Answer
$x=-\dfrac{5}{2}\pm\dfrac{\sqrt{33}}{2}$ satisfies the given conditions
Work Step by Step
$y_{1}=\dfrac{2x}{x+2}$ $,$ $y_{2}=\dfrac{3}{x+4}$ and $y_{1}+y_{2}=1$
If $y_{1}+y_{2}$ is equal to $1$, then $\dfrac{2x}{x+2}+\dfrac{3}{x+4}$ is also equal to $1$.
$\dfrac{2x}{x+2}+\dfrac{3}{x+4}=1$
Evaluate the sum of fractions on the left side and simplify:
$\dfrac{2x(x+4)+3(x+2)}{(x+2)(x+4)}=1$
$\dfrac{2x^{2}+8x+3x+6}{(x+2)(x+4)}=1$
$\dfrac{2x^{2}+11x+6}{(x+2)(x+4)}=1$
Take $(x+2)(x+4)$ to multiply the right side:
$2x^{2}+11x+6=(x+2)(x+4)$
Evaluate the product on the right:
$2x^{2}+11x+6=x^{2}+4x+2x+8$
Take all terms to the left side of the equation and simplify:
$2x^{2}+11x+6-x^{2}-4x-2x-8=0$
$x^{2}+5x-2=0$
Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
In this case, $a=1$, $b=5$ and $c=-2$
Substitute the known values into the formula and evaluate:
$x=\dfrac{-5\pm\sqrt{5^{2}-4(1)(-2)}}{2(1)}=\dfrac{-5\pm\sqrt{25+8}}{2}=...$
$...=\dfrac{-5\pm\sqrt{33}}{2}=-\dfrac{5}{2}\pm\dfrac{\sqrt{33}}{2}$
$x=-\dfrac{5}{2}\pm\dfrac{\sqrt{33}}{2}$ satisfies the given conditions
