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

1. ## 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.

2. solve foe y in each quation and graph them bolth.....

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

dan

3. 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

4. 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.

5. 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

6. 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.

7. 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.

8. 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

9. 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

10. 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.

11. 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! )

12. 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 . x = - 2.73 not -2.72

13. 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 . 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

14. 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.

15. 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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