Question

How do you solve $4\left( x+3 \right)=6x$?

Answer 1

Hint: The equation given in the question $4\left( x+3 \right)=6x$ has a single variable of the highest equal to one. So the given equation is a linear equation. To solve this equation, we have to solve the bracket on the left hand side, that is, multiply $4$ with $\left( x+3 \right)$ using the distributive law. Then, we need to separate the constants and the variables by writing the variables on the left hand side and the constants on the right hand side of the equation. Finally, applying the algebraic operations on the equation, we will obtain the final solution of the given equation.

Complete step by step answer:
The equation is given in the above question as
$4\left( x+3 \right)=6x$
We know from the distributive property of multiplication, we know that $a\left( b+c \right)=ab+ac$. Applying this property on the LHS of the above equation, we can write
$\begin{align}
  & \Rightarrow 4x+4\left( 3 \right)=6x \\
 & \Rightarrow 4x+12=6x \\
\end{align}$
Now for solving the above equation, we have to separate the variable and the constant terms. For this we subtract $12$ from both sides of the above equation to get
\[\begin{align}
  & \Rightarrow 4x+12-12=6x-12 \\
 & \Rightarrow 4x=6x-12 \\
\end{align}\]
Now, we subtract $6x$ from both sides of the above equation to get
$\begin{align}
  & \Rightarrow 4x-6x=6x-12-6x \\
 & \Rightarrow -2x=-12 \\
\end{align}$
Finally, on dividing both sides by $-2$ we get
$\begin{align}
  & \Rightarrow \dfrac{-2x}{-2}=\dfrac{-12}{-2} \\
 & \Rightarrow x=6 \\
\end{align}$

Hence, the required solution of the given equation $4\left( x+3 \right)=6x$ is $x=6$.

Note: After obtaining the solution of the given equation, back substitute it into the given equation to check if it is correct or not. Substituting $x=6$ in $4\left( x+3 \right)=6x$ we will obtain $36=36$ which is obviously true. But in the case the LHS is not coming equal to the RHS after back substituting then it means that the solution is incorrect.