What is the solution set for the equation #|4a + 6| − 4a = 10# ?

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Answer 1 · 2015-10-13T17:02:03

#a = -2#

The first thing to do here is isolate the modulus on onse side of the equation by adding #4a# to both sides

#|4a + 6| - color(red)(cancel(color(black)(4a))) + color(red)(cancel(color(black)(4a))) = 10 + 4a#

#|4a + 6| = 10 + 4a#

Now, by definition, the absolute value of a real number will only return positive values, regardless of the sign of said number.

This means that the first condition that any value of #a# must satisfy in order to be a valid solution will be

#10 + 4a >= 0#

#4a >= -10 implies a >= -5/2#

Keep this in mind. Now, since the absolute value of a number returns a positive value, you can have two possibilities

#4a + 6 < 0 implies |4a + 6| = -(4a+6)#

In this case, the equation becomes

#-(4a + 6) = 10 + 4a#

#-4a - 6 = 10 + 4a#

#8a = - 16 implies a = ((-16))/8 = -2#

#(4a + 6) >=0 implies |4a + 6| = 4a + 6#

This time, the equation becomes

#color(red)(cancel(color(black)(4a))) + 6 = 10 + color(red)(cancel(color(black)(4a)))#

# 6 != 10 implies a in O/#

Therefore, the only valid solution will be #a = -2#. Notice that it satisfies the initial condition #a >= -5/2#.

Do a quick check to make sure that the calculations are correct

#|4 * (-2) + 6| - 4 * (-2) = 10#

#|-2| +8 = 10#

#2 + 8 = 10 color(white)(x)color(green)(sqrt())#