Question

How do you solve for \[d\] in, \[a=\dfrac{d+c}{b}\]?

Answer 1

Hint: In this question, we will use the properties of multiplication and addition. We know that if equals are added to equals the wholes are equal. Similarly, is the case with multiplication if equals are multiplied by a certain equal we get the same answer on LHS (Left Hand Side) and RHS (Right Hand Side).

Complete step by step answer:
According to the question we have to solve for \[d\]
The given equation is to be rearranged keeping in mind the various properties associated with addition, multiplication, subtraction and division.
We have here, \[a=\dfrac{d+c}{b}\]
We know that multiplying a constant on both sides of an equality will also be equal. Now we will multiply both sides by \[b\], we get
\[\Rightarrow a\times b=\left( \dfrac{d+c}{b} \right)\times b\]
We have \[b\] both in the numerator and denominator in the RHS so it will get cancelled and we now have
\[\Rightarrow ab=d+c\]
We know that when added to equals, wholes are equal.
So now add –c on both the sides, which on adding gives us
\[\Rightarrow ab+(-c)=d+c+(-c)\]
We now have \[c\] and \[-c\] in the RHS which will get subtracted and finally we get
\[\Rightarrow d=ab-c\]
We know that \[a=b\] can also be written as \[b=a\] and vice versa.
Therefore, the answer is \[d=ab-c\].

Note:
While solving the question, the properties involved with the various arithmetic operations should not be intermixed with each other and should be carefully done. The entity passing onto the other side of the equality sign whether it is in the numerator or denominator should be written properly keeping in mind the reversal from denominator to numerator when the entity is passed over the equality and vice versa and should not be confused with the same.