Question

If an algebraic expression is given as \[\dfrac{{a + 1}}{b} + 3 = \dfrac{{4a}}{b}\] , then b can be written as
A). \[b = \dfrac{{3a + 1}}{3}\]
B). \[b = \dfrac{{ - 3a + 1}}{3}\]
C). \[b = \dfrac{{3a - 1}}{3}\]
D). \[b = \dfrac{{ - 2a + 1}}{3}\]

Answer 1

Hint: We need to find the value of b in terms of a for that first take the LCM in the left hand side then eliminate b from denominators of both the sides, and finally evaluate the remainings to get the value of b.

Complete step-by-step answer:
Firstly let us try to take the LCM of Left hand side terms i.e., \[\dfrac{{a + 1}}{b} + 3\]
So we will get it as
\[\begin{array}{l}
\therefore \dfrac{{a + 1}}{b} + 3\\
 = \dfrac{{a + 1 + 3 \times b}}{b}\\
 = \dfrac{{a + 3b + 1}}{b}
\end{array}\]
So now the value of \[\dfrac{{a + 1}}{b} + 3 = \dfrac{{4a}}{b}\] becomes \[\dfrac{{a + 3b + 1}}{b} = \dfrac{{4a}}{b}\] So now we can cancel out b from both the sides and we will be left with
\[\begin{array}{l}
\therefore \dfrac{{a + 3b + 1}}{b} = \dfrac{{4a}}{b}\\
 \Rightarrow a + 3b + 1 = 4a
\end{array}\]
Now from here Let us try to find the value of b, which will be
\[\begin{array}{l}
 \Rightarrow a + 3b + 1 = 4a\\
 \Rightarrow 3b + 1 = 4a - a\\
 \Rightarrow 3b + 1 = 3a\\
 \Rightarrow 3b = 3a - 1\\
 \Rightarrow b = \dfrac{{3a - 1}}{3}
\end{array}\]
So from here it is clear that option C is the correct option here.

Note: Beware of the signs while solving the linear equation for the value of b in terms of a, also that when + goes to other side of the equal to sign it becomes - and when \[ \times \] goes to other side of equal to it becomes \[ \div \]