How do you simplify $\dfrac{{3a + 3b}}{{ - a - b}}$ ?

Hint: In this question, we are given a fraction and we have to simplify $\dfrac{{3a + 3b}}{{ - a - b}}$ . We simplify the numerator and the denominator of this fraction by using the distributive property. According to the distributive property, the sum of the product of one number with another number and the product of that same number with some other number is equal to the product of that number with the sum of the other two numbers and vice-versa, that is, $ab + ac = a(b + c)\,or\,a(b + c) = ab + ac\,$ . $3a$ represents the product of 3 and a, and $3b$ represents the product of 3 and b, similarly $ - a$ is equal to the product of -1 and a, and $ - b$ is equal to the product of -1 and b. Thus, we can simplify the given fraction by applying the distributive property.

Complete step-by-step solution:
We are given that $\dfrac{{3a + 3b}}{{ - a - b}}$ , on applying the distributive property in the numerator and the denominator, we get –
$\Rightarrow \dfrac{{3(a + b)}}{{ - 1(a + b)}}$
Now, $a + b$ is present in both the numerator and the denominator, so we cancel it out –
$ \Rightarrow \dfrac{{3a + 3b}}{{ - a - b}} = \dfrac{3}{{ - 1}}$
We know that $\dfrac{a}{{ - b}} = \dfrac{{ - a}}{b}$ , so –
$ \Rightarrow \dfrac{{3a + 3b}}{{ - a - b}} = - 3$
Hence, the simplified form of $\dfrac{{3a + 3b}}{{ - a - b}}$ is -3.

Note: We must note that after applying the distributive property, we have done the prime factorization of the numerator and the denominator, that is, we have expressed the numerator and the denominator as a product of their prime factors –
$
 \Rightarrow 3(a + b) = 3 \times (a + b) \\
  \Rightarrow - 1(a + b) = - 1 \times (a + b) \\
 $
The quantity $(a + b)$ is common in the prime factorization of both the numerator and the denominator, so we cancel it out and get the simplified form of the given fraction.