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anonymous

  • 4 years ago

ok i need a lil clarificationon this problem...the answer i got was marked wrong on my test but i cant seem to come up with the answer they said was correct. heres the question... Write an equation of the line containing the given point and parallel to the given line. (2,-8) ; 2x-7y=3

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  1. anonymous
    • 4 years ago
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    |dw:1327791230103:dw|first i solved the equation for y to put it in slope intercept form

  2. anonymous
    • 4 years ago
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    i know that for my new equation for a parallel line the slope stays the same so i got y=(2/7)x+b then i inserted the point values into the equation for x and y and got -14=b then i plugged this b value into the equation to give me y=(2/7)x-14 but the test said the "correct" answer was y= (2/7)x-(60/7) so can someone please tell me where i went wrong?

  3. anonymous
    • 4 years ago
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    rearranging 7y = 2x - 3 y = (2/7) x - 3/7 so the slope of ur line will be 2/7 it passes thru point (2,-8) so y -(-8) = 2/7 (x - 2) y + 8 = 2/7(x-2) multiply thru by 7 7y + 56 = 2x - 4 7y = 2x - 60

  4. anonymous
    • 4 years ago
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    this can also be written as y= (2/7)x-(60/7)

  5. anonymous
    • 4 years ago
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    you used y = (2/7) x + b -8 = (2/7)*2 + b b = -8 - 4/7 = - 60/7

  6. anonymous
    • 4 years ago
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    the other equation of straight line which i used is y-y1 = m(x-x1) where m = slope and (x1,y1) is the given point

  7. anonymous
    • 4 years ago
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    ok with that?

  8. anonymous
    • 4 years ago
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    i see what i did wrong now. i divided the -8 by 4/7 instead of subtracting it, i must have been having a blonde moment...lol. thank for the help!

  9. anonymous
    • 4 years ago
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    blonde moment lol - i get similar moments - in my case they are 'senior moments'

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