How do you solve #abs(15-3x)=12#?
1 Answer
Feb 21, 2017
Explanation:
There are 2 solutions to the equation.
#"solving "15-3x=color(red)(+-)12#
#color(blue)"first solution"#
#"solve "15-3x=color(red)(+12)# subtract 15 from both sides of the equation.
#cancel(15)cancel(-15)-3x=12-15#
#rArr-3x=-3# divide both sides by - 3
#(cancel(-3) x)/cancel(-3)=(-3)/(-3)#
#rArrx=1#
#color(blue)"second solution"#
#"solve "15-3x=color(red)(-12)#
#rArr-3x=-12-15=-27#
#rArrx=(-27)/(-3)=9#
#color(magenta)"As a check"# Substitute these values into the left side of the equation and if equal to the right side then they are the solutions.
#• x=1to|15-(3xx1)|=|12|=12larr" true"#
#• x=9 to|15-(3xx9)|=|-12|=12larr" true"#
#rArr" solutions are "x=1" or "x=9#