(-6,-1) and parallel to 2x+3y=3

Answer 1

3y=-2x + 3

y = (-2/3)x + 1, -2/3 = slope

y=(-2/3)x + b, where b is the y intercept. Plug in the point (-6,-1) to solve for b

-1 = (-2/3)(-6) + b

b = -1 -4 = -5

y=(-2/3)x -5

Answer 2

We need to find an equation of a line that is parallel to the given equation and contains the point (-6,-1). First, we know that parallel lines have the same slope, so we must determine the slope of our given equation. The typical equation of a line is in the form y = mx + b, where "m" is the slope, so we must put our equation into this form.

2x + 3y = 3 original equation

3y = -2x + 3 subtract 2x from both sides

y = (-2x + 3)/3 divide both sides by 3

y = (-2/3)x + 1 equation in the correct form

As we can see now, the slope, m, is -2/3. Using this information and the (x,y) coordinates given we can solve for the y-intercept, b:

y = mx + b

-1 = (-2/3)(-6) + b

-1 = 4 + b

b = -1 - 4

b = -5

Now we know the slope, m = -2/3, and the y-intercept, b = -5, so we can write the equation of the line as

y = (-2/3)x - 5.

Hope this helps!