Write an equation of the line containing the given point and parallel to the given line.

(-7,6); 8x=9y+5

The equation of the line is y =

First find the slope m of the given line by transforming it to the form:

y=mx+b
In this case, divide both sides by 9 to give m = 8/9.
After that, use the slope-point form given in:
http://www.jiskha.com/display.cgi?id=1284102012

8/9x + 83/9

To find the equation of a line parallel to a given line, we need to determine the slope of the given line.

First, let's rewrite the given equation, 8x = 9y + 5, in slope-intercept form (y = mx + b), where m is the slope and b is the y-intercept.

8x = 9y + 5
9y = 8x - 5
y = (8/9)x - 5/9

From the equation y = (8/9)x - 5/9, we can see that the slope of the given line is 8/9.

Since we want to find a line parallel to the given line, the parallel line will also have a slope of 8/9.

Now, we have the slope (m = 8/9) and a point (-7, 6). We can use the point-slope form of a linear equation to find the equation of the line.

The point-slope form is given by: y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope.

Substituting the given values into the point-slope form:
y - 6 = (8/9)(x - (-7))

Simplifying:
y - 6 = (8/9)(x + 7)
y - 6 = (8/9)x + (8/9)(7)
y - 6 = (8/9)x + 56/9

Adding 6 to both sides to isolate y:
y = (8/9)x + 56/9 + 6
y = (8/9)x + 56/9 + 54/9
y = (8/9)x + 110/9

Thus, the equation of the line containing the point (-7, 6) and parallel to the given line 8x = 9y + 5 is:
y = (8/9)x + 110/9