For $$A = \{(−4, −20), (−3, −9), (−2, −4), (−1, −11), (−1, −3), (1, 2), (1, 5), (2, 10), (2, 14), (3, 6), (4, 8), (4, 12)\}$$ define the relation $R$ on $A$ by $(a, b) R (c, d)$ if $ad = bc$.
On
Leo's answer gives you all that you really need, but it might be helpful to see the computation of one equivalence class; I'll compute $[\langle 4,8\rangle]$. You know that $\langle 4,8\rangle\mathbin{R}\langle c,d\rangle$ if and only if $4d=8c$. This condition is equivalent to the condition $d=2c$, so we're looking for the pairs $\langle c,d\rangle\in A$ such that $d=2c$. The set consisting of precisely those pairs is
$$[\langle 4,8\rangle]=\{\langle -2,-4\rangle,\langle 1,2\rangle,\langle 3,6\rangle,\langle 4,8\rangle\}\;.$$
First of all you have to check the property of reflexivity for all elements of $A$. For instance: $-4 \cdot -20= -20 \cdot -4$.
Secondly you have to check all $aRb$ for symmetry. For instance $(-2,-4)R(3,6)$ because $-2 \cdot 6 = -4 \cdot 3$. Now we have to verify if $(3,6)R(-2,-4)$.
Thirdly you have to check the transitivity: if $aRb$ and $bRc$ then $aRc$. For instance: $(3,6)R(4,8)$. We already know that $(-2,-4)R(3,6)$. You now have to check if $(-2,-4)R(4,8)$.
Hints for parts $b$ en $c$ of your question. To get a good intuitive feel for equivalence classes and equivalence relations I suggest you take a piece of paper and draw some dots on it. Each dot will represent an element $a \in A$. Draw a line between two dots $a$ and $b$ if $aRb$. Also draw a line if it turns out $bRa$. Do you know what your drawing should look like when you are finished? Can you see the equivalence classes? How many are there?
Hope this helps