How do you solve the equation #abs(4/3-2/3x)=3/4#?

Absolute value equations are a little tough. Some teachers say that they "make numbers positive", but that doesn't mean #x# is always a positive number. Absolute value bars concern distance. That's why they "make the number positive"; because there's no such thing as a negative distance (we can't have #-262# feet).

When we have #sqrt(x)=y#, we take the inverse of the square root to "undo" it. But what's the inverse of absolute value bars? Nothing. But, we do know that whatever #x# equals, it can be either positive or negative.

So, the way we solve for absolute value equations is to let one equation be positive and another be negative:

#abs(4/3-2/3x)=3/4#

Positive situation
#(4/3-2/3x)=3/4#

subtract #4/3# on both sides

#-2/3x=-7/12#

divide by #-2/3#

#x=7/8#

Negative solution
#-(4/3-2/3x)=3/4#

divide by #-1# on both sides

#4/3-2/3x=-3/4#

subtract #4/3# on both sides

#-2/3x=-25/12#

divide by #-2/3# on both sides

#x=25/8#

So, our solutions are #x=25/8# and #7/8#. Just to double check, let's grahp our equation:

graph{y=abs(4/3-2/3x)-3/4}

Yep, intercepts at #0.875# and #3.125#, or #7/8# and #25/8#.

#"the value of the "color(blue)"absolute value function"# is always positive.

#"However, the value of the expression inside the bars"#
#"can be positive or negative"#

#"This means there are 2 possible solutions to the equation"#

#color(red)(+-)(4/3-2/3x)=3/4#

#color(blue)"First possible solution"#

#4/3-(2x)/3=3/4larrcolor(red)" positive value"#

#rArr(2x)/3=4/3-3/4=7/12#

#rArr24x=21larrcolor(red)" cross-multiplying"#

#rArrx=21/24=7/8larrcolor(magenta)" first possible solution"#

#color(blue)"Second possible solution"#

#(2x)/3-4/3=3/4larrcolor(red)" negative value"#

#rArr(2x)/3=3/4+4/3=25/12#

#rArr24x=75larrcolor(red)" cross-multiplying"#

#rArrx=75/24=25/8larrcolor(magenta)" second possible solution"#

#color(blue)"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.

#|4/3-(2/3xx7/8)|=|4/3-7/12|=|3/4|=3/4#

#|4/3-(2/3xx25/8)|=|4/3-25/12|=|-3/4|=3/4#

#rArrx=7/8" or " x=25/8" are the solutions"#