Find the values of x where x+3>2/(x-4), what about x+3>2/mod(x-4)?

a) Create an inequality in x: (x+3)(x-4)<2. Expand and find the critical points, x^2-x-14>0. Find the answers to this inequality using quadratic formula and then test for a value between your two critical points x0=(1+sqrt(57))/2 and x1=(1-sqrt(57))/2. Also as x-4=0 at x=4 we need to consider the asympotote at x=4. As 0 in the original equation satisfies the inequality, use x=0 as a test as it is between the two cvs we can conclude that to satisfy the inequality, x1<x4. b)if for x+3>2mod(x-4) the equation is the same for x>4. For x<4 then we need to compare with (x+3)>-2/(x-4), Therefore expand to get x^2 -x-10<0. The roots of this equation are x=(1+-sqrt(41))/2. Then compare with x=0 again and find that x>4, (1+sqrt(41))/2>x>(1-sqrt(41))/2