<s>[INST] Can we solve this without cubic formula? I'm looking for an algebraic solution for $x$. 
$$ \frac{x}{x+2} -3 = \frac{5x}{x^2-4}+x$$ 
My first go at this involved converting this into an expression with a cubic numerator and $(x+2)(x-2)$ as the denominator. To find the roots, I then tried to divide each factor in the denominator into to cubic. No success. 
I've converted this expression to:
$$  x(x-7)= (x+3)(x+2)(x-2)$$
which illustrates the futility of my first approach. There are no common factors. Can I solve this without invoking the cubic formula?

Edit: To clarify, the the Precalculus textbook calls for an algebraic and graphic solution. If you have an algebraic solution that would be accessible to a precalculus student, please provide it. 
 [/INST] Assuming $x^2\neq 4$, you end with the cubic equation
$$x^3+2 x^2+3 x-12=0$$ So, consider the function $$f(x)=x^3+2 x^2+3 x-12 \implies f'(x)=3 x^2+4 x+3$$ The first derivative does not cancel which means that there is only one real root.
Now, use inspection : $f(0)=-12$, $f(1)=-6$, $f(2)=10$. So, the root is between $1$ and $2$. Looking deeper $f(\frac 32)=\frac 38$ telling that the root is slightly below $1.5$.
Make $x=y+\frac 32$ which makes the equation to be
$$g(y)=y^3+\frac{13 y^2}{2}+\frac{63 y}{4}+\frac{3}{8}$$ if we admit that $y$ is small, then $$g(y) \approx \frac{63 y}{4}+\frac{3}{8}=0 \implies y=-\frac{1}{42}$$ So, an approximate solution is 
$$x \approx \frac 32-\frac{1}{42}=\frac{31}{21}\implies f(\frac{31}{21})=\frac{34}{9261}$$ which is now quite small.
Repeat the process making now $x=y+\frac{31}{21}$ giving
$$g(y)=y^3+\frac{45 y^2}{7}+\frac{2270 y}{147}+\frac{34}{9261}$$ then $$g(y) \approx \frac{2270 y}{147}+\frac{34}{9261}=0 \implies y=-\frac{17}{71505}$$
 So, an approximate solution is 
$$x \approx \frac{31}{21}-\frac{17}{71505}=\frac{105538}{71505}\approx 1.47595$$ while the exact solution (solving the cubic) would be $\approx 1.47595$ !
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