Question 13258
Start with your original equation 3x+5y=8. FIRST, you need to solve for y.  3x+5y-8 = 8-8. We have 3x+5y-8=0.  Now move the term with y variable to the other side of the equation.  3x+5y-5y-8 = 0-5y. We have, 3x-8=-5y.  Now divde both sides of the equation by -5.  We have (3/-5)x-(8/-5)=y or (-3/5)x+(8/5)=y.
We now have the slope of this equation -3/5 (the coefficient of the x-variable).  So, we know that any line of this same slope will be parallel to this line.  Your equation will look something like this, (-3/5)x + constant = y.  To get a line to pass through a certain point you need to know the point-slope form of a line.  It is (y-y1) = slope(x-x1). So with a certain point (x1,y1), in this case the point(x1= -1, y1= 7), we have (y-7)= (-3/5)*{x-(-1)}.  Distributing the x on the right hand side of this equation yields (y-7)= (-3/5)x - (3/5).  Now adding 7 to both sides we have (y-7+7)= (-3/5)x - (3/5) + 7.  Note 7 = (35/5) so we have y = (-3/5)x - (3/5) + (35/5). Ultimately the parallel line passing through point (-1,7) is y = (-3/5)x + (32/5)