Show a full solution to the inequality [(x+2)/(x-4)] \leqslant 3 Can someone please tell me what I did wrong?

Question

Show a full solution to the inequality [(x+2)/(x-4)]


`\leqslant`
3

Can someone please tell me what I did wrong?

Answer 1

Your sign chart is a bit off. The first two rows are correct. However keep in mind that you factored out a negative (when you factored -2x+14) so you need to negate the signs of the f(x) row.

Another thing to keep in mind: x = 4 is not allowed as its the vertical asymptote. However, x = 7 is valid. So you'll have an open circle at x = 4 and a close circle at x = 7

The answer in interval notation is (-infinity, 4) U [7, infinity)
which translates to this disjoint set of inequalities: x < 4 or x >= 7