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graphing linear equations?

angels-disabled
1-Newbie

graphing linear equations?

??? how do you do that? here's an example:
8x-y=16 the only thing i think i know is that...i think you only get one point and it wil just make one straight line? we're having a test on it i think next friday....but i have work on it due tomorrow!
6 REPLIES 6

In the case of 8x-y=16, the ordered pair (3,8) satisfies the equation. Can you find another pair that also satisfies it? You then plot the two points and draw the line.
PKA

O.K.? well...we did something like...get rid of x and somehow find y? i don't know..but, what about a problem such as... 2x-5y=12?...i just made that up! so i'm not sure if it will work?

Look, ordered pairs are points in the plane. Lines are collections of points in the plane. These points, (p,q), satisfy the equations of lines, ax+by=c, that is a*p+b*q=c. All I have to do is find two points (u,v) & (t,s) such that a*u+b*x=c & a*t+b*s=c. Two points determine a line! The way we find these points is not fixed. So you may solve for one of the variables, either x or y. But, this is not necessary! To graph a line, all we need is two points.
PKA

??? how do you do that? here's an example:
8x-y=16 the only thing i think i know is that...i think you only get one point and it wil just make one straight line? we're having a test on it i think next friday....but i have work on it due tomorrow!

your question:
first you isolate the "y" to get it by itself ( almost )

8x-y=16
+y +y
8x=16+y

subtract 16

8x-16=y
then you just pick points:
lets say you want s to be 3
substitute:

8(3)-16=y
24-16=y
8=y
so your point is (3,8)

sorry, jackal_1500
this is from Laura

Angel,

When you mentioned you were confused about your teacher asking you to get rid of the x term, I think you were remembering about the y-intercept, which is where a line crosses the y-axis. At that point x = 0.

So in your equation, 8x - y = 16
Set x = 0.

8*0 - y = 16
0 - y = 16

- y = 16

y = -16

So one point on your line is (0,-16).
Now you have two points, with the one found earlier.

For any line, you can try a value for x in the equation, then find a matching value for y to find a point on the line.

Mona
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