The other tutor's solution is incorrect

 +  = 

Since one of the radicals is isolated, we square both sides:

To make things a little easier, 

let A = 
let B = 
let C = 

Then the above equation becomes:

         A + B = C

Squaring both sides:

      (A + B)² = C²

(A + B)(A + B) = C²

 A² + 2AB + B² = C² 

Since A = , then A² = x+4
Since B = , then B² = 2x-1
let C = , then C² = 7x+1

So now we have

 x+4 + 2AB + 2x-1 = 7x+1

Simplify and solve for 2AB

 3x + 3 + 2AB = 7x + 1
          2AB = 4x - 2

Divide through by 2 since all coefficients are even:

           AB = 2x - 1

Square both sides again:

        (AB)² = (2x - 1)²

         A²B² = (2x - 1)(2x - 1)              

         A²B² = 4x² - 4x + 1

Now substitute A² = x+4  and B² = 2x-1 from above:

 (x+4)(2x-1) = 4x² - 4x + 1 

 2x² + 7x - 4 = 4x² - 4x + 1

 -2x² + 11x - 5 = 0

  2x² - 11x + 5 = 0

(x - 5)(2x - 1) = 0

x - 5 = 0;  2x - 1 = 0

    x = 5;      2x = 1

                 x = 

Check each solution as they may be extraneous: 
             
Checking x = 5

 +  = 
 +  = 
 +  = 
3 + 3 = 6
    6 = 6

That checks so x = 5 is a solution.

Checking x = 

 +  = 
 +  = 
 +  = 
 +  = 
 + 0 = 
 = 

That checks also, so x =  is also a solution.

Edwin