1) x < 4 the answer is <span> Option C: x E(-infinity,4) (it is strict) 2) </span>2x +6 < 8, and <span>2x < 8 - 6 = 2, and x<2/2=1 the answer is </span> <span>D. x E (-infinity, 1) </span>3) <span>X greater than or equal to 8 and x < - 4 </span>X greater than 8 means x ≥ 8 so we have x ≥ 8 and <span> x < - 4 so the answer is </span> <span>B. x E (-4,8] </span> 4) 7 > x + 6 or x -2 greater than or equal to 3, <span>7 > x + 6 or x -2 ≥ 3 </span> <span>7 -6> x, and 1>x </span> x -2 ≥ 3 x<span>≥ 3+2=5 finally </span>x<1 and x≥5
the answer is <span>C.x E (1,5] 5) </span> |x+3| > 12, |x+3| = { -(x+3) if <span>x+3<0 or </span><span>(x+3) if </span><span>x+3>0}</span>
so, it is -(x+3)>12 is equivalent to (x+3)< -12 or (x+3)>12 the answer is <span>B. x + 3 > 12 or x + 3 <-12</span>
