# Thread: Graphical solution of simultaneous linear equations.

1. ## Graphical solution of simultaneous linear equations.

Question is...

29. Solve the following system of equations graphically.

x + y = 2

2y = -5x + 10

So, i'm pretty sure I do something like turn it into y = 2 - x then... Ah, I don't know.
I'm having a hard time grasping this for some reason.

2. Originally Posted by rocky123
Question is...

29. Solve the following system of equations graphically.

x + y = 2

2y = -5x + 10

So, i'm pretty sure I do something like turn it into y = 2 - x then... Ah, I don't know.
I'm having a hard time grasping this for some reason.
You have the right idea. The equations are both lines. Their intersection is a point (x,y) that satisfies both equations. So graph both of them on the same graph and see where they intersect.

3. I don't know how to graph it though?

4. Originally Posted by rocky123
I don't know how to graph it though?
On paper, one way that will work for these lines is to find the y-intercept and x-intercept and connect them. On a graphing calculator, solve for y and type them in one after the other (usually Y1 and Y2 or some such).

5. Could you show me how to do it by just using slope and y-intercept method?

This is mainly grade 9, which I don't remember much of.

Just a random guess with numbers out of nowhere but...

y = 2 - x

... and...
I can't even take a guess, funny how this is suppose to be the "easiest" question, yet this is the only question I can't do easily.

6. Originally Posted by rocky123
Could you show me how to do it by just using slope and y-intercept method?

This is mainly grade 9, which I don't remember much of.

Just a random guess with numbers out of nowhere but...

y = 2 - x

... and...
I can't even take a guess, funny how this is suppose to be the "easiest" question, yet this is the only question I can't do easily.
Try to know, rather than guess.

You did the first one right. Solving

x + y = 2

for y, we can subtract x from both sides to get

y = 2 - x

You could also write as

y = -x + 2

to get it to look more like

y = mx + b

here m=-1 and b=2, where m is slope and b is y-intercept.

It's a bit of work for me to get an image up, so maybe you can follow my directions instead. Draw the x and y axes. Then mark point (0,2) which is y-intercept. Then go over one and down one to point (1,1) and mark it. Then again to mark (2,0). Etc. Connect them to make a line. That's what slope means, rise over run, we can write rise is -1 and run is 1.

Do the same for the second line.

2y = -5x + 10

y = -(5/2)x + 5

Start at (0,5). Go over two and down five and mark point (2,0). Etc.

Make sense?

7. Well, I kind of understand, you get what you get from y = 2 - x.

Then from that you turn it into the graph.

Then 2y = -5x + 10

Is just a simple equasion...
But I don't see where you're getting the numbers like go down 5, and to the right once and... Yeah, where?

8. Well, I kind of understand, you get what you get from y = 2 - x.

Then from that you turn it into the graph.

Then 2y = -5x + 10

Is just a simple equasion...
But I don't see where you're getting the numbers like go down 5, and to the right once and... Yeah, where?

9. Originally Posted by rocky123
Well, I kind of understand, you get what you get from y = 2 - x.

Then from that you turn it into the graph.

Then 2y = -5x + 10

Is just a simple equasion...
But I don't see where you're getting the numbers like go down 5, and to the right once and... Yeah, where?
Start from the y-intercept, then use the slope to get more points on the line. Suppose the slope is a/b. That tells you that if (x,y) is a point on the line, then (x+b,y+a) is also a point on the line. This is "rise over run."

10. But slope is y2 - y1/x2-x1
Isn't it? :S.

Not quite sure I understand your example, sorry this is really confusing me for some reason. Even though i'm pretty sure i've done this all before.

Do you just match a bunch of combinations? For example, xy, yx, mb,bm, xm?

I'm so, so lost. D:

11. Oh, I see now. Wouldn't my answer be on the graph as

(0,2) and (0,5) ?

How do you get the others? Or is that it?

12. Originally Posted by rocky123
Oh, I see now. Wouldn't my answer be on the graph as

(0,2) and (0,5) ?

How do you get the others? Or is that it?
Hmm I had some other responsibilities to attend to.. the slope part is clear now? I'm not sure what you mean about (0,2) and (0,5). The final answer to the problem will be a single point, the point of intersection of the two lines. Since you're presumably doing this on graph paper with a pencil and ruler (or similar), it may be hard to tell exactly what point this is, but if the answer seems to be a pair of integers then you'll be able to check that answer by plugging values into the equation.

If you're still confused, let me know.. I can make some images, it'll just take some time.

13. Oh, I see what the answer is suppose to be.

It's just one point?

But that one point is the intersection.

So wouldn't it be (-5,2)?

Or (2,-5) ?
If not, yes please I could really use some visual help.
Thank you so much for the effort.