How do you solve the equation #abs(2(1/3+1/2x))=1#?
2 Answers
Explanation:
#"distribute the factor"#
#rArr|(2(1/3+1/2x)|=|2/3+x|#
#"the value inside the absolute value function can be"#
#color(blue)"positive or negative"#
#color(blue)"first solution"#
#2/3+x=1rArrx=1-2/3=1/3#
#color(blue)"second solution"#
#-(2/3+x)=1#
#rArr-2/3-x=1#
#rArr-x=1+2/3=5/3rArrx=-5/3#
#color(blue)"As a check"#
#x=1/3"#
#rArr|2/3+1/3|=|1|=1#
#x=-5/3#
#rArr|2/3-5/3|=|-1|=1#
Use the piecewise definition of the absolute value function to separate the equation into two equations and then solve each equation.
Explanation:
The piecewise definition of the absolute value function is:
In this case
Substitute into the definition:
Simplify the domain restrictions:
Separate the given equation into two equations with its respective domain restriction:
Multiply the second equation by -1:
Distribute the two in both equations:
Subtract
The domain restrictions can be dropped, because neither equation violates them: