Solve the inequality x/(x+2) ≤ 4/(x-3) for x ≠ -2 or 3

First we need to find the critical values. To find the critical values we must rearrange the equation and then replace the inequality symbol with an equals sign. The ≤ can be treated as an equality as long as both sides are not divided or multiplied by a negative number. So we multiply both sides by ((x+2)^2)(x-3)^2. We square the denominator so as to ensure it is not a negative number. Now we have (x/(x+2))((x+2)^2)(x-3)^2 ≤ (4/(x-3))((x+2)^2)(x-3)^2 which we can cancel and collect the terms on the LHS to get x(x+2)(x-3)^2 - 4(x-3)*(x+2)^2 ≤ 0.

Now rearrange to see that (x+2)(x-3)(x(x-3)-4(x+2)) ≤ 0 (x+2)(x-3)(x^2-3x-4x-8) ≤ 0 (x+2)(x-3)(x-7x-8) ≤ 0 (x+2)(x-3)(x-8)(x+1) ≤ 0 So the critical values are x = -2,3,8 and -1 and these points are where the graph crosses the x axes and therefore where the function is equal to zero.

Finally we need to check the 5 regions of numbers given by the 4 critical values. These are (-∞,-2),(-2,-1),(-1,3),(3,8) and (8,∞). An easy way to do this is to pick an easy number in-between the end points and plug this number into the rearranged LHS equation we have above, counting the negative and postive brackets. Call this equation F(x) = (x+2)(x-3)(x-8)(x+1) then F(-5) is greater than zero F(-1.5) is less than zero F(1) is greater than zero F(5) is less than zero and F(10) is greater than zero

so x/(x+2) ≤ 4/(x-3) when -2 < x ≤ -1 and when 3 < x ≤ 8