How do you solve #7-3abs(4d-7)=4#?

1 Answer
Aug 2, 2015

Answer:

#d=2# or #d=3/2#

Explanation:

Your absolute value equation looks like this

#7 - 3|4d-7| = 4#

Start by isolating the modulus on one side of the equation

#color(red)(cancel(color(black)(7))) - color(red)(cancel(color(black)(7))) - 3|4d-7| = 4 - 7#

#(color(red)(cancel(color(black)(-3))) * |4d-7|)/color(red)(cancel(color(black)(-3))) = (-3)/(-3)#

#|4d-7| = 1#

Now, this equation will produce two values for #d#, depending on which condition is true

  • If #(4d-7)>0#, then you have

#|4d-7| = 4d-7#

and the equation becomes

#4d-7 = 1 => d = 8/4 = color(green)(2)#

  • If #(4d-7)<0#, then you have

#|4d-7| = -(4d-7) = -4d + 7#

This will get you

#-4d+7 = 1 => d = 6/4 = color(green)(3/2)#