PDA

View Full Version : System of Equation

amai
02-12-2007, 09:44 PM
Solve this system by graphing:
y= x - 1
y= 9 - x

I understand the first one; y= x - 1. My first point would be at -1, then I rise 1 point and run 1 point (x always has a 1 in front of it), and that's my second point.

I get that.

I don't get the second one though; y= 9 - x

Is x even suppose to be there? Do I move it? Switch it around with 9?

Here's what I think. Leaving the equation just as it is; my first point on the graph would be at -1, then I would rise 9 and run 1, that'll be my second point.

BUT!

When I graph that my solution is (0,1), but it's suppose to be (5,4).

So I've gone wrong some where...any ideas how to do this!

Much appreciated.

Epsilon=One
02-13-2007, 08:10 AM
Is x even suppose to be there? Do I move it? Switch it around with 9?Doesn't matter; if you don't change any signs.

Here's what I think. Leaving the equation just as it is; my first point on the graph would be at -1...What would be the second coordinate of the first point?

...then I would rise 9 and run 1, that'll be my second point.No.

Clue: The slope of your graphed lines are opposite and will cross at: 5,4.

HallsofIvy
02-14-2007, 10:32 AM
"y= x - 1
y= 9 - x

I understand the first one; y= x - 1. My first point would be at -1"
I presume you mean x= -1. y, of course, is -1-1= -2.

", then I rise 1 point and run 1 point (x always has a 1 in front of it), and that's my second point.

I get that."
Good

"I don't get the second one though; y= 9 - x"
Same thing. Choose some value for x- I'm not sure why you chose x= -1 before (I always prefer x= 0) but that will work. When x= -1, y= 9-(-1)= 10 so one point is (-1,10). Now you can use "run= 1, rise= -1" (because of the -1 in front of the x).

A more fundamental method is to just find two points- you know the graphs are straight lines and any straight line is determined by two points.

For example, for y= x- 1, you've already determined that when x=-1, y= -2 so (-1 -2) is a point on the graph. Now choose x to be any other number, say x= 1. When x= 1, y= 1- 1= 0 so (1, 0) is also a point on the graph. Mark those two points and draw the line through them.

For y= 9- x, choose any two values of x you like. When x= 0, for example, y= 9- 0= 9. Mark the point (0, 9). When x= 5, y= 9- 5= 4. Mark the point (5,4). Draw the line through those two points. That's all you need to do.

Mr. Robin Parsons
02-14-2007, 03:15 PM
Same thing. Choose some value for x- I'm not sure why you chose x= -1 before (I always prefer x= 0) but that will work. When x= -1, y= 9-(-1)= 10 so one point is (-1,10). Now you can use "run= 1, rise= -1" (because of the -1 in front of the x).
A more fundamental method is to just find two points- you know the graphs are straight lines and any straight line is determined by two points.
For example, for y= x- 1, you've already determined that when x=-1, y= -2 so (-1 -2) is a point on the graph. Now choose x to be any other number, say x= 1. When x= 1, y= 1- 1= 0 so (1, 0) is also a point on the graph. Mark those two points and draw the line through them.
For y= 9- x, choose any two values of x you like. When x= 0, for example, y= 9- 0= 9. Mark the point (0, 9). When x= 5, y= 9- 5= 4. Mark the point (5,4). Draw the line through those two points. That's all you need to do.Like (I) have said, an intelligent person at math...et al :cool:

stonebro
03-16-2007, 04:14 AM
Solve this system by graphing:
y= x - 1
y= 9 - x

For simplicity just write

y = x - 1
y = -x + 9

Since y = 9 - x and y = -x + 9 obviously are the same. To draw a line you need either a) two points on the line or b) one point and the slope. You can find both easily in the equations.

For a), substitue x = 0 in both equations so that you get y(0) = 0 - 1 = -1 for the first, now you know that the first line passes through the point (x, y) = (0, -1). For the second line you get y(0) = 9 - x = 9, so this line passes through (0, 9). Mark these points in your coordinate system. Now do the same thing with a second point for both equations, and draw the lines. Extend in any direction as far as you need to for them to cross, the crossing point is of course the solution.

For b), just substitute x=0 in both to get a point, then the slope of the line is just the constant that's multiplied with the free variable, x in your equations. So the first line has a slope of +1, the second a slope of -1. This means that the first line will rise 1 unit in y for 1 unit in x. You can now draw the line with a ruler. Rinse and repeat for the second.

Solve analytically to confirm your answer by noting that the lines must have the same (x,y) values in the solution point so you can just put the two y's equal to eachother:

x - 1 = -x + 9 -> 2x = 10 -> x = 5

But x = 5 means that y = 5-1 = 4 so that solution is indeed (4,5).