Answers of x

$\frac{5}{x}+\frac{2}{x+3}=\frac{4}{x+1}+\frac{3}{x+2}$

$\left( \frac{5}{x} +1 \right) +\left(\frac{2}{x+3} +1 \right) = \left( \frac{4}{x+1}+1 \right) +\left(\frac{3}{x+2}+1 \right)$

$\frac{5+x}{x}+\frac{5+x}{x+3}=\frac{5+x}{x+1}+\frac{5+x}{x+2}$

$(5+x)\left(\frac{1}{x}+\frac{1}{x+3}-\frac{1}{x+1}-\frac{1}{x+2}\right)=0$

$\Rightarrow 5+x =0$ or $\displaystyle\frac{1}{x}+\frac{1}{x+3}-\frac{1}{x+1}-\frac{1}{x+2}=0$

$1^{st}$ term

$5+x=0 \Rightarrow x=\boxed{-5}$

$2^{nd}$ term

$\frac{1}{x}+\frac{1}{x+3}-\frac{1}{x+1}-\frac{1}{x+2}=0$

$\frac{2x+3}{x(x+3)}=\frac{2x+3}{(x+1)(x+2)}$

$(2x+3) \left(\frac{1}{x^2+3x}-\frac{1}{x^2+3x+2}\right)=0$

$\Rightarrow 2x+3=0 \Rightarrow x=\boxed{-\frac{3}{2}}$ or $\displaystyle \frac{1}{x^2+3x}-\frac{1}{x^2+3x+2}=0$

• $\displaystyle \frac{1}{x^2+3x}-\frac{1}{x^2+3x+2}=0$ has no real roots for it.

Therefore, the sum of all real roots are $-5+\left(-\frac{3}{2}\right)= \boxed{-6.5}$

$\begin{aligned}\dfrac5x+\dfrac2{x+3}=&\ \dfrac4{x+1}+\dfrac3{x+2}\\\dfrac{5(x+3)+2x}{x(x+3)}=&\ \dfrac{4(x+2)+3(x+1)}{(x+1)(x+2)}\\\dfrac{5x+15+2x}{x^2+3x}=&\ \dfrac{4x+8+3x+3}{x^2+3x+2}\\\dfrac{7x+15}{x^2+3x}=&\ \dfrac{7x+11}{x^2+3x+2}\\(7x+15)(x^2+3x+2)=&\ (7x+11)(x^2+3x)\\7x^3+36x^2+59x+30=&\ 7x^3+32x^2+33x\\4x^2+26x+30=&\ 0\\2x^2+13x+15=&\ 0\\(2x+3)(x+5)=&\ 0\\x=&\ -\dfrac32,-5\end{aligned}$

Thus, sum of all roots in above equation is $\left(-\dfrac32\right)+(-5)=-\dfrac{13}2=\boxed{6.5}$

If you find a common denominator for both sides of the equation, you're left with

(7x+15)/(x*(x+3)) = (7x+11)/((x+1)(x+2)) Cross multiplying: 7x^3 +32x^2+33x=7x^3+36x^2+59x+30 32x^2 +33x = 36x^2+59x+30 4x^2+26x+30=0 Quadratic formula : = (-26+/- sqrt(26^2-4(4)(30))/(2(4)) (-26 +/- sqrt(676-480))/(8) (-26 +/- 14)/(8) (-13+/- 7)/4 There are only two solutions to a quadratic ---> x = -3/2, -5

-10/2 - 3/2 = -13/2 = -6.5

From solving above equation we get, 4x^2 + 26x + 30 = 0 so by applying vieta's formula we get sum of all real roots of this equation, i.e. -26/4 = -6.5