Question Number 114768 by ZiYangLee last updated on 21/Sep/20

Solve for ∣x−2∣+∣1−x∣=4

Answered by bobhans last updated on 21/Sep/20

(1) x<1 ⇒ 2−x+1−x=4                             3−2x = 4 ; x=−(1/2)  (2) 1≤x<2⇒2−x+x−1=4                                 1=4 ←no solution  (3)x≥2 ⇒x−2+x−1=4                          2x = 7; x=(7/2)  ∴ x = −(1/2) or x = (7/2)

Answered by mathmax by abdo last updated on 22/Sep/20

⇒f(x)=∣x−2∣+∣1−x∣−4 =0    x                    −∞                1               2                  +∞  ∣x−2∣                     −x+2       −x+2     x−2  ∣x−1∣                   −x+1       x−1            x−1  f(x)                        −2x−1      −3          2x−7  ⇒f(x) = { ((−2x−1 if x≤1)),((−3   if 1≤x≤2     )) :}                       { 2x−7  if    x≥2  so     x≤1   we get f(x)=0 ⇒−2x−1 =0 ⇒x =−(1/2)  1≤x≤2     f(x)=0 ⇒−3 =0 imposdible  x≥2            f(x)=0 ⇒2x−7 =0 ⇒x =(7/2)