How do you solve #5/(8x)-1/2=7/(6x)# and check for extraneous solutions?
1 Answer
Jul 7, 2017
Explanation:
#"multiply all terms by " 24x#
#rArr15-12x=28#
#rArr-12x=13rArrx=-13/12#
#color(blue)"As a check"#
#"left side "=5/(8xx-13/12)-1/2=-15/26-13/26=-14/13#
#"right side "=7/(6xx-13/12)=-14/13#
#rArrx=-13/12" is the solution"#