d1: -2x + 3y + 1 = 0

d2: 2x - 3y +1 = 0

For d1:

d1: -2x + 3y +1 = 0

==> y = (2/3)x + 1/2

**Then the slope is m = 2/3**

The y-intercept is where the function intersect with y-axis, then x should be 0:

==> y= (2/3)*0 + 1/3 = 1/3

**Then y-intercept = 1/3**

** **

For d2:

d2= 2x+3y +1 = 0

==> y= (-2/3)x -1/3

Then the slope m = -2/3

The y-intercept = -1/3

We'll put the equation of d1 in the standard form:

y=mx + n, where the coefficients represent:

- m - the slope

- n - y intercept

-2x+3y+1=0

We'll isolate 3y to the left side:

3y=2x-1

We'll divide by 3 both sides:

y=(2x/3) - 1/3

The standard form of the equation is:

**y=(2x/3) - 1/3**

**The slope of the line d1 is m = 2/3 and the y intercept is n = -1/3.**

** **

We'll put the equation of d2 in the standard form:

y=mx + n, where the coefficients represent:

- m - the slope

- n - y intercept

2x-3y+1=0

We'll isolate -3y to the left side:

-3y=-2x-1

We'll divide by -3 both sides:

y=(2x/3) + 1/3

The standard form of the equation is:

**y=(2x/3) + 1/3**

**The slope of the line d1 is m = 2/3 and the y intercept is n = 1/3.**

We can convert each equation to slope and intercept form:

y =mx+c , where m is slope and c is the y intercept of the line.

d1: -2x+3y+1 = 0 . add 2x-1 and rewrite as:

3y = 2x-1. Divide by 3.

y = (2/3) -1/3. So the slope of d1 is 2/3 and y intercept is -1/3.

d2:

2x-3y +1 = 0. Subtract 2x+1 .

-3y = -(2x+1) = -2x-1. Divide by -3.

y = (-2/3)x +(-1/3). So slope = -2/3 and y intercept -1/3.

We can express the equation ax+by+c=0 in the form y=(-a/b)x-c/b where -a/b is the slope and -c/b is the y intercept.

For the line d1 with the equation -2x+3y+1=0 the slope works out to 2/3 and the y-intercept is -1/3

For the line d2 with the equation 2x-3y+1=0 the slope works out to 2/3 and the y-intercept is 1/3