#### Answer

$x=-9$ and $x=\dfrac{3}{4}$ satisfy the given conditions

#### Work Step by Step

$y_{1}=6\Big(\dfrac{2x}{x-3}\Big)^{2}$, $y_{2}=5\Big(\dfrac{2x}{x-3}\Big)$, and $y_{1}$ exceeds $y_{2}$ by $6$
If $y_{1}$ exceeds $y_{2}$ by $6$, then $y_{2}+6=y_{1}$ and $y_{1}-y_{2}=6$
If $y_{1}-y_{2}=6$, then $6\Big(\dfrac{2x}{x-3}\Big)^{2}-5\Big(\dfrac{2x}{x-3}\Big)$ is also equal to $6$:
$6\Big(\dfrac{2x}{x-3}\Big)^{2}-5\Big(\dfrac{2x}{x-3}\Big)=6$
Evaluate all operations indicated on the left side:
$6\Big[\dfrac{4x^{2}}{(x-3)^{2}}\Big]-\dfrac{10x}{x-3}=6$
$\dfrac{24x^{2}}{(x-3)^{2}}-\dfrac{10x}{x-3}=6$
Multiply the whole equation by $(x-3)^{2}$
$(x-3)^{2}\Big[\dfrac{24x^{2}}{(x-3)^{2}}-\dfrac{10x}{x-3}=6\Big]$
$24x^{2}-10x(x-3)=6(x-3)^{2}$
Evaluate all indicated operations:
$24x^{2}-10x^{2}+30x=6(x^{2}-6x+9)$
$24x^{2}-10x^{2}+30x=6x^{2}-36x+54$
Take all terms to the left side of the equation and simplify:
$24x^{2}-10x^{2}+30x-6x^{2}+36x-54=0$
$8x^{2}+66x-54=0$
Solve by factoring:
$2(x+9)(4x-3)=0$
Take the $2$ to divide the right side:
$(x+9)(4x-3)=\dfrac{0}{2}$
$(x+9)(4x-3)=0$
Set both factors equal to $0$ and solve each individual equation for $x$:
$x+9=0$
$x=-9$
$4x-3=0$
$4x=3$
$x=\dfrac{3}{4}$
$x=-9$ and $x=\dfrac{3}{4}$ satisfy the given conditions