Changing $y=mx+b$ equation into $ax+by=c$ I'm stuck on this question and I'm not totally sure how to transform an equation of the form $y=mx+b$ into an equation of the form $ax+by=c$. This is how far I have gotten 
$$y=-\frac{1}{3}x+\frac{29}{3}$$
How do I transform this into an $ax+by=c$ equation?
 A: Typically when changing from the the $y = mx+b$ form (slope-intercept form)  to the $ax+by=c$ form (standard form) you want $a$ and $b$ to be integers. First we can multiply both sides by three (because this is the denominator of the slope $m$), then just move the term with $x$ to the other side of the equation by adding $x$ to both sides:
$$\begin{align}
  y&=-\frac{1}{3}x+\frac{29}{3}\\
  3(y)&=3\left(-\frac{1}{3}x+\frac{29}{3}\right)\\
  3y&=-x+29\\\\
  x + 3y &= 29
\end{align}$$
More generally, if you wanted $a$,$b$, and $c$ to be integers in the standard form, you would first have to take the denominators of $m$ and $b$ from the slope-intercept form and multiply through by their least common multiple (in our case here, that least common multiple happens to just be $3$).
A: You transform it into
$$
\frac{1}{3} x + y = \frac{29}{3}
$$
by adding $(1/3) x$ to both sides of the equation. So you got $a = 1/3$ and $b = 1$ and $c = 29/3$.
In fact for $\lambda \ne 0$ the set of solutions $(x,y)$ does not change, if we multiply both sides by $\lambda$:
$$
\frac{\lambda}{3}x + \lambda y = \frac{29\lambda}{3}
$$
Among all those equations one can decide for a nice looking one, in this case $\lambda = 3$ looks a bit less complex (no division):
$$
x + 3y = 29
$$
A: Generally when you  write it in the format $ax +by =c$ you want the coefficients to be integers for simplicity. But dividing the coefficients won't affect the properties of the line in anyway. The following equations represent the same line.
$y= \frac{-1}{3} x +\frac{29}{3}$
$ \frac{1}{3} x+y=\frac{29}{3}$
$x+ 3y=29$
