Question

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

Answer 1

first i solved the equation for y to put it in slope intercept form

Answer 2

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?

Answer 3

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

Answer 4

this can also be written as y= (2/7)x-(60/7)

Answer 5

you used
y = (2/7) x + b
-8 = (2/7)*2 + b
b = -8 - 4/7 = - 60/7

Answer 6

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

Answer 7

ok with that?

Answer 8

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!

Answer 9

blonde moment lol - i get similar moments - in my case they are 'senior moments'