Question 75
Solve the following:
m−m−12=1−m−23
Given, m−m−12=1−m−23 ⇒2m−(m−1)2=3−(m−2)3⇒3(2m−m+1)=2(3−m+2) [By cross multiplication]⇒3(m+1)=2(5−m)⇒3m+3=10−2m⇒3m+2m=10−3⇒5m=7∴m=75
Solve: m−(m−1)2=1−(m−2)3.