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    Q) write the equation of a line that is parallel to the line 2x - 4y = 8 and goes through the point (3,0)

    I rearranged in the form y=mx + c and got y= 1/2x - 2

    Then I work out the equation of the parallel line by doing this right? y-y1 = 1/2 (x-x1)

    so its y-0 = 1/2 (x-3)

    and I got y= 2x - 3

    and its wrong
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    (Original post by samantham999)
    Q) write the equation of a line that is parallel to the line 2x - 4y = 8 and goes through the point (3,0)

    I rearranged in the form y=mx + c and got y= 1/2x - 2

    Then I work out the equation of the parallel line by doing this right? y-y1 = 1/2 (x-x1)

    so its y-0 = 1/2 (x-3)

    and I got y= 2x - 3

    and its wrong
    It was all perfect until the very last bit! y = 1/2 ( x - 3 ) becomes:y = x/2 - 3/2
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    (Original post by wightsnowolf)
    It was all perfect until the very last bit! y = 1/2 ( x - 3 ) becomes:y = x/2 - 3/2
    don't we x by 2 and expand the bracket?
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    (Original post by samantham999)
    don't we x by 2 and expand the bracket?
    That was the expansion of the bracket. The 1/2 got multiplied by all elements inside the bracket
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    (Original post by wightsnowolf)
    That was the expansion of the bracket. The 1/2 got multiplied by all elements inside the bracket
    so did you get the answer by x outside number by inside numbers? 1/2(x-3)

    1/2x - 3/2 ?
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    (Original post by samantham999)
    don't we x by 2 and expand the bracket?
    It's the same thing but it can just make it easier in some cases by allowing you to work with whole numbers rather than fractions. If you did multiply by 2 the answer should be 2y=x-3.
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    (Original post by Samii123)
    It's the same thing but it can just make it easier in some cases by allowing you to work with whole numbers rather than fractions. If you did multiply by 2 the answer should be 2y=x-3.
    wouldn't i multiply the whole equation not just 2y?
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    (Original post by samantham999)
    wouldn't i multiply the whole equation not just 2y?
    That's what I meant. If you double y-0=1/2(x-3) it comes out to 2y=1(x-3) and obviously you don't need to expand it so it just becomes 2y=x-3.
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    for this question why did we x by 3 then?

    find the equation of a line parallel to the line 3y-2x-6=0 and passing through the point (-1,-2)
    y=2/3x + 2

    then we use y-y1 = mx (x-x1)

    y--2 = 2/3 (x--1)

    and then we X by 3

    we get: 3y + 6 = 2x + 2

    then divide by 3 and -6

    and we get y= 2/3x - 4/3
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    (Original post by Samii123)
    That's what I meant. If you double y-0=1/2(x-3) it comes out to 2y=1(x-3) and obviously you don't need to expand it so it just becomes 2y=x-3.
    I get it so we can do it either way, I just made the mistake by x 2 by x when it was 1/2
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    (Original post by samantham999)
    I get it so we can do it either way, I just made the mistake by x 2 by x when it was 1/2
    Yh sometimes the question will ask for integers so you can't always leave it as a fraction.
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    (Original post by Samii123)
    Yh sometimes the question will ask for integers so you can't always leave it as a fraction.
    we just divide so we get rid of the fraction?
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    (Original post by samantham999)
    we just divide so we get rid of the fraction?
    X the whole equation by the denominator so when you had the one with 1/2x you x it all by 2.
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    (Original post by Samii123)
    X the whole equation by the denominator so when you had the one with 1/2x you x it all by 2.
    ah I keep saying divide!!! I meant x the equation to get rid of the fraction (i.e the opposite )
    thanks
 
 
 
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