How do you solve #abs(5-2x)=13#?

1 Answer
Mar 2, 2018

#x=-4" or "x=9#

Explanation:

#"the value inside the "color(blue)"absolute value bars "" can be"#
#"positive or negative "#

#rArr"there are 2 possible solutions"#

#5-2x=13larrcolor(red)"positive inside bars"#

#"subtract 5 from both sides"#

#cancel(5)cancel(-5)-2x=13-5#

#rArr-2x=8#

#"divide both sides by "-2#

#(cancel(-2) x)/cancel(-2)=8/(-2)#

#rArrx=-4larrcolor(blue)"first solution"#

#-(5-2x)=13larrcolor(red)"negative inside bars"#

#rArr-5+2x=13#

#"add 5 to both sides"#

#cancel(-5)cancel(+5)+2x=13+5#

#rArr2x=18#

#"divide both sides by 2"#

#(cancel(2) x)/cancel(2)=18/2#

#rArrx=9larrcolor(blue)"second solution"#

#color(blue)"As a check"#

#"Substitute these 2 possible solutions into the left side"#
#"of the equation and if equal to right side then they are"#
#"the solutions"#

#x=-4rArr|5+8|=|13|=13larr" True"#

#x=9rArr|5-18|=|=13|=13larr" True"#

#rArrx=-4" or "x=9" are the solutions"#