# How Can I Solve Linear Equations Graphically, Not Algebretically??

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• Sep 26th 2006, 05:13 PM
NemesysEvolved
How Can I Solve Linear Equations Graphically, Not Algebretically??
Ive been given a graph to plot my 2 lines, and find the point of intersection. I have 2 equations naturally, and they go like this.

2x + 5y= 10
3x + 2y=-2

How do I go about this? Im also assuming once I solve these algebretically the point of intersection on the graph will be the points I get from the algebretic part. Much thx peeps I really need this.
• Sep 26th 2006, 05:37 PM
dan
solve foe y in each quation and graph them bolth.....

if you solve algebreticlly first you will have a more precice anwer...

dan
• Sep 26th 2006, 05:47 PM
dan
graph the equations y=(10 -2x)/5 and y=(-2 - 3x)/3

the graphs cross at (-4.444 , 3.777)

have a nice day^_^

dan
• Sep 26th 2006, 05:52 PM
NemesysEvolved
ya but if I solve them algebretically I only get 1x value and 1 y value,
x=30/-11 and y=34/11, which are both numbers with like 10 decimals.

Im suppose to graph each equation as a line then get the point of intersection.
• Sep 26th 2006, 05:56 PM
dan
Quote:

Originally Posted by NemesysEvolved
ya but if I solve them algebretically I only get 1x value and 1 y value,
x=30/-11 and y=34/11, which are both numbers with like 10 decimals.

Im suppose to graph each equation as a line then get the point of intersection.

go ahed and graph them...plot some values for each equation. the quations are liner and should only cross once...

dan
• Sep 26th 2006, 05:57 PM
NemesysEvolved
Quote:

Originally Posted by dan
graph the equations y=(10 -2x)/5 and y=(-2 - 3x)/3

the graphs cross at (-4.444 , 3.777)

have a nice day^_^

dan

ok but how can I possibly put those on a point on a piece of graph paper, their not even whole numbers there an undefined value in it being x. I need to have a point for x and a point for y on each equation so I can make 2 straight lines from each equation and get the point of intersection.
• Sep 26th 2006, 05:58 PM
OReilly
Quote:

Originally Posted by NemesysEvolved
Ive been given a graph to plot my 2 lines, and find the point of intersection. I have 2 equations naturally, and they go like this.

2x + 5y= 10
3x + 2y=-2

How do I go about this? Im also assuming once I solve these algebretically the point of intersection on the graph will be the points I get from the algebretic part. Much thx peeps I really need this.

Solve both equations for y:
1) y=(10-2x)/5
2) y=(-2-3x)/2

Blue line represents equation 2x + 5y= 10 and red line 3x + 2y=-2.

Dan has right answer (-4.444 , 3.777), but my graph doesn't show that x = -4.444.
I am puzzled by that, graph was from MS Excel. In graph clearly can be seen that x is between -2 and -4. That is error, maybe bug in MS Excel.
• Sep 26th 2006, 06:01 PM
dan
Quote:

Originally Posted by OReilly
Solve both equations for y:
1) y=(10-2x)/5
2) y=(-2-3x)/2

Blue line represents equation 2x + 5y= 10 and red line 3x + 2y=-2.

Dan has right answer (-4.444 , 3.777), but my graph doesn't show that x = -4.444.
I am puzzled by that, graph was from MS Excel. In graph clearly can be seen that x is between -2 and -4. That is error, maybe bug in MS Excel.

my calc got a defferent graph... i'm doing the graphs on my laptop and writing you on a frends computer so i cant show you but there must be a bug...

dan
• Sep 26th 2006, 06:06 PM
dan
Quote:

Originally Posted by NemesysEvolved
ok but how can I possibly put those on a point on a piece of graph paper, their not even whole numbers there an undefined value in it being x. I need to have a point for x and a point for y on each equation so I can make 2 straight lines from each equation and get the point of intersection.

you don't need to plot a whole number... just plot where the graphs intersect and lable the point (-4.44444,3.7777777)

if you are still having truble let us know...

dan
• Sep 26th 2006, 06:11 PM
OReilly
Quote:

Originally Posted by dan
my calc got a defferent graph... i'm doing the graphs on my laptop and writing you on a frends computer so i cant show you but there must be a bug...

dan

2x + 5y= 10 ---> x=(10-5y)/2
3x + 2y=-2 ---> x=(-2-2y)/3

(2-2y)/3=(10-5y)/2
2(-2-2y)=3(10-5y)
-4-4y=30-15y
11y=34
y=34/11
y=3.09

3x+2*3.09=-2
3x+6.18=-2
3x=-2-6.18
x=-8.18/3
x=-2.72

So correct answer is (-2.72, 3.09).

MS Excel graph that I putted is correct.
• Sep 26th 2006, 06:14 PM
topsquark
You guys need to check your algebra. I'm getting:
(x,y) = (-30/11, 34/11) = (-2.73, 3.09)
which matches the graph.

-Dan

(Ya beat me! :) )
• Sep 26th 2006, 06:21 PM
OReilly
Quote:

Originally Posted by topsquark
You guys need to check your algebra. I'm getting:
(x,y) = (-30/11, 34/11) = (-2.73, 3.09)
which matches the graph.

-Dan

(Ya beat me! :) )

I have doubted first in MS Excel graph, trusting dan more than Bill Gates, but Bill won this time!

And I forgot to round x :D . x = - 2.73 not -2.72
• Sep 26th 2006, 06:36 PM
NemesysEvolved
Quote:

Originally Posted by OReilly
I have doubted first in MS Excel graph, trusting dan more than Bill Gates, but Bill won this time!

And I forgot to round x :D . x = - 2.73 not -2.72

ok so how do I plot numbers into each equation when each equation is put into x=10-5y/2, and x=(-2-2y)/3, becaue I just made my own values for y then used to coordinates for xand y that i got from each equation and put them on my graph, but the point of intersection is nowhere near the x=-2.73 and the other u mentioned.So what am I doing wrong here?

What I did here was put y=4 into the 1st equation, and got the coordinates (-5,4) and for the second I put y=6 into it, and got (-4.6666667,6), when I put those on the graph my points of intersection were (-3.8, 0.9) roughly
• Sep 26th 2006, 06:45 PM
OReilly
Quote:

Originally Posted by NemesysEvolved
ok so how do I plot numbers into each equation when each equation is put into x=10-5y/2, and x=(-2-2y)/3, becaue I just made my own values for y then used to coordinates for xand y that i got from each equation and put them on my graph, but the point of intersection is nowhere near the x=-2.73 and the other u mentioned.So what am I doing wrong here?

What I did here was put y=4 into the 1st equation, and got the coordinates (-5,4) and for the second I put y=6 into it, and got (-4.6666667,6), when I put those on the graph my points of intersection were (-3.8, 0.9) roughly

I am not sure what you did but you should take for both equations values lets say from -5 to 5 for y and calculate x and you get coordinates you can put on graph.

So, take first equation x=10-5y/2 and calculate all values x for y={-5,-4,-3,...5} and draw line on graph and do again for second equation and you will get intersection.
• Sep 26th 2006, 07:23 PM
NemesysEvolved
can you tell me if im solving this right, there is 2 different ways I can enter this in my calculator and I get 2 different answers.

x=10-5y/2
im gonna sub y for 1
x=10-5(1)/2 so that equals 2.5, so the coordinates are (2.5,1)
now if you calculate the value of y using the same equation but subbling in y=2, I get the coordinates (0,2), then if I sub in y=3 i get the coordinates (-2.5,3) That line does not line up on the graph!! Im lost completely it seems so simple but isnt.
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