Fractional Equations 3

After simplification, we see it is a 4 degree equation. So it has 4 roots or 4 values of x satisfies the equation.

The condition of x is $x\neq -1$ , we set $y=\frac{13-x}{x+1}$ , then we have this system $\begin{cases} xy(x+y)=42\\x+(x+1)y=13\end{cases}$ $\Leftrightarrow\begin{cases} xy(x+y)=42\\ x+y+xy=13\end{cases}$ Now we set $t=x+y$ and $u=xy$ , the the system becomes $\begin{cases} t+u=13\\t.u=42\end{cases}$ Solving and we get $(t;u)=(7;6);(6;7)$ as the answer So now we have two systems $(I)\begin{cases}x+y=7\\x.y=6\end{cases}$ $(II)\begin{cases} x+y=6\\x.y=7\end{cases}$ Solving them gives you $x=\{1;6;3\pm\sqrt{2}\}$ as the answer

No need to simplify simply first take x common in 1st part then u get 1 sol then the rest is 1 sol then simplify the part in the bracket to get a quadratic eq so we get 1+1+2=3