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Please help me with this maths question, i'm really baffed!!! Watch

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    Find the points of intersection of the line 2X + 3Y = 12 with the axes

    Find also the gradient of this line

    I know one of the intersections is (0,4), but how do I work out the other one.

    And my workings made -2/3 the gradient, but apparently it is -4/6, why is this?
    why would I lose marks for answering this?

    Please help.........I could really do with the help
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    -2/3 is just -4/6 simplified.

    For the intersection of the line with the Y-axis, make y=0 then solve for x.
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    rearrange to make y the subject : y= -2/3x + 4
    using y = mx + c (0,4) is the y intercept
    x intercepts occur when y = 0

    0=-2/3x + 4
    2/3x = 4
    x = 6

    so axis intercepts are (0,4) & (6,0)
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    (Original post by davidmarsh01)
    -2/3 is just -4/6 simplified.

    For the intersection of the line with the Y-axis, make y=0 then solve for x.
    but the mark scheme said that -2/3 is wrong
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    (Original post by lost15)
    but the mark scheme said that -2/3 is wrong
    -2/3 = -4/6

    now usually in gradient type questions you don't simplify ..but...in the end..it is the same
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    tip: when an equation is in the form :  y= mx + n then that  n will always be your y-axis intercept.
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    Hey Guys thanks, I get that now!!!

    But does anybody understand the second part of this question

    Solve the equation 2X² +3X=0

    i got X= 0 or = -3/2

    Part 2, I do not understand 2X² +3X - k =0 has no real roots

    Please help me with this, I really appresciate it. You people are way nicer than my maths teacher


    Find the set of values of k for which the equation
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    (Original post by lost15)
    Hey Guys thanks, I get that now!!!

    But does anybody understand the second part of this question

    Solve the equation 2X² +3X=0

    i got X= 0 or = -3/2

    Part 2, I do not understand 2X² +3X - k =0 has no real roots

    Please help me with this, I really appresciate it. You people are way nicer than my maths teacher


    Find the set of values of k for which the equation
    For the equation to have no real roots, b^2 -4ac < 0. Sub in the values and solve.

    What level are you? This question seems like Scottish Higher (AS), but the one above seems more like GCSE-level. I suppose the English system might just be different.
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    No real roots so b^2-4ac<0. So 9-8k<0. So k>\frac98
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    (Original post by MostCompetitive)
    No real roots so b^2-4ac<0. So 9-8k<0.
    This is true, but you shouldn't give full solutions, just try to show the person what approach to take. This actually applies to someone else above as well.
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    (Original post by derangedyoshi)
    For the equation to have no real roots, b^2 -4ac < 0. Sub in the values and solve.

    What level are you? This question seems like Scottish Higher (AS), but the one above seems more like GCSE-level. I suppose the English system might just be different.
    This is all AS level mathematics (these topics are covered in the first AS unit), and y=mx+c is GCSE level but still in the AS C1 books
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    (Original post by MostCompetitive)
    No real roots so b^2-4ac<0. So 9-8k<0. So k>\frac98
    I'll have you know that I got 8A*s at GCSE and I fully intend to get A's in all of my A Levels.
    Do you know why I do well, because when I need help I'm not afraid to ask for it. Thank you for your cynicism and negativity, I 'll harness it for good.
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    (Original post by MostCompetitive)
    This is AS level mathematics (these topics are covered in the first AS unit).
    OK. I'm doing Higher.
 
 
 
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