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    Ok guys I have ton of maths questions I am stuck on, so please if you can help with any say what number it is and tell me please. I have inattentive adhd and I find it very hard to concentrate in class so I don't know all of this stuff.

    1. Find the equation of a line which passes through the point A(3,-2) and is parallel to line 3y-2x=5

    2.Find the value of 3^-2

    3. Express the following with a rational denominator ... 5/square root 10

    4. Simplify the square root of 108+ the square root of 12

    Any ideas would be awesome!
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    (Original post by stephielola)
    1. Find the equation of a line which passes through the point A(3,-2) and is parallel to line 3y-2x=5
    If we start with the first question, the best thing would be to rearrange  3y - 2x = 5 into the form of  y = mx + c so that it's easier to see the gradient.

    If a line is parallel to this, what does it mean?
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    (Original post by KaylaB)
    If we start with the first question, the best thing would be to rearrange  3y - 2x = 5 into the form of  y = mx + c so that it's easier to see the gradient.

    If a line is parallel to this, what does it mean?
    The line will never meet x
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    (Original post by stephielola)
    The line will never meet x
    Not really, if two lines are parallel, then they have the same gradient this can be seen below

    But you are correct that they will never meet each other

    So have you rearranged the equation into the form  y = mx + c ? :holmes:
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    (Original post by KaylaB)
    Not really, if two lines are parallel, then they have the same gradient this can be seen below

    But you are correct that they will never meet each other

    So have you rearranged the equation into the form  y = mx + c ? :holmes:
    Yeah I got 3y=-2x+5
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    (Original post by stephielola)
    Yeah I got 3y=-2x+5
    Can you divide it by 3 so you get the equation for just y

    Once you have that, we know that the x coefficient represents the gradient of the straight line.

    So, now that we know the gradient of this line, and we know that the gradient of the line parallel to it is the same we can find the equation of the new line by substituting the point A (3,2)
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    (Original post by KaylaB)
    Can you divide it by 3 so you get the equation for just y

    Once you have that, we know that the x coefficient represents the gradient of the straight line.

    So, now that we know the gradient of this line, and we know that the gradient of the line parallel to it is the same we can find the equation of the new line by substituting the point A (3,2)
    y=-2/3x +5/3
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    (Original post by stephielola)
    y=-2/3x +5/3
    It should be positive  \frac{2} {3}x
    Spoiler:
    Show




     3y-2x=5

3y=2x+5

y = \dfrac{2} {3}x + \frac{5} {3}





    So now for the equation of the parallel line, we currently have  y=\dfrac{2} {3}x + c
    And we know that point A (3,2) lies on this line.

    How could we find out the value of c using point A? :holmes:
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    (Original post by KaylaB)
    It should be positive  \frac{2} {3}x
    Spoiler:
    Show





     3y-2x=5

3y=2x+5

y = \dfrac{2} {3}x + \frac{5} {3}






    So now for the equation of the parallel line, we currently have  y=\dfrac{2} {3}x + c
    And we know that point A (3,2) lies on this line.

    How could we find out the value of c using point A? :holmes:
    Honestly I have no clue x
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    (Original post by KaylaB)
    If we start with the first question, the best thing would be to rearrange  3y - 2x = 5 into the form of  y = mx + c so that it's easier to see the gradient
    OP: don't worry about the spoiler yet - I don't want to confuse you by suggesting another method:

    Spoiler:
    Show

    Actually, the quickest thing to do is to *keep* the 3y-2x part unchanged. This determines the gradient of the line, changing the constant 'shifts' the line without changing direction. So any parallel line will have the form 3y-2x = C (for some C). Substitute in the point the line has to go through to find the necessary value of C.

    <META>: Since I know you've been following the "thoughts on the forum" discussion, this is where having a lot more maths experience is a double edged sword. We're a lot quicker to see "actually, the *best* way of doing this is probably...", and at the same time, it's easy for us to suggest things that may be outside a comfort zone. This method would have been a bit of a leap for me when I first sat my A-level, I think.</META>
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    (Original post by stephielola)
    Honestly I have no clue x
    We know that point A definitely is on this line, so if you substitute in the x and y values at point A into the equation it should work

    So  y= \dfrac{2} {3}x + c



A(3,2) x=3, y=2



2= \dfrac{2} {3}(3) + c

    Could you tell me what c would be?
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    (Original post by KaylaB)
    We know that point A definitely is on this line, so if you substitute in the x and y values at point A into the equation it should work

    So  y= \dfrac{2} {3}x + c



A(3,2) x=3, y=2



2= \dfrac{2} {3}(3) + c

    Could you tell me what c would be?
    Nope this makes no sense to me
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    (Original post by DFranklin)
    Actually, the quickest thing to do is to *keep* the 3y-2x part unchanged. This determines the gradient of the line, changing the constant 'shifts' the line without changing direction. So any parallel line will have the form 3y-2x = C (for some C). Substitute in the point the line has to go through to find the necessary value of C.

    Since I know you've been following the "thoughts on the forum" discussion, this is where having a lot more maths experience is a double edged sword. We're a lot quicker to see "actually, the *best* way of doing this is probably...", and at the same time, it's easy for us to suggest things that may be outside a comfort zone. This method would have been a bit of a leap for me when I first sat my A-level, I think.
    I have personally never done it that way before! It does actually sound easier though :beard:, so I'll probably use it in the future - so cheers for that! :hat2: This is the method I've done even before GCSE :dontknow:
    But I do definitely understand what you mean, and this is why it's great to have a place where there are lots of people on here with other methods and teachings so they can share tips
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    (Original post by stephielola)
    Nope this makes no sense to me
    I'm super sorry, it is quite confusing isn't it :beard:

    I have actually just been made aware of an easier way to approach this question which should make far more sense.

    So if we go back to the start, we have the equation  3y-2x=5

    For the parallel line, we can keep the  3y-2x the same as parallel line have the same gradient

    Then we can substitute the values of point A (3,2) into  3y-2x

    This will find the constant and you will have the equation of the line

    Does this make slightly more sense?
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    (Original post by kaylab)
    i'm super sorry, it is quite confusing isn't it :beard:

    I have actually just been made aware of an easier way to approach this question which should make far more sense.

    So if we go back to the start, we have the equation  3y-2x=5

    for the parallel line, we can keep the  3y-2x the same as parallel line have the same gradient

    then we can substitute the values of point a (3,2) into  3y-2x

    this will find the constant and you will have the equation of the line

    does this make slightly more sense? :h
    thankyoouuu so much
 
 
 
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