Question 828701
<pre>
Since the sides must be positive integers, 
4x will always be greater than 2x-3.
So we have three cases to consider:

Case 1:  {{{x+10 <= 2x-3 < 4x}}}
         {{{x+10 <= 2x-3}}}
             {{{13<=x}}}

Since these are the sides of a triangle, the sum of the
two shorter sides must by greater than the longest side

         {{{x+10+2x-3>4x}}}
           {{{3x+7>4x}}} 
             {{{7>x}}} 

So there are no solutions in case 1, since those 
inequalities contradict each other.         

Case 2:  {{{2x-3 <= x+10 <= 4x}}}
         {{{2x-3 <= x+10}}} and {{{x+10<=4x}}}
         {{{x <= 13}}} and {{{10<=3x}}}
                    {{{3&1/3<=x}}}
                    {{{4<=x}}}   
         

Since these are the sides of a triangle, the sum of the
two smaller sides must be greater than the largest side

         {{{2x-3+x+10>4x}}}
           {{{3x+7>4x}}} 
             {{{7>x}}}
             {{{6>=x}}}   

So we have the possibility {{{4<=x<=6}}}

For x=4  {{{2x-3}}},{{{x+10}}},{{{4x}}}

2x-3=2(4)-3=8-3=5
x+10=4+10=14
4x=4(4)=16

The sides are 5,14,16   

The sum of the two longer sides is 30, which 
is divisible by the shortest side, 5.

So one solution is x=4 

For x=5  {{{2x-3}}},{{{x+10}}},{{{4x}}}

2x-3=2(5)-3=10-3=7
x+10=5+10=15
4x=4(5)=20

The sides are 7,15,20

The sum of the two longer sides is 35, which 
is divisible by the shortest side, 7.

So a second solution is x=5 

For x=6  {{{2x-3}}},{{{x+10}}},{{{4x}}}
          9,16,24

The sum of the two longer sides is 40, which 
is NOT divisible by the shortest side, 9. So x=6
is NOT a solution.       

So from case 1, we have exactly two solutions, x=4 and x=5

Case 3:  {{{2x-3 < 4x <= x+10}}}
         {{{4x<=x+10}}}
         {{{3x<=10}}}
        {{{x<=3&1/3}}}
          {{{x<=3}}}   
         
Since these are the sides of a triangle, the sum of the
two shorter sides must be greater than the longest side

         {{{2x-3+4x>x+10}}}
           {{{6x-3>x+10}}} 
             {{{5x>13}}}
               {{{x>2&3/5}}}
             {{{x>=3}}}   

So we have one possibility x=3

{{{2x-3}}},{{{4x}}},{{{x+10}}}

2x-3=2(3)-3=6-3=3
4x=4(3)=12
x+10=3+10=13

The sides are 3,12,13

The sum of the two longer sides is 25, which 
is NOT divisible by the shortest side, 3. So x=3
is NOT a solution. 

------------

This exhausts all the possibilities, so the answer is 

There are exactly two possibilities:

x=4 which yields sides 5, 14, 16

x=5 which yields sides 7, 15, 20 

Edwin</pre>