Question 32056
Let's see:

{{{(1/(x+2))+(2/(x-3))<=1}}} Add the fractions on the left.
{{{((x-3)+2(x+2))/((x+2)(x-3)) <=1}}} Multiply both sides by{{{(x+2)(x-3)}}}
{{{x-3+2x+4 <= (x+2)(x-3)}}} Simplify.
{{{3x+1<=x^2-x-6}}} Subtract{{{(3x+1)}}} from both sides.
{{{0<=x^2-4x-7}}} or {{{x^2-4x-7>=0}}}
Begin by solving the quadratic equation{{{x^2-4x-7 = 0}}}Solve for x using the quadratic formula{{{x=(-b+-sqrt(b^2-4ac))/2a}}}.
{{{x=(4+-sqrt(16+28))/2}}} Simplify.
{{{x=(4+-sqrt(44))/2}}}
{{{x=(4+-2sqrt(11))/2}}}
{{{x=2+-sqrt(11)}}}
The x-intercepts are:
{{{x = 2+sqrt(11)}}}
{{{x = 2-sqrt(11)}}}

But since we are solving:{{{x^2-4x-7>=0}}} we are looking for x-values with positve y-values. Looking at the graph of the quadratic equation:
{{{graph(300,200,-3,6,-12,5,x^2-4x-7)}}}

The solution set is {{{x<=2-sqrt(11)orx>=2+sqrt(11)}}}