Question

How do you solve $4a-3\left( a-2 \right)=2\left( 3a-2 \right)$ ?

Answer 1

Hint: To solve the above equation which is given as: $4a-3\left( a-2 \right)=2\left( 3a-2 \right)$. We are going to simplify L.H.S and R.H.S of the equation separately and then we will arrange the equation in such a way so that $''a''$ will be on one side of the equation and the constants on another side of the equation.

Complete step by step answer:
The equation in $''a''$ which we have to find the solution of is as follows:
$4a-3\left( a-2 \right)=2\left( 3a-2 \right)$
In the above equation, on L.H.S of the above equation, we are going to multiply 3 by $\left( a-2 \right)$ and on R.H.S of the above equation, we are going to multiply 2 by $\left( 3a-2 \right)$ we get,
$\begin{align}
  & 4a-3a+6=6a-4 \\
 & \Rightarrow a+6=6a-4 \\
\end{align}$
Now, arranging the above equation in such a manner so that $''a''$ terms on one side of the equation and the constants on another side and we get,
Subtracting $''a''$ on both the sides we get,
$\begin{align}
  & a-a+6=6a-a-4 \\
 & \Rightarrow 6=5a-4 \\
\end{align}$
Now, adding 4 on both the sides we get,
$\begin{align}
  & 6+4=5a \\
 & \Rightarrow 10=5a \\
\end{align}$
Dividing 5 on both the sides we get,
$\begin{align}
  & \dfrac{10}{5}=a \\
 & \Rightarrow a=2 \\
\end{align}$
From the above, we have solved the value of $''a''$ as 2. Hence, the solution of the given equation is $a=2$.

Note: The other way of solving the above problem is as follows:
The equation given in the above problem of which we have to find the solution is:
$4a-3\left( a-2 \right)=2\left( 3a-2 \right)$
Adding $3\left( a-2 \right)$ on both the sides we get,
$\begin{align}
  & 4a-3\left( a-2 \right)+3\left( a-2 \right)=2\left( 3a-2 \right)+3\left( a-2 \right) \\
 & \Rightarrow 4a=6a-4+3a-6 \\
 & \Rightarrow 4a=9a-10 \\
\end{align}$
Now, writing $''a''$ terms on one side of the equation and the constants on another side of the equation we get by adding 10 on both the sides.
$\begin{align}
  & 4a+10=9a-10+10 \\
 & \Rightarrow 4a+10=9a \\
\end{align}$
Subtracting $4a$ from both the sides we get,
\[\begin{align}
  & 4a-4a+10=9a-4a \\
 & \Rightarrow 10=5a \\
 & \Rightarrow \dfrac{10}{5}=a \\
 & \Rightarrow 2=a \\
\end{align}\]
As you can see that in the above solution also we are getting the same value of $''a''$ which is equal to 2. Through this method, we have also verified that the value of $''a''$ that we have solved in the above solution is correct.