Algebra problem help!!

write the equation in slope-intercept form for the line through (-7,2) and parallel to 3x+5y=6

y - y1 = m(x - x1) is the standard line equation, where y1 is our y-coordinate (2) and x1 is our x-coordinate (-7).

Plug it in:
y - 2 = m(x - -7)
y - 2 = m(x + 7)

Now, to find m:
Solve 3x + 5y = 6 for y:
3x + 5y = 6
5y = 6 - 3x
y = (6/5) - (3/5)x
or
y = -(3/5)x + (6/5)

Since this is slope-form (y = mx + b), we have m = -3/5.
Plug this into our equation, and:

y - 2 = (-3/5)(x + 7)

Distribute and solve for y:
y - 2 = (-3/5)x - (21/5)
y = (-3/5)x - (21/5) + 2
y = (-3/5)x - (21/5) + (10/5)
y = (-3/5)x - (11/5)

3x+5y=6
5y = -3x + 6
y = (-3/5) + 6/5
Slope of this line = -3/5
Slope of our line = -3/5
Equation
y-2 = (-3/5)(x+7)
5y-10 = -3x - 21
5y = -3x - 11
y = (-3/5)x - 11/5

y = (-3/5)x - (11/5)