SOLUTION: I always get stuck on this one part of this type of equation and was hoping for an explanation as to how to get past this one point. (3,-8); 6x + 7y = 3 I do fine until I get t

Question 419661: I always get stuck on this one part of this type of equation and was hoping for an explanation as to how to get past this one point.
(3,-8); 6x + 7y = 3
I do fine until I get to...
y-(-8) = -7/6x -21/6
the instructions say to add 8 to both sides
which leaves y = -7/6, but am unable to finish the equation as neither adding or subtracting -8 nor 8 makes the answer correct or make sense to me.
I'm at a loss and am wondering if you could explain to me in simple terms what I am doing wrong or how to finish this equation.
Thank you for any help.
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
(3,-8); 6x + 7y = 3
I do fine until I get to...
y-(-8) = -7/6x -21/6
the instructions say to add 8 to both sides
.
You don't say, but I think your original problem is asking you for the formula that crosses a certain point and is parallel to a line.
(3,-8); 6x + 7y = 3
.
first find slope of:
6x + 7y = 3
7y = -6x + 3
y = (-6/7)x + 3/7
slope is -6/7
.
A line parallel must have the same slope.
So, now we have
slope is -6/7
and
one point:
(3,-8)
.
Plug into the "point-slope" form:
y - y1 = m(x - x1)
y - (-8) = (-6/7)(x - 3)
y + 8 = (-6/7)x + 18/7
Now, we subtract 8 from both sides:
y = (-6/7)x + 18/7 - 8
y = (-6/7)x + 18/7 - 56/7
y = (-6/7)x - 38/7 (this is what they're looking for)

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