Answer:
Number 7: Solution [tex][-5,-\frac{1}{9}][/tex]U[tex](3,\infty)[/tex]
Number 8: solution [tex](-6,-5) U(-1,5) [/tex]
Explanation:
Number 7: Given inequality is [tex] \frac{x-6}{x+5} - \frac{x+4}{x-3}\leq 0[/tex]
First we simplify the equation and then we find critical points.
[tex] \frac{(x-6)(x-3)-(x+4)(x+5)}{(x+5)(x-3)} \leq 0[/tex]
[tex] \frac{x^2-9x+18-x^2-9x-20}{(x+5)(x-3)} \leq 0[/tex]
[tex] \frac{-18x-2}{(x+5)(x-3)} \leq 0[/tex]
Critical points are : [tex] x=-\frac{1}{9},3,-5[/tex]
Now we plot the points on number line and find the true and false solution. Please see the attach file fig 1 for number line.
We will check each interval for solution.
Let we take any interval [tex](-\frac{1}{9},3)[/tex] and take any point between the interval and substitute into equation. We get [tex]\frac{2}{15}[/tex] >0 which is greater than 0.
This interval would be false.
Similarly, we will check each interval .
We get solution, [tex][-5,-\frac{1}{9}][/tex]U[tex](3,\infty)[/tex]
Number 8: Given inequality is [tex] \frac{x^2-4x-5}{x^2+11x+30} < 0[/tex]
First we simplify the equation and then we find critical points. Factor numerator and denominator. We get
[tex] \frac{(x-5)(x+1)}{(x+5)(x+6)} < 0[/tex]
Critical points are : [tex] x=-6,-5,-1,5[/tex]
Now we plot the points on number line and find the true and false solution. Please see the attach file fig 2 for number line.
We will check each interval for solution.
Let we take any interval [tex](-1,5)[/tex] and take any point x=0 between the interval and substitute into equation. We get [tex]-\frac{1}{6}[/tex] <0 which is less than 0.
This interval would be true.
Similarly, we will check each interval .
We get solution, [tex](-6,-5) U(-1,5) [/tex]
