| x + 6 | + x x − 1 > 3 {\displaystyle {\frac {|x+6|+x}{x-1}}>3}
שלב א': תחום ההגדרה x ≠ 1 {\displaystyle x\neq 1}
שלב ב':
( | x + 6 | + x ) ( x − 1 ) x − 1 > 3 ( x − 1 ) 2 {\displaystyle {\frac {(|x+6|+x)(x-1)}{x-1}}>3(x-1)^{2}}
( 2 x + 6 ) ( x − 1 ) > 3 ( x − 1 ) 2 {\displaystyle (2x+6)(x-1)>3(x-1)^{2}}
( 2 x + 6 ) ( x − 1 ) − 3 ( x − 1 ) 2 > 0 {\displaystyle (2x+6)(x-1)-3(x-1)^{2}>0}
( x − 1 ) [ ( 2 x + 6 ) − 3 ( x − 1 ) ] > 0 {\displaystyle (x-1)[(2x+6)-3(x-1)]>0}
( x − 1 ) [ 2 x + 6 − 3 x + 3 ] > 0 {\displaystyle (x-1)[2x+6-3x+3]>0}
( x − 1 ) [ 9 − x ] > 0 {\displaystyle (x-1)[9-x]>0}
פתרונות: 1 < x < 9 {\displaystyle 1<x<9}