word problems and turning them into algebraic equation to slove

 Let D represent the donkey and H the horse:

The donkeys statement looks like this:

1/2(H-1)=D+1

1/2H-1/2=D+1

1/2H=D+1 1/2

H=2(D+1 1/2)

H=2D+3

The horses statement looks like this:

3(D-1)=H+1

3D-3-1=H

H=3D-4

Since H=H therefore

3D-4=2D+3

D=7 The donkey carried 7 bales

Now plug 7 into one of the original equation to solve for H:

H=2(7)+3=17
The Horse carried 17 bales.

Test the answer:

If the horse gets one more bale from the donkey he has 18, which would be 3 times 6.

3(7-1)=17+1

3(6)=17+1

18=18

Let's assume that a donkey is carrying "x" bales and a horse is carrying "y" bales.

(1.) If a horse will give to a donkey 1 bale, then a horse will be carrying (y - 1) bales and a donkey will be carrying (x + 1) bales, which will be half as much as horse's. 

So, we can write first equation:
             1
x + 1 = ---(y - 1)
             2 

(2.) From other side, if a donkey will give 1 bale to a horse, then a donkey will be carrying (x - 1) bales and a horse will be carrying (y + 1) bales, which is 3 times as much as donkey's.

Thereby, we can write second equation:

y + 1 = 3(x - 1)

Now, let's solve system of two equations:

x + 1 = 0.5(y - 1)
y + 1 = 3(x - 1)

Let's open parentheses and move all variables on the left side of each equation and numbers on the right side:  

x + 1 = 0.5y - 0.5
y + 1 = 3x - 3

 x - 0.5y = - 1 - 0.5
-3x + y = - 1 - 3

     2x - y = -3
+
   - 3x + y = - 4  
---------------------
  - x = - 7
    x = 7

Let's plug in the value of "x" into second original equation

y + 1 = 3(7 - 1)
y = 18 - 1 = 17

Originally, a donkey was carrying 7 bales and a horse was carrying 17 bales.