View Full Version : graph the system: y = x - 4 and y = 2 - x

kathy's

07-04-2006, 06:01 PM

I am given this system of equations:

. . .(a) y = x - 4

. . .(b) y = 2 - x

I have to graph this manually and find the intersection. I know how to graph it right with this formula: y=mx+b

But is the slope how can you tell and what number is the y intercept? I believe it goes as follows:

. . .for (a)

. . .m(slope) = 1/1

. . .b(y intercept) = -4

. . .for (b)

. . .m = -1/1

. . .b = 2

I think I am doing something wrong, because I don't know how to greaph when they only give me one number.

stapel

07-04-2006, 06:32 PM

But is the slope how can you tell and what number is the y intercept?

I don't know how to greaph when they only give me one number.

I'm sorry, but I don't understand what you mean in these sentences.

You say you know how to graph the lines. So graph the two lines. They're asking you to find the intersection point. Once you've graphed the lines, look at the graph, and find (by looking) the point where the two lines cross. This will be the solution point for the system of equations.

Eliz.

skeeter

07-05-2006, 09:02 AM

you want these two given equations ...

(a) y = x - 4

(b) y = 2 - x

in y = mx + b form

for (a) ...

y = 1x + (-4) , m = 1 and b = -4

for (b) ...

y = 2 - x

rewrite it as

y = -1x + 2 , m = -1 and b = +2

can you do your graphs now?

jonboy

07-06-2006, 08:13 AM

I hope that you do not have to plot a million points till you find the intersection point. I would solve the system and find the intersection point, then plot it, and then plot the line the two lines that makes the intersection.

\L\bold y=x-4

\L\bold y=-x+2

Since both equations are equal to y, then just set both of the equations equal to each other.

Given: \L\bold x-4=-x+2

Addition Property: \L\bold x=-x+6

Addition Propery: \L\bold 2x=6

Division Property: \L\bold x=3

So fill \L\bold x in one of our original equations to find \L\bold y (any will work since both are equal to \L\bold y).

Graph this point and graph some points representing the two lines.

jonboy

07-06-2006, 12:25 PM

Now once you fill in your values for x and find the y you start to notice a pattern so it makes it very simple. Your graph should look something like this:

http://img317.imageshack.us/img317/2402/goodgraph5tc.png

Powered by vBulletin® Version 4.2.2 Copyright © 2014 vBulletin Solutions, Inc. All rights reserved.