Which has to be the value of the parameter a so the lines d1 and d2 to be parallels?d1: 2x-5y+3=0 d2: (a+1)x-3y-1=0

For d1 to be parallel to d2, the slope of d1 has to be equal to the slope of d2:

m1 = m2

Let's find out the slope of d1:

2x-5y+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 -5y to the left side:

-5y = -2x - 3

We'll multiply by -1:

5y = 2x+3

We'll divide both sides by 5:

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

So, the slope of d1 is m1 = 2/5

Now, we'll find out the slope of d2:

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

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

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

We'll divide by -3, both sides:

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

The slope of d2 is m2 = (a+1)/3

But m1=m2, so we'll get:

2/5 = (a+1)/3

We'll cross multiply:

5(a+1) = 6

5a+5-6=0

5a-1=0

We'll add 1 both sides:

5a = 1

We'll divide by 5, both sides:

a = 1/5