Find the slope of the line 3x-(t+1)y-1=0 if is parallel with the line 5x-2y+3=0?

For lines to be parallel, the slope of d1:3x-(t+1)y-1=0 has to be equal to the slope of d2:5x-2y+3=0.

m1 = m2

Let's find out the slope of d2:

5x-2y+3=0

We'll re-write the equation into the form: y=mx+n, where m is the slope of the line.

For this reason, we'll isolate -2y to the left side:

-2y = -5x - 3

We'll multiply by -1:

2y = 5x + 3

We'll divide both sides by 2:

y = (5/2)*x + (3/2)

So, the slope of d2 is m2 = 5/2

Now, we'll find out the slope of d1:

3x-(t+1)y-1=0

We'll isolate -(t+1)y to the left side:

-(t+1)y = -3x + 1

We'll divide by -(t+1), both sides:

y = 3*x/(t+1) - 1/(t+1)

The slope of d1 is m1 = 3/(t+1)

But m1=m2, so we'll get:

5/2 =  3/(t+1)

We'll cross multiply:

5(t+1) = 6

5t+5-6=0

5t-1=0

We'll add 1 both sides:

5t = 1

We'll divide by 5, both sides:

t = 1/5

So, the slope of 3x-(t+1)y-1=0 is m1 = 3/(t+1)

m1 = 3/(1/5 + 1) = 3*5/6 = 5/2

m1 = 5/2