Question

Solve the following:
\[3a - \dfrac{1}{5} = \dfrac{a}{5} + 5\dfrac{2}{5}\]
A. 2
B. 4
C. 6
D. 3

Answer 1

Hint: In the given problem we need to solve this for ‘a’. Here we need to be careful in the RHS. Here we have a mixed fraction, first, we need to convert them into improper fractions. We can solve this using the transposition method. The common transposition method is to do the same thing (mathematically) to both sides of the equation, with the aim of bringing like terms together and isolating the variable (or the unknown quantity). That is we group the ‘a’ terms on one side and constants on the other side of the equation.

Complete step by step answer:
Given, \[3a - \dfrac{1}{5} = \dfrac{a}{5} + 5\dfrac{2}{5}\].
First convert the mixed fraction,
\[5\dfrac{2}{5} = 5 + \dfrac{2}{5}\]
\[5\dfrac{2}{5} = \dfrac{{25 + 2}}{5}\]
\[5\dfrac{2}{5} = \dfrac{{27}}{5}\].
Above problem becomes,
\[3a - \dfrac{1}{5} = \dfrac{a}{5} + \dfrac{{27}}{5}\]
We transpose \[ - \dfrac{1}{5}\] which is present in the left-hand side of the equation to the right hand side of the equation by adding \[\dfrac{1}{5}\] on the right hand side of the equation.
\[3a = \dfrac{a}{5} + \dfrac{{27}}{5} + \dfrac{1}{5}\]
Similarly, we transpose \[\dfrac{a}{5}\] to the left-hand side of the equation by subtracting \[\dfrac{a}{5}\] on the left hand side of the equation.
\[3a - \dfrac{a}{5} = \dfrac{{27}}{5} + \dfrac{1}{5}\]
Taking LCM and simplifying we have
\[\dfrac{{15a - a}}{5} = \dfrac{{27 + 1}}{5}\]
\[\dfrac{{14a}}{5} = \dfrac{{28}}{5}\]
Multiply 5 on both sides of the equations,
\[14a = 28\]
Divide the whole equation by 14
\[a = \dfrac{{28}}{{14}}\]
\[ \Rightarrow a = 2\].

Note:
We can solve this by substituting the given options in the given problem. After simplifying if we have LHS is equal to RHS then our substituted option is the correct one. But for some problems, this method won’t work.
In the above, we did the transpose of addition and subtraction. Similarly, if we have multiplication, we use division to transpose. If we have division, we use multiplication to transpose. Follow the same procedure for these kinds of problems.