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Math Help - Having difficulty with linear equations

  1. #1
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    Having difficulty with linear equations

    Hello,

    I've been out of school for over 6 years so these questions are probably relatively simple, but I'm just not grasping the concept here. I'm doing a bridging course for a university degree, and my high-school math has just gone out the window. I don't remember any of it.

    Anyway, I'm being asked the following question.

    If the line passing through points (1 , a) and (4, -2) is parallel to the line passing through the points (2, 8) and (-7, a+4), what is the value of a?

    I don't even have any idea on where to begin to work this out .

    Any help would be appreciated.

    Cheers,
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  2. #2
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    Re: Having difficulty with linear equations

    Quote Originally Posted by astuart View Post
    Hello,

    I've been out of school for over 6 years so these questions are probably relatively simple, but I'm just not grasping the concept here. I'm doing a bridging course for a university degree, and my high-school math has just gone out the window. I don't remember any of it.

    Anyway, I'm being asked the following question.

    If the line passing through points (1 , a) and (4, -2) is parallel to the line passing through the points (2, 8) and (-7, a+4), what is the value of a?

    I don't even have any idea on where to begin to work this out .

    Any help would be appreciated.

    Cheers,
    slope between any two points (x_1,y_1) , (x_2,y_2) is

    m = \frac{y_1-y_2}{x_1-x_2}

    note that parallel lines have the same slope.
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    Re: Having difficulty with linear equations

    Quote Originally Posted by skeeter View Post
    slope between any two points (x_1,y_1) , (x_2,y_2) is

    m = \frac{y_1-y_2}{x_1-x_2}

    note that parallel lines have the same slope.
    Yep, I understand that, but I don't know how to solve for (a) because it's a variable so from what I can gather the equation would be

    m=\frac{(-2)-a}{4-1}.

    This means that the other point would be

    m = \frac{(a+4)-8}{-7-2}

    However, from here, how do I solve for a?

    A tip in the right direction is suitable, I'm really wanting to solve this myself and hopefully it just 'clicks'
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    Re: Having difficulty with linear equations

    Quote Originally Posted by astuart View Post
    Yep, I understand that, but I don't know how to solve for (a) because it's a variable so from what I can gather the equation would be

    m=\frac{(-2)-a}{4-1}.

    This means that the other point would be

    m = \frac{(a+4)-8}{-7-2}

    However, from here, how do I solve for a?

    A tip in the right direction is suitable, I'm really wanting to solve this myself and hopefully it just 'clicks'
    \frac{(-2)-a}{4-1}= \frac{(a+4)-8}{-7-2} solve for a
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    Re: Having difficulty with linear equations

    so, I see you do know where to start ...

    \frac{-2-a}{4-1} = \frac{(a+4) - 8}{-7-2}

    \frac{-2-a}{3} = \frac{a-4}{-9}

    cross multiply ...

    -9(-2-a) = 3(a-4)

    finish solving for a.
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    Re: Having difficulty with linear equations

    Quote Originally Posted by Plato View Post
    \frac{(-2)-a}{4-1}= \frac{(a+4)-8}{-7-2} solve for a
    This is where I'm falling down. From what I think I understand, I need to isolate a. Isolating a on one of these equations would give me the answer to the other side, correct (due to them being equal)?
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    Re: Having difficulty with linear equations

    Quote Originally Posted by skeeter View Post
    so, I see you do know where to start ...

    \frac{-2-a}{4-1} = \frac{(a+4) - 8}{-7-2}

    \frac{-2-a}{3} = \frac{a-4}{-9}

    cross multiply ...

    -9(-2-a) = 3(a-4)

    finish solving for a.
    Still not getting it . I can't believe It's not clicking - I think I need to re-visit some basic algebra.

    Would the next step of the equation be

    -9(-2-a) = 3(a-4)

    -18-9a=3a-12

    Is this heading in the right direction?
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  8. #8
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    Re: Having difficulty with linear equations

    Quote Originally Posted by astuart View Post
    Is this heading in the right direction?
    You got the idea, yes. But let's be careful with our distribution:

    -9(-2-a) = 3(a-4)

    \Rightarrow(-9)(-2)+(-9)(-a) = 3a - 3(4)

    \Rightarrow18+9a = 3a - 12.

    Remember that any time you multiply two numbers with the same sign (both positive or both negative), the answer will be positive.
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    Re: Having difficulty with linear equations

    Quote Originally Posted by Reckoner View Post
    You got the idea, yes. But let's be careful with our distribution:

    -9(-2-a) = 3(a-4)

    \Rightarrow(-9)(-2)+(-9)(-a) = 3a - 3(4)

    \Rightarrow18+9a = 3a - 12.

    Remember that any time you multiply two numbers with the same sign (both positive or both negative), the answer will be positive.
    Right, so after having a break for an hour or so, it's a little bit clearer.

    \Rightarrow18+9a = 3a - 12. (subtract 3a)

    \Rightarrow18+6a = - 12. (divide by 6)

    \Rightarrow3+a = - 2. (subtract 3)

    \Rightarrow a = - 5.

    It's pretty simple really. Thanks for everybody's patience. Great forum too!
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