Answer
$x=4$ and $x=\dfrac{2}{3}$ satisfy the given conditions.
Work Step by Step
$y_{1}=\dfrac{3}{x-1}$ $,$ $y_{2}=\dfrac{8}{x}$ and $y_{1}+y_{2}=3$
If $y_{1}+y_{2}$ is equal to $3$, then $\dfrac{3}{x-1}+\dfrac{8}{x}$ is also equal to $3$:
$\dfrac{3}{x-1}+\dfrac{8}{x}=3$
Evaluate the sum of fractions on the left side of the equation and simplify:
$\dfrac{3x+8(x-1)}{x(x-1)}=3$
$\dfrac{3x+8x-8}{x(x-1)}=3$
$\dfrac{11x-8}{x(x-1)}=3$
Take $x(x-1)$ to multiply the right side:
$11x-8=3x(x-1)$
$11x-8=3x^{2}-3x$
Take all terms to the right side and simplify:
$0=3x^{2}-3x-11x+8$
$0=3x^{2}-14x+8$
Rearrange:
$3x^{2}-14x+8=0$
Solve by factoring:
$(x-4)(3x-2)=0$
Set both factors equal to $0$ and solve each individual equation for $x$:
$x-4=0$
$x=4$
$3x-2=0$
$3x=2$
$x=\dfrac{2}{3}$
$x=4$ and $x=\dfrac{2}{3}$ satisfy the given conditions.
