jmc

Combining the two terms on the left-hand side under a common denominator, we get $$\frac{x^2+11x+28-7(x+4)}{x+4}=x,$$or $$\frac{x^2+4x}{x+4}=x.$$If $x\ne -4,$ then the left-hand side reduces to $\frac{x(x+4)}{x+4}=x,$ so the equation is always true. And if $x=-4,$ then the denominator of the left-hand side is zero, so the equation is not true. Hence, the solution set consists of all $x$ such that $x\ne -4.$ In interval notation, this is $$x\in \boxed{(-\infty ,-4)\cup (-4,\infty )}.$$