Question 1066867
Since we want to solve for 'x', its best to get all the terms involving x onto one side:

          {{{ p^2(1-x)-2pqx = q^2(1+x) }}}
          {{{  p^2 - p^2x - 2pqx = q^2+q^2x }}}
Now bring all 'x' related terms to the left, and everything else to the right:
          {{{  -p^2x - 2pqx - q^2x = q^2 - p^2 }}}  
Multiply both sides by -1, this has the effect of changing the left to all +, and swaps the signs on the right:
        {{{  p^2x + 2pqx + q^2x = p^2 - q^2 }}}
          {{{   x (p^2 + 2pq + q^2) = p^2 - q^2 }}}
          {{{    x =  (p^2 - q^2) / (p^2 + 2pq + q^2 ) }}}

Not done yet, there is further simplification:

  Numerator:    {{{ p^2 - q^2 = (p-q)(p+q) }}}
  Denominator:  {{{ p^2 + 2pq + q^2 = (p+q)^2 }}}

  So      {{{    x =  ((p-q)(p+q)) / ((p+q)(p+q)) }}}  

Canceling a "p+q" from numerator and denominator:
            {{{  highlight(x = (p-q)/(p+q) )}}}