Question

The solution of the equation \[\dfrac{3a-2}{3}+\dfrac{2a+3}{2}=a+\dfrac{7}{6}\] is
1) \[a=2\]
2) \[a=\dfrac{1}{3}\]
3) \[a=-2\]
4) \[a=\dfrac{1}{2}\]

Answer 1

Hint: Here, in this type of problem, we always approach the sum to solve by taking LCM then by cross multiplying with LHS and RHS and by separating the constant on one side and variable on another side we get the value of ‘a’.

Complete step-by-step solution:
According to the question it is given that \[\dfrac{3a-2}{3}+\dfrac{2a+3}{2}=a+\dfrac{7}{6}\]
Here, we have to find the solution that means we have to find the value of ‘a’.
For that first of all we need to take the LCM which is 6 on LHS then we get:
\[\Rightarrow \dfrac{2\left( 3a-2 \right)+3\left( 2a+3 \right)}{6}=a+\dfrac{7}{6}\]
We also need to take the LCM which is 6 on RHS then we get:
\[\Rightarrow \dfrac{2\left( 3a-2 \right)+3\left( 2a+3 \right)}{6}=\dfrac{6a+7}{6}\]
If you observe the above equation then you can see that 6 gets cancelled then we write the equation as
\[\Rightarrow 2\left( 3a-2 \right)+3\left( 2a+3 \right)=6a+7\]
By simplifying further this above equation we get:
\[\Rightarrow 6a-4+6a+9=6a+7\]
Here in this above equation, we can see that one \[6a\] gets cancelled we get:
\[\Rightarrow 6a-4+9=7\]
By rearranging the term, we get:
\[\Rightarrow 6a=7-5\]
By solving the further we get:
\[\Rightarrow 6a=2\]
Here, if you see then 2 and 6 gets cancelled and remain 3 on LHS we get:
\[\Rightarrow 3a=1\]
By dividing by 3 on both sides we get:
 \[\Rightarrow a=\dfrac{1}{3}\]
So, the correct option is “option B”.

Note: While solving this type of problem always remember that variables are always on one side and constants on other sides which makes the problem more comfortable to solve and also avoid silly mistakes in this type of problem. In this step \[6a-4+6a+9=6a+7\] you can also solve LHS first that is \[12a-4+9=6a+7\] and shift variable on one side and constants on other side and after simplifying then also you will get the same answer as in the above solution.