How do I change an $Ax+By=C$ equation to a $y=mx+b$ equation where $m$ is the slope and b is the $y-$ intercept like $4x+2y=8$? \[\]

Hint: We recall the three forms of writing a linear equation: the general form $Ax+By+C=0$, the slope intercept form $y=mx+b$ and the standard form $Ax+By=C$. We take the term with which $x$ is multiplied to the right hand side and then divide both sides of the given equation $Ax+By=C$ by a coefficient of $y$ to convert it into slope-intercept form. We use obtained $m,b$ in terms of $A,B,C$ to get the slope point from of $4x+2y=8$.\[\]

Complete step by step answer:
We know from the Cartesian coordinate system that every linear equation $Ax+By+C=0$ can be represented as a line. If the line is inclined with positive $x-$axis at an angle $\theta $ then its slope is given by $m=\tan \theta $ and if it cuts $y-$axis at a point $\left( 0,b \right)$ from the origin the $y-$intercept is given by $b$. The slope-intercept form of equation is given by
\[y=mx+b....\left( 1 \right)\]
We know that the standard form of linear equation otherwise also known as intercept form is written with constant $C$ on the right side of equality sign as
\[Ax+By=C...\left( 2 \right)\]
 Let us subtract $Ax$ from both sides of the above equation to have;
\[By=-Ax+C\]
We divided both side of above equation by $B$ to have
\[y=\dfrac{-A}{B}x+\dfrac{C}{B}.....\left( 3 \right)\]
We see that the above equation is in the slope-intercept form. We compare equation (1) and (3) to have
\[m=\dfrac{-A}{B},b=\dfrac{C}{B}\]
We are given the equation$4x+2y=8$. Here we have $A=4,B=2,C=8$. The slope-point form of the equation $4x+2y=8$ with slope $m=\dfrac{-A}{B}=\dfrac{-4}{2}=-2$ and intercept $b=\dfrac{C}{B}=\dfrac{8}{2}=4$ is
\[\begin{align}
  & y=\left( -2 \right)x+4 \\
 & \Rightarrow y=-2x+4 \\
\end{align}\]

Note: We note that if $A=0$ we have $m=\dfrac{-0}{B}=0$ and the line is parallel to the $x-$axis. If $B=0$ then we have $m=\dfrac{-A}{0}=\infty $ and the line is perpendicular to the $x-$axis. If we have intercept $C=0$ then the line passes through the origin. If two lines ${{A}_{1}}x+{{B}_{1}}y={{C}_{1}},{{A}_{2}}x+{{B}_{2}}y={{C}_{2}}$ are parallel the their slopes are equal which means $\dfrac{-{{A}_{1}}}{{{B}_{1}}}=\dfrac{-{{A}_{2}}}{{{B}_{2}}}$ where we can use alternedo to have $\dfrac{{{A}_{1}}}{{{A}_{2}}}=\dfrac{{{B}_{1}}}{{{B}_{2}}}$.