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How do you write the equation $y+6=\dfrac{3}{2}\left( x-4 \right)$ in standard form?

Answer
VerifiedVerified
535.5k+ views
Hint: Here in this question we have been asked to write the standard form of the given straight line equation $y+6=\dfrac{3}{2}\left( x-4 \right)$ . From the basic concepts of straight line we know that the standard form of a straight line equation is given as $ax+by=c$ where $\dfrac{-a}{b}$ is the slope of the straight line.

Complete step by step answer:
Now considering from the question we have been asked to write the standard form of the given straight line equation $y+6=\dfrac{3}{2}\left( x-4 \right)$ .
From the basic concepts of straight line we know that the standard form of a straight line equation is given as $ax+by=c$ where $\dfrac{-a}{b}$ is the slope of the straight line.
For doing that, we will simplify the given equation using arithmetic operations. Now we will multiply both sides of the expression with two. After doing that we will have $\Rightarrow 2\left( y+6 \right)=3\left( x-4 \right)$ .
Now we will further simplify this expression
$\begin{align}
  & \Rightarrow 2y+12=3x-12 \\
 & \Rightarrow 3x-12-2y-12=0 \\
 & \Rightarrow 3x-24-2y=0 \\
 & \Rightarrow 3x-2y=24 \\
\end{align}$

Therefore we can conclude that the standard form of the given straight line equation $y+6=\dfrac{3}{2}\left( x-4 \right)$ is given as $3x-2y=24$

Note: While answering questions of this type we should be sure with the straight line equations concepts that we are going to apply in between the process and the calculations that we are going to perform in between the steps. This is a very simple and easy question and can be answered accurately in a short span of time. Very few mistakes are possible in questions of this type. If some had confused and made a mistake during the calculation and written the equation as
 $\begin{align}
  & \Rightarrow 2y=3x-24 \\
 & \Rightarrow 2y=3\left( x-8 \right) \\
 & \Rightarrow y=\dfrac{3}{2}\left( x-8 \right) \\
\end{align}$
then we will end up having a wrong conclusion.