Text Solve, stating the conditions of existence: 4x−1=x+2.\frac{4}{x-1}=x+2.x−14=x+2. Solution Condition: x≠1x\neq 1x=1. Multiply by (x−1)(x-1)(x−1): 4=(x+2)(x−1)=x2+x−2 ⇒ x2+x−6=0.4=(x+2)(x-1)=x^{2}+x-2\;\Rightarrow\; x^{2}+x-6=0.4=(x+2)(x−1)=x2+x−2⇒x2+x−6=0. (x−2)(x+3)=0 ⇒ x=2 or x=−3(both acceptable).(x-2)(x+3)=0\;\Rightarrow\; x=2\ \text{or}\ x=-3\quad(\text{both acceptable}).(x−2)(x+3)=0⇒x=2 or x=−3(both acceptable). x1=2,x2=−3 \boxed{\,x_1=2,\quad x_2=-3\,}x1=2,x2=−3