# Finding the zeros and intercepts---help / check answer?

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• Sep 15th 2008, 09:02 PM
tiar
Quote:

Originally Posted by 11rdc11
$2x - 3y = C$

$-3y = -2x + C$

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

Ok now put in whatever value for C

The only thing that is happening is the graph is moving up or down. It is easier to see what I'm trying to explain. Try plooting it and you will see what I mean.

Sorry, I took a break and did another section of math which I've completed for the most part.
Okay, I am still stuck on this question.
Let's plug in "2" for C...how would I proceed in solving it? I cannot add 2x + 2 together...I'm really trying to remember how to do this...
and how would I plot it?
• Sep 15th 2008, 09:08 PM
11rdc11
What are you trying to solve for?
• Sep 15th 2008, 09:11 PM
tiar
Wait...I'm not sure. Am I solving for anything? I think I need to solve for x and y if I want to plot the correct corresponding lines to the number I put in for C?
Ahh I'm so confused...
• Sep 15th 2008, 09:15 PM
11rdc11
If you put in 2 for C you have

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

Are they asking you to find the x and y intercepts of this equation? Thats the only thing I can see the problem is asking to solve for?
• Sep 15th 2008, 09:22 PM
tiar
I guess so?

If C is an arbitrary real constant, an equation such as 2x - 3y = C is said to define a family of lines. Choose four different values of C and plot the corresponding lines on the same coordinate axes. What is true about hte lines that are members of this family?

Do you need the intercepts to plot those lines? If you do I am not sure how to solve them. Thanks again.
• Sep 15th 2008, 10:10 PM
11rdc11
It would help. Plot them and the answer will be obvious(Happy)
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