A line parallel to 4x=3y-13?

I came up with m=4/3
y-3=4/3 + 16/3
after working problem got ans
y= 4/3x+25/3

confused???

To find a line parallel to the equation 4x=3y-13, we need to determine its slope.

The given equation is in the form of Ax + By = C, where A is the coefficient of x, B is the coefficient of y, and C is a constant term.

In this case, the equation 4x=3y-13 can be rewritten as 4x - 3y = -13.

Now, we can compare this equation to the general form Ax + By = C.

The coefficient of x is 4, and the coefficient of y is -3.

The slope of a line in the form Ax + By = C is found by dividing the coefficient of x by the coefficient of y, so the slope is 4/-3, which simplifies to -4/3.

Therefore, any line parallel to 4x=3y-13 will have a slope of -4/3.

Now, we can use the point-slope form of a linear equation (y - y1 = m(x - x1)) to find the equation of a line parallel to 4x=3y-13 passing through a given point (x1, y1).

Let's use the point (0,0) in this example.

Using the slope m = -4/3 and the point (0,0), we can substitute the values into the point-slope form:

y - 0 = (-4/3)(x - 0)

Simplifying, we have:

y = (-4/3)x

Therefore, the equation of the line parallel to 4x=3y-13 is y = (-4/3)x.

I apologize, but I don't see where the equation y-3=4/3 + 16/3 comes from in your question. If you could provide more information or context, I'd be happy to help further clarify.