solving simultaneous linear equations graphically
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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.
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I got like two opposite points and the line doesn't go across them.
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That's because one of the dots should be (0,-1) instead of (0,1). Check the y axis!
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.
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.
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Original post by z_o_e
You did it!
Original post by TheOtherSide.
You did it!
Are you sure? I don't believe it!
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Original post by z_o_e
Pretty sure that's correct. You could always check by solving the simultaneous equations algebraically.
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?
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Original post by z_o_e
You mean what you're using to plot the next graph? These all look fine.
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
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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
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.
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.
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!
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Original post by z_o_e
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!
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.
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Original post by z_o_e
No problem.
Okay. Feel free to tag me if you need any more help.
Original post by TheOtherSide.
No problem.
Okay. Feel free to tag me if you need any more help.
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?
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Original post by z_o_e
Is it okay if you can please do an explanation example for the second question?
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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
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.
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.
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
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).
(edited 7 years ago)
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.
Everything else you've done is correct as far as I can tell.
Still got the same.
With your points.
:*(
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Original post by z_o_e
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.
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.
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!!!!
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