Am I graphing this equation correctly?

**Somatic** · September 30th 2009, 12:22 PM

Hello everyone!
I'm not sure if I'm doing this right. My instructions were to "Graph the given equations:"

$\frac{y}{3}-\frac{x}{6} = -2$

I'm using these numbers for x.
-2, -1, 0, 1, 2

This is what I've been doing:

(1) $\frac{y}{3}-\frac{-2}{6} = -2$

(2)Use LCD to get rid of fraction? *
I used: $\frac{3}{1}$ on both sides. then I got:
$y - \frac{-2}{2} = -6$

(3) $y - (-1) = -6$ *

(4)Then I added -1 to both sides.
$y = -7$

So that means my first point would be:
$(-2,-7)$

Then when I try to do it when $x = -1$
I get $y = -2$

With x = -1, I think I mess up on step 4.
$y - \frac{-1}{2} = - 6$

I subtract both sides with $\frac{-1}{2}$
And I keep getting $y = -2$

Help please and thanks in advance.

* = Not sure about this step.

**e^(i*pi)** · September 30th 2009, 12:35 PM

Originally Posted by Somatic

Hello everyone!
I'm not sure if I'm doing this right. My instructions were to "Graph the given equations:"

$\frac{y}{3}-\frac{x}{6} = -2$

I'm using these numbers for x.
-2, -1, 0, 1, 2

This is what I've been doing:

(1) $\frac{y}{3}-\frac{-2}{6} = -2$

(2)Use LCD to get rid of fraction? *
I used: $\frac{3}{1}$ on both sides. then I got:
$y - \frac{-2}{2} = -6$

(3) $y - (-1) = -6$ *

(4)Then I added -1 to both sides.
$y = -7$

So that means my first point would be:
$(-2,-7)$

Then when I try to do it when $x = -1$
I get $y = -2$

With x = -1, I think I mess up on step 4.
$y - \frac{-1}{2} = - 6$

I subtract both sides with $\frac{-1}{2}$
And I keep getting $y = -2$

Help please and thanks in advance.

* = Not sure about this step.

Convert the equation into the form $y=mx+c$

$<br /> \frac{y}{3}-\frac{x}{6} = -2<br />$

$\frac{y}{3} = -2 + \frac{x}{6}$

Multiply by 3:

$y = \frac{x}{2} - 6 = \frac{x-12}{2}$

Now is should be easier to see that $f(-1)$ is $-\frac{13}{2}$ and $f(-2) = -7$

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