You are Here: Home >< Maths

# solving simultaneous linear equations graphically

Announcements Posted on
Why bother with a post grad course - waste of time? 17-10-2016
1. (Original post by z_o_e)
I don't understand this!!!!!!!

I got like two opposite points and the line doesn't go across them.

Posted from TSR Mobile
That's because one of the dots should be (0,-1) instead of (0,1). Check the y axis!
2. (Original post by TheOtherSide.)
With '2x - y = 9' the line should go through the points (0,-9) and (4.5,0). Try changing the numbers on the axes.

And the other line doesn't look completely diagonal - it should pass through the points (1,1), (2,2) and so on.
My final try. I'm officially giving up if this is incorrect.

Posted from TSR Mobile
Attached Images

3. (Original post by z_o_e)
My final try. I'm officially giving up if this is incorrect.

Posted from TSR Mobile
You did it!
4. (Original post by TheOtherSide.)
You did it!
Are you sure? I don't believe it!

Posted from TSR Mobile
5. (Original post by z_o_e)
Are you sure? I don't believe it!

Posted from TSR Mobile
Pretty sure that's correct. You could always check by solving the simultaneous equations algebraically.
6. (Original post by TheOtherSide.)
Pretty sure that's correct. You could always check by solving the simultaneous equations algebraically.
Yasss I got 5!!!!
Thank you so much !!!

How's the working out for this one?

Posted from TSR Mobile
Attached Images

7. (Original post by z_o_e)
Yasss I got 5!!!!
Thank you so much !!!

How's the working out for this one?

Posted from TSR Mobile
You mean what you're using to plot the next graph? These all look fine.
8. (Original post by TheOtherSide.)
You mean what you're using to plot the next graph? These all look fine.
Yeah but I got the solution as (3,3)

And it doesn't fit the equation...
X + y=0

3+3=6

Posted from TSR Mobile
Attached Images

9. (Original post by z_o_e)
Yeah but I got the solution as (3,3)

And it doesn't fit the equation...
X + y=0

3+3=6

Posted from TSR Mobile
Remember that the value for y is -9, as you already worked out, so the line for the second equation should cross the y axis a lot further down.

Plus the line x + y = 0 should have a negative gradient, which I now realise wasn't obvious from the method you used to plot the graph. When there is a 0 involved, I'd suggest rearranging the equation so that you get y = something, so that you know whether the line has a positive or negative gradient.

In this case, y = -x, which tells you that the line x + y = 0 is actually perpendicular to what you just drew.
10. (Original post by TheOtherSide.)
Remember that the value for y is -9, as you already worked out, so the line for the second equation should cross the y axis a lot further down.

Plus the line x + y = 0 should have a negative gradient, which I now realise wasn't obvious from the method you used to plot the graph. When there is a 0 involved, I'd suggest rearranging the equation so that you get y = something, so that you know whether the line has a positive or negative gradient.

In this case, y = -x, which tells you that the line x + y = 0 is actually perpendicular to what you just drew.
Oh no!

Posted from TSR Mobile
11. Don't worry! It was just the negatives you got confused about. Make sure you check those thoroughly in the exam, and you should be fine!
12. (Original post by TheOtherSide.)
Don't worry! It was just the negatives you got confused about. Make sure you check those thoroughly in the exam, and you should be fine!
Thank you!!

Will try to carry on with this tomorrow.

Posted from TSR Mobile
13. (Original post by z_o_e)
Thank you!!

Will try to carry on with this tomorrow.

Posted from TSR Mobile
No problem.

Okay. Feel free to tag me if you need any more help.
14. (Original post by TheOtherSide.)
No problem.

Okay. Feel free to tag me if you need any more help.
Is it okay if you can please do an explanation example for the second question?

Posted from TSR Mobile
15. (Original post by z_o_e)
Is it okay if you can please do an explanation example for the second question?

Posted from TSR Mobile
I'll try to!

So with the equation 2x - y = 9:
When x = 0
2*0 - y = 9
0 - - 9 = 9, so y has to be -9 and the coordinates for this point are (0,-9)
When y = 0
2x - 0 = 9
2x = 9
x = 4.5 and the coordinates for this point are (4.5,0)

Plot these two points on the graph and join them with a line, then extend the line.

With the equation x + y = 0:
Rearrange this so that you get y on one side.
x + y = 0
y = -x
This tells you that the line passes through the origin and has a gradient of -1 and to work out what points this line goes through, you can substitute in values for y.
For example, when x = 1, y = -1 and when x = 5, y = -5, which give you two coordinates which you can plot: (1,-1) and (5,-5)

Plot these points or another two and join them with a line, then extend the line.

(I realise that I got confused by the orientation of your graph in the quote below and the line for x + y = 0 was correct there)
(Original post by z_o_e)
I got this for the second questionPosted from TSR Mobile
16. (Original post by TheOtherSide.)
I think you plotted x - y = 1 wrong - if you rearrange this equation, you get y = x + 1, which tells you that the y-intercept is at (0,1) but the other line is correct!
This is incorrect. If you substitute x=0 into the first you get y=-1, but if you substitute x=0 into the second you get y=1. The correct rearrangement is y = x - 1. The y intercept is (0,-1).
Everything else you've done is correct as far as I can tell.
17. (Original post by morgan8002)
This is incorrect. If you substitute x=0 into the first you get y=-1, but if you substitute x=0 into the second you get y=1. The correct rearrangement is y = x - 1. The y intercept is (0,-1).
Everything else you've done is correct as far as I can tell.
Yep, I realised that afterwards, hence the addition to my post above.

Spoiler:
Show
The orientation of the photo might have thrown me off.

Edit: Oh wait, no, different equation. I think I made a stupid mistake, but then ended up saying that the y intercept was in fact (0,-1).
18. (Original post by morgan8002)
This is incorrect. If you substitute x=0 into the first you get y=-1, but if you substitute x=0 into the second you get y=1. The correct rearrangement is y = x - 1. The y intercept is (0,-1).
Everything else you've done is correct as far as I can tell.
Still got the same.

:*(

Posted from TSR Mobile
Attached Images

19. (Original post by z_o_e)
Still got the same.

:*(

Posted from TSR Mobile
Morgan was talking about the first graph.

Remember what I told you earlier:
The y intercept is at -9, not 9, and the gradient of the other graph is -1, so it will go through the points (1,-1),(2,-2),(3,-3) and so on.
20. (Original post by TheOtherSide.)
Morgan was talking about the first graph.

Remember what I told you earlier:
The y intercept is at -9, not 9, and the gradient of the other graph is -1, so it will go through the points (1,-1),(2,-2),(3,-3) and so on.
Oh!

I tried Again!!!!

Posted from TSR Mobile

## Register

Thanks for posting! You just need to create an account in order to submit the post
1. this can't be left blank
2. this can't be left blank
3. this can't be left blank

6 characters or longer with both numbers and letters is safer

4. this can't be left empty
1. Oops, you need to agree to our Ts&Cs to register

Updated: June 6, 2016
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Today on TSR

### How does exam reform affect you?

From GCSE to A level, it's all changing

Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read here first

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams