Tri ABC ,de parallel to bc …

Given, In {tex}\triangle ABC,{/tex}, {tex}DE || BC{/tex},

By Thales theorem , we get 
{tex}\frac{AD}{DB}=\frac{AE}{EC}{/tex} 
{tex}\Rightarrow \frac{4x-3}{3x-1}=\frac{8x-7}{5x-3}{/tex} 
{tex}\Rightarrow (4x-3)(5x-3)= (8x-7)(3x-1){/tex} 
{tex}\Rightarrow 20x^2-12x+9-15x=24x^2-21x-8x+7{/tex}
{tex}\Rightarrow + 24x^2 - 20x^2 -21x + 12x+ 15 x-8x+7- 9 = 0{/tex} 
{tex}\Rightarrow 4x^2-2x-2=0{/tex} 
{tex}\Rightarrow 2x^2-x-1=0{/tex} 
{tex}\Rightarrow 2x^2-2x+x-1=0{/tex} 
{tex}\Rightarrow 2x(x-1)+x(x-1)=0{/tex} 
{tex}\Rightarrow (2x+1)(x-1)=0{/tex}
{tex}\Rightarrow (2x+1)=0{/tex} or {tex}(x-1)=0{/tex}
{tex}\Rightarrow 2x = -1{/tex} or {tex}x = 1{/tex} 
{tex}x=-\frac{1}{2}{/tex}  or 1
If {tex}x=-\frac{1}{2}{/tex} , then {tex}AD=4\times \frac{-1}{2}-3=-5{/tex} < 0 [not possible]
Hence, x = 1 is the required value.