I have a few more questions.No problem, happy to help!

[mathdad]

Find an equation of the line that passes through (6, -3) and has y-intercept 8.

The y-intercept 8 can be expressed as the point (0, 8).

Correct?

I then find the slope of (6, -3) and (0, 8).

Yes?

The next step is to plug one of the above points and the slope into the point-slope formula and solve for y.

Is any of this right?

 

[Greg]

Find an equation of the line that passes through (6, -3) and has y-intercept 8.

The y-intercept 8 can be expressed as the point (0, 8).

Correct?

Correct.

I then find the slope of (6, -3) and (0, 8).

Yes?

Yes.

The next step is to plug one of the above points and the slope into the point-slope formula and solve for y.

An equation for a line is y = mx + b where m is slope and b is the y-intercept. To write the equation of your line in this form we need the slope and y-intercept. For the y-intercept, what does the fact that (0, 8) is on the line tell you when you substitute for x and y in your y = mx + b equation?

 

[mathdad]

Correct.
Yes.
An equation for a line is y = mx + b where m is slope and b is the y-intercept. To write the equation of your line in this form we need the slope and y-intercept. For the y-intercept, what does the fact that (0, 8) is on the line tell you when you substitute for x and y in your y = mx + b equation?

When I substitute (0,8) into y = mx + b, the answer is b = 8.
This means the graph crosses the y-axis at the point (0,8).

 

[mathdad]

(6,-3) & (0,8)

m = (8-(-3))/(0-6)

m = (8+3)/(-6)

m = -11/6

I will use (0,8).

y - 8 = (-11/6)(x - 0)

y - 8 = (-11x/6)

y = (-11x/6) + 8

Yes?

 

[MarkFL]

We are given that the line has $y$-intercept 8, so your line may be written as:

\(\displaystyle y=mx+8\)

Now, we are given the point on the line $(6,-3)$, and so substituting for $x$ and $y$, we have:

\(\displaystyle -3=m(6)+8\)

Solving for $m$, we find:

\(\displaystyle m=-\frac{11}{6}\)

And so our line is:

\(\displaystyle y=-\frac{11}{6}x+8\)

This agrees with your result. (Yes)

 

[mathdad]

We are given that the line has $y$-intercept 8, so your line may be written as:

\(\displaystyle y=mx+8\)

Now, we are given the point on the line $(6,-3)$, and so substituting for $x$ and $y$, we have:

\(\displaystyle -3=m(6)+8\)

Solving for $m$, we find:

\(\displaystyle m=-\frac{11}{6}\)

And so our line is:

\(\displaystyle y=-\frac{11}{6}x+8\)

This agrees with your result. (Yes)

Always good to know more than one method. BTW, thank you for answering my PM questions. I will reply in full later today...