# A Graph Problem :?

• May 15th 2007, 03:42 AM
IHATECHONGAS
A Graph Problem :?
Graph the line x+y= 5 on the graph provided. Then graph the results.

x|y
5|0
0|?
?|?

Graph the line x-y=1 on the graph.

x|y
?|?
?|?
?|?

What is the point of intersection of the two lines?
What is the slope of line y=3/4x-1?
What is the y-intercept of line 2x+3y=6?

Find an equation of the line with slope -4/3 and y-intercept 3. Leave equation in slope-intercept form.

Find an equation of the line that goes thru the point (2,1) and has a slope of 4. Leave equation in point-slope form.

I know it's a lot but I really need a lot of help. Thank you in advance!
• May 15th 2007, 04:24 AM
r_maths
x+y= 5
change this so it becomes y = 5 -x
This way, u can choose any x value and subtitute it in to get y.
ie. x = 0
y = 5 - 0
y = 5
Point (0, 5)
Choose at least 3 other points to draw the line.
http://img175.imageshack.us/img175/6301/xy5jf5.gif

x-y=1
Do the same.
y = x -1
Choose at least 3 other points to draw the line.
http://img175.imageshack.us/img175/8241/xy1of9.gif

What is the point of intersection of the two lines?
In the graph I made the dots red, where the lines intersects x and y axes.

What is the slope of line y=3/4x-1?
This is in the form of y = mx + c
m is the slope so slope = 3/4

What is the y-intercept of line 2x+3y=6?
Y intercept is when x = 0
Substitute this in and you get: 2*0 + 3y = 6
3y = 6
y = 2
Y-intercept = (0, 2)

Find an equation of the line with slope -4/3 and y-intercept 3. Leave equation in slope-intercept form.
I don't know if this is what they are looking for.
3 = -4/3x + c
Y - Int. x = 0
y = -4/3 + 3

Find an equation of the line that goes thru the point (2,1) and has a slope of 4. Leave equation in point-slope form.
I don't know if this is what they are looking for.
1 = 4x + c
1 = 8 + c
c = -7
• May 15th 2007, 05:58 AM
IHATECHONGAS
Thank you so much for all your help! It is much appreciated! Can I ask you..how do I figure out how to fill out the input/output table? And i'm afraid the last two questions aren't what they are looking for but I sooo appreciate all your help! Thanks a mil! :)
• May 15th 2007, 06:18 AM
r_maths
Quote:

Originally Posted by IHATECHONGAS
Can I ask you..how do I figure out how to fill out the input/output table?

x+y= 5
change this so it becomes y = 5 -x

x|y
5|0
0|?
?|?

You have the x value. Subtitute this into
y = 5 -x

y = 5 - 5 = 0
y = 5 - 0 = 5
y = 5 - ? = (in this table, the x's are going down in 5s so the next value should be -5.)
y = 5 - (-5) = 10
• May 15th 2007, 10:51 AM
IHATECHONGAS
Sooo..on this table it would be backwards???

Graph the line x-y=1 on the graph.

x|y
?|?
?|?
?|?
• May 15th 2007, 11:23 AM
r_maths
Quote:

Originally Posted by r_maths
x+y= 5
x-y=1
Do the same.
y = x -1
Choose at least 3 other points to draw the line.
http://img175.imageshack.us/img175/8241/xy1of9.gif

Make up 3 x points and substitute into equation y = x -1
For example. if you look at the line I drew for you.
When x = 0 y = -1 hence the point (0, -1)