Question & Answer

solve the given equation $\dfrac{a+1}{b}+3=\dfrac{4a}{b};$ and make ‘b’ the subject of the formula.

          (A) b=$\dfrac{3a+1}{3}$
          (B) b=$\dfrac{3a-1}{3}$
          (C) b=$\dfrac{5a-1}{3}$
          (D) none of the above

ANSWER Verified Verified
Hint: We will be using the concept of changing the subject of the formula. It is a variable which is expressed in terms of the other variables involved in the formula. Formulas are written as a single variable, the subject of the formula is on the left hand side of the equation and everything else goes on the right-hand side of the equation.

Complete step-by-step solution -
Now we have to make ‘b’ as the subject of the formula in $\dfrac{a+1}{b}+3=\dfrac{4a}{b}$
Subject of the formula means that we have given an equation in terms of the variables. To change the subject of a formula, we’ll begin with the variable to become the new subject, and apply inverse operation as for solving equations within the opposite order to the order conventions.
Now in the question we will be making ‘b’ as the subject of the formula.
So, we will start with the equation given to us $\dfrac{a+1}{b}+3=\dfrac{4a}{b}$ ………… (1)
Now we have to make ‘b’ as the subject of the formula so to do so we will be applying inverse operations on the equation (1).
Now we will be taking b as LCM in left hand side in (1)
Now we will be eliminating the b from the denominator in the equation. By doing this we will get
Now we will be doing inverse operation in the LHS to eliminate all other variables and constant in the left-hand side except b.
So, we will subtract a+1 on both sides on the equation.
Now dividing the whole equation by 3 on both the sides. We will get
So, option (b) $b=\dfrac{3a-1}{3}$ is the correct option.

Note: These types of questions are usually calculation-intensive so calculation mistakes must be kept in mind while solving such questions also while applying inverse operations care must be taken as to what inverse operation has to be done in order to isolate the subject.