how do I solve: find the equation with the line parallel to the line with 4/3x-2y=6 passing through (-3,-2)?

Question

passing through the point I mean

doesn't 6 need to be divided by 2 also?

yeah, it was the reiprocal divided by 2, but the 6 was divided also.

Answer 1

since the lines are parallel, you know that they will retain the same slope...... m = 4/3

also, you must take the x- and y-coordinates that you have and use them in the slope-intercept form equation: y = mx + b

the only thing you are truly lacking in this problem is your y-intercept: b

so, put the coordinates in the slope-intercept equation like so: -2 = (4/3)(-3) + b
b = -2 - (-12/3)
b = -2 + 4
b = 2

now, you must put the equation of the parallel line in slope-intercept form using "b" and the slope: y = (4/3)x + 2

Answer 2

4/3x-2y=6 Answer: y=2/3x
-2y= -4/3x+6
y=2/3x-3
Plug in the point
-2=2/3(-3)+b use slope intercept form: y=mx+b, m is the slope, slope is 2/3.
-2= -2+b
b=0
he forgot do divide the six by two

Answer 3

First, solve for y:

4x/3 - 2y = 6
-2y = 6 - 4x/3
y = 2x/3 - 6

So, the slope is 2/3.

y = 2x/3 + b
-2 = 2(-3)/3 + b
-2 = -2 + b
b = 0

y = 2x/3

Answer 4

(4/3).x - 2y = 6
2y = (4/3).x - 6
y = (2/3).x - 3
m = (2/3)
Gradient of parallel line is also (2/3)
y + 2 = (2/3).(x + 3)
y + 2 = (2/3).x + 2
y = (2/3).x is required line.