In $△\mathrm{ADE} \mathrm{and} △\mathrm{ABC}, \mathrm{DE}\parallel \mathrm{BC}$. Find the value of x.

In $△\mathrm{ADE} \mathrm{and} △\mathrm{ABC}, \mathrm{DE}\parallel \mathrm{BC}$.

Thales’s Theorem states that if a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio.

By using Thales' theorem

$\frac{AD}{DB}=\frac{AE}{EC}$

$\frac{x+1}{3x-1}=\frac{x}{4x-3}$

$(x+1)(4 x-3)=x(3 x-1)$

$4x^2-3x+4x-3=3x^2-x$

$x^2+2x-3=0$

$x^2+3x-x-3=0$

$x(x+3)-1(x+3)=0$

$(x+3)(x-1)=0$

$x=-3,1$

We can ignore the negative value of x as it cannot be negative.

$x=1$

Therefore, the value of x is 1.