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    (Original post by djmans)
    if its f'(x) what should i do?
    If it's asking you to find f'(x) that means differentiate it.
    If it gives you f'(x) and it asks you to find the curve, then integrate it.




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    (Original post by djmans)
    so f its f'(x) i must integrate
    f'(x) is the derivative with repsect to x of f(x) so to find f(x) you must 'undo' the differentiation, so you must integrate.
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    could you help me with this (b) its an easy question but i am struggling to get the answers.
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    (Original post by djmans)
    could you help me with this (b) its an easy question but i am struggling to get the answers.
    Re-arrange one of the equations in the form y = mx + c and then sub this into the other equation.
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    (Original post by Zacken)
    Re-arrange one of the equations in the form y = mx + c and then sub this into the other equation.
    that what i been trying for than last hour but cant get the answer, could you please help
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    (Original post by djmans)
    that what i been trying for than last hour but cant get the answer, could you please help
    Put up a picture of your working.
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    (Original post by Zacken)
    Put up a picture of your working.
    i cant upload it, its too big
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    (Original post by djmans)
    i cant upload it, its too big
    Line L1: m = \frac{12-3}{11--1} = \frac{9}{12} = \frac{3}{4}, so y -12 = \frac{3}{4}(x-11) \Rightarrow y = 12 + \frac{3}{4}(x-11).

    Plug this into the other equation:

    3\left(\frac{3}{4}(x-11) + 12\right) + 4x = 30 and solve this.
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    (Original post by Zacken)
    Line L1: m = \frac{12-3}{11--1} = \frac{9}{12} = \frac{3}{4}, so y -12 = \frac{3}{4}(x-11) \Rightarrow y = 12 + \frac{3}{4}(x-11).

    Plug this into the other equation:

    3\left(\frac{3}{4}(x-11) + 12\right) + 4x = 30 and solve this.
    thanks i got the answer but before i was trying the same method by rearranging the first equation and substituting into eq 2 but got something completely different
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    (Original post by djmans)
    thanks i got the answer but before i was trying the same method by rearranging the first equation and substituting into eq 2 but got something completely different
    See if you can spot your mistake yourself, otherwise you'll need to put up your working in some way for me to check. But, make sure you understand what I did.
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    (Original post by Zacken)
    See if you can spot your mistake yourself, otherwise you'll need to put up your working in some way for me to check. But, make sure you understand what I did.
    why did you sub directly using this eq from (a)

    than using this eq, its the same thing but you get a different answer.
    4y – 3x –15 = 0
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    (Original post by djmans)
    why did you sub directly using this eq from (a)

    than using this eq, its the same thing but you get a different answer.
    4y – 3x –15 = 0
    No, you don't get a different answer; you're just doing it wrong somehow, but I can't check unless you show me your working.
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    (Original post by Zacken)
    No, you don't get a different answer; you're just doing it wrong somehow, but I can't check unless you show me your working.
    you were right, i tried the question again today i got it.
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    (Original post by djmans)
    you were right, i tried the question again today i got it.
    Awesome.
 
 
 
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