Inequality

I subtracted 1 from both sides of the inequality, which resulted in $\frac{-13x+50}{(x+10)(x-4)}<0$ . Note that the sign of entire expression depends only on the signs of $-13x+50,x+10,$ and $x-4$ . Each sub-expression changes sign only as you cross the points where it equals zero (due to the continuity of the factors). Therefore, I can simply use a test point method in each of the following intervals: $(-\infty,-10),(-10,50/13),(50/13,4),$ and $(4,\infty)$ . For the first interval, I choose $-12$ . This gives $-13x+50=-13(-12)+50>0,x+10=-12+10<0,$ and $x-4=-12-10<0$ . This leads to a positive value for the entire expression. Therefore, the first interval is not part of the solution set, since we need it to be negative. Next I choose 0 to test the second interval: $-13x+50=0+50>0,x+10=0+10>0,$ and $x-4=0-4<0$ . Since this produces a negative value in the entire expression, $(-10,50/13)$ is in the solution set. Next, I will choose $51/13$ to test the third interval: $-13x+50=-13(51/13)+50=-51+50=-1<0,x+10=-51/13+10=(-51+130)/10>0,$ and $x-4=51/13-4=-1/13<0$ . Since we have two negatives and a positive, the entire expression is not negative in this case, and, so the third interval is not in the solution set. To test the final interval, I substitute $100$ for $x$ : $-13x+50=-13(100)+50<0, 100+10>0,$ and $100-4>0$ , which satisfies the original inequality, and, so $(4,\infty)$ is in the solution set. Hence, the solution set is $(-10,50/13)\cup(4,\infty)$ .

The LHS of the above inequality can be written as:

$\frac{x^2 - 7x + 10}{x^2 + 6x - 40} = 1 - \frac{13x - 50}{(x+10)(x-4)}$

Clearly, $x \neq -10, 4$ which prevents division by zero. Also, $x \neq \frac{50}{13}$ which prevents equality with the RHS of our original inequality. Finally, as $x \rightarrow \infty$ the rational expression $\frac{13x - 50}{(x+10)(x-4)} \rightarrow 0$ which still satisfies the original inequality.

Hence the solution set is $x \in (-10, \frac{50}{13}) \cup (4, \infty).$