Question

What value of $x$ will make $DE\parallel AB$ in the given figure?
$AD=8x+9$, $CD=x+3$, $BE=3x+4$, $CE=x$

Answer 1

In a system of parallel lines,

$\frac{AD}{DC}=\frac{BE}{EC}\Rightarrow \frac{8x+9}{x+3}=\frac{3x+4}{x}\Rightarrow 8x^2+9x={3x}^2+13x+12\Rightarrow 5x^2-4x-12=0\Rightarrow 5x^2-10x+6x-12=0\Rightarrow 5x(x-2)+6(x-2)=0\Rightarrow (5x+6)(x-2)=0\Rightarrow x=-\frac{6}{5},2$

Since length cannot be negative, value of $x$ should be $2.$