Glencoe Algebra 2, Chapter 2, 2.3, Section 2.3, Problem 47

Given the line passes through (2, 1), parallel to graph of 2x+3y =6
as the slope of the line 2x+3y =6 is as follows
2x+3y =6
=> 3y = -2x+6
=> y = (-2/3) x +(6/3)
=> y=(-2/3) x +2
this is of the form y= mx +c where m is the slope
so the slope of the line 2x+3y =6 is m = -2/3
so the line passing through the point (x,y)=(2,1) is also -2/3 as both the lines are parallel
then the line equation passing through (2,1) is follows
y= mx+c
1 = (-2/3)*2+c
1= -4/3 +c
c = 1+4/3 = 7/3
so the desired equation of the line is
y = (-2/3)x + 7/3
and it is plotted and in the attachments and the point (2,1) is plotted with a blue dot