How do you write the equation \[y+7=\dfrac{1}{2}\left( x+2 \right)\] in the standard form?

Hint: The degree of the equation is the highest power to which the variable is raised. The degree of the equation decides if the equation is linear, quadratic, cubic, etc. The standard form of a straight-line equation is \[ax+by+c=0\], here \[a,b,c\in \] Real numbers. To convert a straight-line equation to its standard form, we have to take all the terms to one side of the equation.

Complete step by step answer:
We are given the equation \[y+7=\dfrac{1}{2}\left( x+2 \right)\] and have to convert it to standard form. The highest power of the variable in the equation is 1, hence the degree is 1. It means that the equation is linear and it represents a straight line. The standard form of the straight-line equation is \[ax+by+c=0\]. To convert a straight-line equation to its standard form, we have to take all the terms to one side of the equation. Doing this for the given equation, we get
\[y+7=\dfrac{1}{2}\left( x+2 \right)\]
Expanding the bracket on RHS of equation, we get
\[\Rightarrow y+7=\dfrac{1}{2}x+\dfrac{1}{2}\times 2\]
\[\Rightarrow y+7=\dfrac{1}{2}x+1\]
Subtracting 1 from both sides of the equation, we get
\[\Rightarrow y+7-1=\dfrac{1}{2}x+1-1\]
Subtracting \[\dfrac{1}{2}x\] from both sides of the equation, we get
\[\begin{align}
  & \Rightarrow y+6-\dfrac{1}{2}x=\dfrac{1}{2}x-\dfrac{1}{2}x \\
 & \Rightarrow -\dfrac{1}{2}x+y+6=0 \\
\end{align}\]
We have taken all terms to the LHS of the equation; hence it is the standard form equation. Comparing with the general standard form of the straight-line equation. We get, \[a=-\dfrac{1}{2},b=1,\] and \[c=6\].

Note:
 If we multiply both sides of the standard form of the equation by 2, we get
\[\begin{align}
  & \Rightarrow \left( -\dfrac{1}{2}x+y+6 \right)\times 2=0\times 2 \\
 & \Rightarrow \left( -\dfrac{1}{2}x \right)\times 2+y\times 2+6\times 2=0 \\
 & \Rightarrow -x+2y+12=0 \\
\end{align}\]
This equation also satisfies the properties of standard form. Hence, we can say this to be standard form as it will not change the characteristics of line like slope, and intercepts.