfrac{1}{{10}} - \dfrac{2}{5}\]?

Hint: Fractions are a big part of our daily life and the mathematical world, so we must understand how to perform mathematical operations on two different fractions. In the given question, we have to add the two given fractions. The fractions having the same denominator can be added easily but when the denominators are different, then we first find the LCM of the terms in the denominator and then add the fractions. Using this approach, we can find out the correct answer.

Complete step by step solution:
We have, \[ - \dfrac{1}{{10}} - \dfrac{2}{5}\].
That is,
\[ \Rightarrow - \left( {\dfrac{1}{{10}} + \dfrac{2}{5}} \right)\]
To solve this we need to find the LCM of 10 and 5. We use factors of the 10 and 5 to find the LCM.
The factors of 10 are 1, 2 and 5. That is \[10 = 1 \times 2 \times 5\]
The factors of 5 are 1 and 5. That is \[5 = 1 \times 5\].
We can see that the least common multiple of 10 and 5 is 10. That is \[ \Rightarrow 1 \times 2 \times 5 = 10\]
Now we need to multiply 10 and divide by 10 to the given problem, we get:
\[ \dfrac{{ \Rightarrow - \left( {\dfrac{1}{{10}} + \dfrac{2}{5}} \right) \times 10}}{{10}}\]
Now multiplying 10 for each fraction in the numerator we get,
\[ \Rightarrow \dfrac{{ - \left( {\dfrac{1}{{10}} \times 10 + \dfrac{2}{5} \times 10} \right)}}{{10}}\]
Simplifying on the numerator we have,
\[ \Rightarrow \dfrac{{ - \left( {1 + (2 \times 2)} \right)}}{{10}}\]
\[ \Rightarrow \dfrac{{ - \left( {1 + 4} \right)}}{{10}}\]
\[\Rightarrow \dfrac{{ - 5}}{{10}}\].
\[\Rightarrow - \dfrac{1}{2}\]

Thus we have \[ \Rightarrow - \dfrac{1}{{10}} - \dfrac{2}{5} = - \dfrac{1}{2}\]

Note: If we have two primes and we need to find the LCM of this. Then we need to multiply these numbers, which gives us the LCM. That is The LCM of two or more prime numbers is equal to their product. We can also find the HCF of the 10 and 5. Since we have the factors \[10 = 1 \times 2 \times 5\] and \[5 = 1 \times 5\]. We can see that the highest common multiple (HCF) is 5.