Great answer Ninja,
Grreat use of LaTex too! (I answered your latex question for you as well)
I'm going to do it a different way (and I will practice some LaTex while I am at it)
3y-5x=-13
Ifx=0then3y=−13soy=−133=−413(0,−413)Ify=0then−5x=−13sox=135=235(235,0)$x$0235$y$−4130
FOR PEOPLE LEARNING LATEX
This is the code that I have used.
\\If\; x=0\;\; then \;\;3y=-13 \;\;so \;\;y=\frac{-13}{3}=-4\frac{1}{3}\;\;(0,-4\frac{1}{3})\\\\
If\; y=0\;\; then \;\;-5x=-13 \;\;so \;\;x=\frac{13}{5}=2\frac{3}{5}\;\;\;(2\frac{3}{5},0)\\\\
\begin{array}{|c|c|c|}\hline
x&0&2\frac{3}{5}\\\hline
y&-4\frac{1}{3}&0\\\hline
\end{array}\\
I drew the picture with GeoGebra
Are you wondering how you graph this?
To do this, you have to switch it to the slope intercept form.
The slope intercept form is y = mx + b, where m = slope and b = y intercept.
Let's work this out.
3y−5x=−13y−5/3x=−13/3y=5/3x−13/3y=5/3x−413
We can plot the point (0, -4 1/3) now, because the -4 1/3 tells us the y intercept.
Then, from that point, we can go up 5 and over 3 to the right. We do this because 5/3 tells us our slope, and it is written as up/over. I know to go 3 to the right because this is a positive slope.
If you had something like y = -5/3x + 2, you would go over 3 to the left, because that would be a negative slope.
Anyway, we go up 5 and over 3 to the right, then plot that point (put a dot there.) Now we have two points so we grab a ruler and draw a line through those two points. That's our graphed line!
Here's what this one will look like:
Great answer Ninja,
Grreat use of LaTex too! (I answered your latex question for you as well)
I'm going to do it a different way (and I will practice some LaTex while I am at it)
3y-5x=-13
Ifx=0then3y=−13soy=−133=−413(0,−413)Ify=0then−5x=−13sox=135=235(235,0)$x$0235$y$−4130
FOR PEOPLE LEARNING LATEX
This is the code that I have used.
\\If\; x=0\;\; then \;\;3y=-13 \;\;so \;\;y=\frac{-13}{3}=-4\frac{1}{3}\;\;(0,-4\frac{1}{3})\\\\
If\; y=0\;\; then \;\;-5x=-13 \;\;so \;\;x=\frac{13}{5}=2\frac{3}{5}\;\;\;(2\frac{3}{5},0)\\\\
\begin{array}{|c|c|c|}\hline
x&0&2\frac{3}{5}\\\hline
y&-4\frac{1}{3}&0\\\hline
\end{array}\\
I drew the picture with GeoGebra